评估一串简单的数学式[封闭]

https://stackoverflow.com/questions/928563

06-09-2019
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题

挑战

这里的挑战是(我自己发明的，虽然我也不会感到惊讶，如果它先前已经出现了其他地方的上网)。

编写一个函数，需要一个单一的参数是一个串表示一个简单的数学表达和评估它作为一个浮点值。一个 "简单的表达"可以包括任何的如下：正的或负的十进制数字， +, -, *, /, (, ).表达形式使用(通常的) 缀符号.经营者应该是评估在为了他们的出现，即不如同在 BODMAS, 虽然括号应该是正确的观察到的，当然。功能应该回正确的结果任何尽可能表达这种形式。然而，职能没有处理格式错误表现形式(即那些坏syntax).

例的表现形式：
1 + 3 / -8                            = -0.5       (No BODMAS)
2*3*4*5+99                            = 219
4 * (9 - 4) / (2 * 6 - 2) + 8         = 10
1 + ((123 * 3 - 69) / 100)            = 4
2.45/8.5*9.27+(5*0.0023)              = 2.68...

规则

我预见某些形式的"欺骗"/狡猾，因此，请让我预先警告反对！通过作弊，我指的是使用 eval 或者等同功能在动态的语言，如JavaScript或PHP，或者同样的编制和执行代码。(我认为我的规范的"不BODMAS"之间有很大的保证，这不过。) 除此之外，没有任何限制。我预计几Regex解决方案，但它将很高兴见到超过这一点。

现在，我主要兴趣在C#/.净的解决方案，但是任何其他语言将是完全可以接受过(特别是F#和蟒蛇的功能/混合方法).我还没有决定是否我会接受的最短或最巧妙的解决方案(至少对于语言)作为回答，但我将欢迎 任何形式的解决方案中的任何语言, 除了什么我只是禁止上面!

我的解决方案

我现在已经张贴了我的C#解决方案在这里， (403个字符). 更新： 我的新的解决方案有殴打一个老大 294个字符, 帮助一位可爱的regex!我怀疑，这将很容易得到殴打一些语文出有较轻的语法(特别是funcional/态的人)和已被证明是正确的，但我很好奇如果有人可以打败这个在C#静。

更新

我已经看到了一些非常狡猾的解决方案。感谢大家已经发布的一种。虽然我没有测试过任何人但我会相信的人和为他们至少工作的所有给出的例子。

只是注意到，重新入(即线的安全)是不要求的功能，虽然这是一个奖金。

格式

请发表的所有答案在下列格式用于该目的的简单比较：

语言

数字：???

完全模糊处理的功能：
(code here)
清除/半的模糊功能：
(code here)
任何的笔记算法/聪明快捷方式的需要。

解决方案

的Perl（无EVAL）

<强>的字符数： ~~167 106 （见下面对于106字符版本）~~

全混淆功能：（167个字符如果加入这三行成一个）

sub e{my$_="($_[0])";s/\s//g;$n=q"(-?\d++(\.\d+)?+)"; @a=(sub{$1},1,sub{$3*$6},sub{$3+$6},4,sub{$3-$6},6,sub{$3/$6}); while(s:$$n$|(?<=$)$n(.)$n:$a[7&ord$5]():e){}$_}

清除/版本反混淆：

sub e { my $_ = "($_[0])"; s/\s//g; $n=q"(-?\d++(\.\d+)?+)"; # a regex for "number", including capturing groups # q"foo" in perl means the same as 'foo' # Note the use of ++ and ?+ to tell perl # "no backtracking" @a=(sub{$1}, # 0 - no operator found 1, # placeholder sub{$3*$6}, # 2 - ord('*') = 052 sub{$3+$6}, # 3 - ord('+') = 053 4, # placeholder sub{$3-$6}, # 5 - ord('-') = 055 6, # placeholder sub{$3/$6}); # 7 - ord('/') = 057 # The (?<=... bit means "find a NUM WHATEVER NUM sequence that happens # immediately after a left paren", without including the left # paren. The while loop repeatedly replaces "(" NUM WHATEVER NUM with # "(" RESULT and "(" NUM ")" with NUM. The while loop keeps going # so long as those replacements can be made. while(s:\($n$|(?<=$)$n(.)$n:$a[7&ord$5]():e){} # A perl function returns the value of the last statement $_ }

我最初误解了规则，所以我提交了一个版本的“EVAL”。这里有一个版本，没有它。

当我意识到，在+，-，/和*字符代码的最后一个八进制数字是不同的，这ord(undef)为0。这让我建立了调度表@a为洞察最新位来阵列，并且只是调用代码在该位置7 & ord($3)。

有一个明显的斑点剃掉一个或多个字符 - 变化q""成'' - 但是这将使其难以剪切并粘贴到壳

甚至更短的

字符的数： ~~124 106~~

以通过 ephemient 编辑考虑在内，它现在下降到124个字符（两行连接到一个）

sub e{$_=$_[0];s/\s//g;$n=q"(-?\d++(\.\d+)?+)"; 1while s:\($n$|$n(.)$n:($1,1,$3*$6,$3+$6,4,$3-$6,6,$6&&$3/$6)[7&ord$5]:e;$_}

较短仍然

字符的数： ~~110 106~~

下面向下红宝石溶液进一步推我，虽然我无法到达其104个字符：

sub e{($_)=@_;$n='( *-?[.\d]++ *)'; s:$$n$|$n(.)$n:(($1,$2-$4,$4&&$2/$4,$2*$4,$2+$4)x9)[.8*ord$3]:e?e($_):$_}

我不得不放弃并使用''。那红宝石send窍门是给这个问题真的很有用。

从石头挤水

字符的号：106

一个小的扭曲，以避免除以零检查。

sub e{($_)=@_;$n='( *-?[.\d]++ *)'; s:$$n$|$n(.)$n:($1,0,$2*$4,$2+$4,0,$2-$4)[7&ord$3]//$2/$4:e?e($_):$_}

这里的测试工具用于该功能：

perl -le 'sub e{($_)=@_;$n='\''( *-?[.\d]++ *)'\'';s:$$n$|$n(.)$n:($1,0,$2*$4,$2+$4,0,$2-$4)[7&ord$3]//$2/$4:e?e($_):$_}' -e 'print e($_) for @ARGV' '1 + 3' '1 + ((123 * 3 - 69) / 100)' '4 * (9 - 4) / (2 * 6 - 2) + 8' '2*3*4*5+99' '2.45/8.5*9.27+(5*0.0023) ' '1 + 3 / -8'

其他提示

汇编

<强> 427字节

混淆，组装有优良 A86 成一个。com可执行文件：

dd 0db9b1f89h, 081bee3h, 0e8af789h, 0d9080080h, 0bdac7674h, 013b40286h dd 07400463ah, 0ccfe4508h, 08ce9f675h, 02fc8000h, 013b0057eh, 0feaac42ah dd 0bedf75c9h, 0ba680081h, 04de801h, 04874f73bh, 04474103ch, 0e8e8b60fh dd 08e8a003fh, 0e880290h, 0de0153h, 08b57e6ebh, 0d902a93eh, 046d891dh dd 08906c783h, 05f02a93eh, 03cffcee8h, 057197510h, 02a93e8bh, 08b06ef83h dd 05d9046dh, 02a93e89h, 03bc9d95fh, 0ac0174f7h, 074f73bc3h, 0f3cac24h dd 0eed9c474h, 0197f0b3ch, 07cc4940fh, 074f73b09h, 0103cac09h, 0a3ce274h dd 0e40a537eh, 0e0d90274h, 02a3bac3h, 021cd09b4h, 03e8b20cdh, 0ff8102a9h dd 0ed7502abh, 0474103ch, 0e57d0b3ch, 0be02a3bfh, 014d903a3h, 0800344f6h dd 02db00574h, 0d9e0d9aah, 0d9029f2eh, 0bb34dfc0h, 08a0009h, 01c75f0a8h dd 020750fa8h, 0b0f3794bh, 021e9aa30h, 0de607400h, 08802990eh, 0de07df07h dd 0c392ebc1h, 0e8c0008ah, 0aa300404h, 0f24008ah, 04baa3004h, 02eb0ee79h dd 03005c6aah, 0c0d90ab1h, 0e9defcd9h, 02a116deh, 0e480e0dfh, 040fc8045h dd 0ede1274h, 0c0d90299h, 015dffcd9h, 047300580h, 0de75c9feh, 0303d804fh dd 03d80fa74h, 04f01752eh, 0240145c6h, 0dfff52e9h, 0d9029906h, 0f73b025fh dd 03caca174h, 07fed740ah, 0df07889ah, 0277d807h, 047d9c1deh, 0990ede02h dd 025fd902h, 03130e0ebh, 035343332h, 039383736h, 02f2b2d2eh, 02029282ah dd 0e9000a09h, 07fc9f9c1h, 04500000fh, 0726f7272h db 024h, 0abh, 02h

修改是非模糊源：

mov [bx],bx finit mov si,81h mov di,si mov cl,[80h] or cl,bl jz ret l1: lodsb mov bp,d1 mov ah,19 l2: cmp al,[bp] je l3 inc bp dec ah jne l2 jmp exit l3: cmp ah,2 jle l4 mov al,19 sub al,ah stosb l4: dec cl jnz l1 mov si,81h push done decode: l5: call l7 l50: cmp si,di je ret cmp al,16 je ret db 0fh, 0b6h, 0e8h ; movzx bp,al call l7 mov cl,[bp+op-11] mov byte ptr [sm1],cl db 0deh sm1:db ? jmp l50 open: push di mov di,word ptr [s] fstp dword ptr [di] mov [di+4],bp add di,6 mov word ptr [s],di pop di call decode cmp al,16 jne ret push di mov di,word ptr [s] sub di,6 mov bp,[di+4] fld dword ptr [di] mov word ptr [s],di pop di fxch st(1) cmp si,di je ret lodsb ret l7: cmp si,di je exit lodsb cmp al,15 je open fldz cmp al,11 jg exit db 0fh, 94h, 0c4h ; sete ah jl l10 l9: cmp si,di je l12 lodsb cmp al,16 je ret l10: cmp al,10 jle l12i l12: or ah,ah je l13 fchs l13: ret exit: mov dx,offset res mov ah,9 int 21h int 20h done: mov di,word ptr [s] cmp di,(offset s)+2 jne exit cmp al,16 je ok cmp al,11 jge exit ok: mov di,res mov si,res+100h fst dword ptr [si] test byte ptr [si+3],80h jz pos mov al,'-' stosb fchs pos: fldcw word ptr [cw] fld st(0) fbstp [si] mov bx,9 l1000: mov al,[si+bx] test al,0f0h jne startu test al,0fh jne startl dec bx jns l1000 mov al,'0' stosb jmp frac l12i: je l11 fimul word ptr [d3] mov [bx],al fild word ptr [bx] faddp jmp l9 ret startu: mov al,[si+bx] shr al,4 add al,'0' stosb startl: mov al,[si+bx] and al,0fh add al,'0' stosb dec bx jns startu frac: mov al,'.' stosb mov byte ptr [di],'0' mov cl,10 fld st(0) frndint frac1: fsubp st(1) ficom word ptr [zero] fstsw ax and ah,045h cmp ah,040h je finished fimul word ptr [d3] fld st(0) frndint fist word ptr [di] add byte ptr [di],'0' inc di dec cl jnz frac1 finished: dec di cmp byte ptr [di],'0' je finished cmp byte ptr [di],'.' jne f2 dec di f2: mov byte ptr [di+1],'$' exit2: jmp exit l11: fild word ptr [d3] fstp dword ptr [bx+2] l111: cmp si,di je ret lodsb cmp al,10 je exit2 jg ret mov [bx],al fild word ptr [bx] fdiv dword ptr [bx+2] faddp fld dword ptr [bx+2] fimul word ptr [d3] fstp dword ptr [bx+2] jmp l111 d1: db '0123456789.-+/*()', 32, 9 d3: dw 10 op: db 0e9h, 0c1h, 0f9h, 0c9h cw: dw 0f7fh zero: dw 0 res:db 'Error$' s: dw (offset s)+2

红宝石

字符的号：103

N='( *-?[\d.]+ *)' def e x x.sub!(/$#{N}$|#{N}([^.\d])#{N}/){$1or(e$2).send$3,e($4)}?e(x):x.to_f end

这是一个~~非递归~~恶跳蚤的溶液版本。括号的子表达式进行求值自下而上代替自上而下。

修改：（尽管递归不是语义上必要）转换的“而”给条件+尾递归节省了几个字符，所以它不再是非递归

修改：借款合并正则表达式的丹尼尔·马丁的想法可以节省另外11个字符

修改：这递归甚至比我首先想到的更多有用！ x.to_f可以改写为e(x)，如果x恰好包含一个单一的数字。

修改：使用而不是 'or' '||' 允许一对括号的被丢弃

<强>长版本：

# Decimal number, as a capturing group, for substitution # in the main regexp below. N='( *-?[\d.]+ *)' # The evaluation function def e(x) matched = x.sub!(/$#{N}$|#{N}([^\d.])#{N}/) do # Group 1 is a numeric literal in parentheses. If this is present then # just return it. if $1 $1 # Otherwise, $3 is an operator symbol and $2 and $4 are the operands else # Recursively call e to parse the operands (we already know from the # regexp that they are numeric literals, and this is slightly shorter # than using :to_f) e($2).send($3, e($4)) # We could have converted $3 to a symbol ($3.to_s) or converted the # result back to string form, but both are done automatically anyway end end if matched then # We did one reduction. Now recurse back and look for more. e(x) else # If the string doesn't look like a non-trivial expression, assume it is a # string representation of a real number and attempt to parse it x.to_f end end

C（VS2005）

汉字的号：1360

预处理器的滥用和乐趣代码布局警告（向下滚动以查看）：

#include <stdio.h> #include <stdlib.h> #include <string.h> #define b main #define c(a) b(a,0) #define d -1 #define e -2 #define g break #define h case #define hh h #define hhh h #define w(i) case i #define i return #define j switch #define k float #define l realloc #define m sscanf #define n int _ #define o char #define t(u) #u #define q(r) "%f" t(r) "n" #define s while #define v default #define ex exit #define W printf #define x fn() #define y strcat #define z strcpy #define Z strlen char*p =0 ;k *b (n,o** a){k*f ;j(_){ hh e: i* p==40? (++p,c (d )) :( f= l( 0, 4) ,m (p ,q (% ), f,&_), p+=_ ,f ); hh d:f=c( e);s (1 ){ j( *p ++ ){ hh 0: hh 41 :i f; hh 43 :* f+=*c( e) ;g ;h 45:*f= *f-*c( e);g;h 42 :* f= *f**c( e);g;h 47:*f /=*c (e); g; v: c(0);} }w(1): if(p&& printf (q (( "\\")) ,* c( d) )) g; hh 0: ex (W (x )) ;v :p =( p?y: z)(l(p ,Z(1[ a] )+ (p ?Z(p )+ 1:1)) ,1 [a ]) ;b (_ -1 ,a +1 ); g; }i 0;};fn () {n =42,p= 43 ;i "Er" "ro" t( r) "\n";}

视觉Basic.NET

字符的编号：9759

我更投球手的自己。

注意：不采取嵌套括弧考虑。此外，未经检验的，但我敢肯定它的工作原理。

Imports Microsoft.VisualBasic Imports System.Text Imports System.Collections.Generic Public Class Main Public Shared Function DoArithmaticFunctionFromStringInput(ByVal MathematicalString As String) As Double Dim numberList As New List(Of Number) Dim operationsList As New List(Of IOperatable) Dim currentNumber As New Number Dim currentParentheticalStatement As New Parenthetical Dim isInParentheticalMode As Boolean = False Dim allCharactersInString() As Char = MathematicalString.ToCharArray For Each mathChar In allCharactersInString If mathChar = Number.ZERO_STRING_REPRESENTATION Then currentNumber.UpdateNumber(mathChar) ElseIf mathChar = Number.ONE_STRING_REPRESENTATION Then currentNumber.UpdateNumber(mathChar) ElseIf mathChar = Number.TWO_STRING_REPRESENTATION Then currentNumber.UpdateNumber(mathChar) ElseIf mathChar = Number.THREE_STRING_REPRESENTATION Then currentNumber.UpdateNumber(mathChar) ElseIf mathChar = Number.FOUR_STRING_REPRESENTATION Then currentNumber.UpdateNumber(mathChar) ElseIf mathChar = Number.FIVE_STRING_REPRESENTATION Then currentNumber.UpdateNumber(mathChar) ElseIf mathChar = Number.SIX_STRING_REPRESENTATION Then currentNumber.UpdateNumber(mathChar) ElseIf mathChar = Number.SEVEN_STRING_REPRESENTATION Then currentNumber.UpdateNumber(mathChar) ElseIf mathChar = Number.EIGHT_STRING_REPRESENTATION Then currentNumber.UpdateNumber(mathChar) ElseIf mathChar = Number.NINE_STRING_REPRESENTATION Then currentNumber.UpdateNumber(mathChar) ElseIf mathChar = Number.DECIMAL_POINT_STRING_REPRESENTATION Then currentNumber.UpdateNumber(mathChar) ElseIf mathChar = Addition.ADDITION_STRING_REPRESENTATION Then Dim addition As New Addition If Not isInParentheticalMode Then operationsList.Add(addition) numberList.Add(currentNumber) Else currentParentheticalStatement.AllNumbers.Add(currentNumber) currentParentheticalStatement.AllOperators.Add(addition) End If currentNumber = New Number ElseIf mathChar = Number.NEGATIVE_NUMBER_STRING_REPRESENTATION Then If currentNumber.StringOfNumbers.Length > 0 Then currentNumber.UpdateNumber(mathChar) Dim subtraction As New Addition If Not isInParentheticalMode Then operationsList.Add(subtraction) numberList.Add(currentNumber) Else currentParentheticalStatement.AllNumbers.Add(currentNumber) currentParentheticalStatement.AllOperators.Add(subtraction) End If currentNumber = New Number Else currentNumber.UpdateNumber(mathChar) End If ElseIf mathChar = Multiplication.MULTIPLICATION_STRING_REPRESENTATION Then Dim multiplication As New Multiplication If Not isInParentheticalMode Then operationsList.Add(multiplication) numberList.Add(currentNumber) Else currentParentheticalStatement.AllNumbers.Add(currentNumber) currentParentheticalStatement.AllOperators.Add(multiplication) End If currentNumber = New Number ElseIf mathChar = Division.DIVISION_STRING_REPRESENTATION Then Dim division As New Division If Not isInParentheticalMode Then operationsList.Add(division) numberList.Add(currentNumber) Else currentParentheticalStatement.AllNumbers.Add(currentNumber) currentParentheticalStatement.AllOperators.Add(division) End If currentNumber = New Number ElseIf mathChar = Parenthetical.LEFT_PARENTHESIS_STRING_REPRESENTATION Then isInParentheticalMode = True ElseIf mathChar = Parenthetical.RIGHT_PARENTHESIS_STRING_REPRESENTATION Then currentNumber = currentParentheticalStatement.EvaluateParentheticalStatement numberList.Add(currentNumber) isInParentheticalMode = False End If Next Dim result As Double = 0 Dim operationIndex As Integer = 0 For Each numberOnWhichToPerformOperations As Number In numberList result = operationsList(operationIndex).PerformOperation(result, numberOnWhichToPerformOperations) operationIndex = operationIndex + 1 Next Return result End Function Public Class Number Public Const DECIMAL_POINT_STRING_REPRESENTATION As Char = "." Public Const NEGATIVE_NUMBER_STRING_REPRESENTATION As Char = "-" Public Const ZERO_STRING_REPRESENTATION As Char = "0" Public Const ONE_STRING_REPRESENTATION As Char = "1" Public Const TWO_STRING_REPRESENTATION As Char = "2" Public Const THREE_STRING_REPRESENTATION As Char = "3" Public Const FOUR_STRING_REPRESENTATION As Char = "4" Public Const FIVE_STRING_REPRESENTATION As Char = "5" Public Const SIX_STRING_REPRESENTATION As Char = "6" Public Const SEVEN_STRING_REPRESENTATION As Char = "7" Public Const EIGHT_STRING_REPRESENTATION As Char = "8" Public Const NINE_STRING_REPRESENTATION As Char = "9" Private _isNegative As Boolean Public ReadOnly Property IsNegative() As Boolean Get Return _isNegative End Get End Property Public ReadOnly Property ActualNumber() As Double Get Dim result As String = "" If HasDecimal Then If DecimalIndex = StringOfNumbers.Length - 1 Then result = StringOfNumbers.ToString Else result = StringOfNumbers.Insert(DecimalIndex, DECIMAL_POINT_STRING_REPRESENTATION).ToString End If Else result = StringOfNumbers.ToString End If If IsNegative Then result = NEGATIVE_NUMBER_STRING_REPRESENTATION & result End If Return CType(result, Double) End Get End Property Private _hasDecimal As Boolean Public ReadOnly Property HasDecimal() As Boolean Get Return _hasDecimal End Get End Property Private _decimalIndex As Integer Public ReadOnly Property DecimalIndex() As Integer Get Return _decimalIndex End Get End Property Private _stringOfNumbers As New StringBuilder Public ReadOnly Property StringOfNumbers() As StringBuilder Get Return _stringOfNumbers End Get End Property Public Sub UpdateNumber(ByVal theDigitToAppend As Char) If IsNumeric(theDigitToAppend) Then Me._stringOfNumbers.Append(theDigitToAppend) ElseIf theDigitToAppend = DECIMAL_POINT_STRING_REPRESENTATION Then Me._hasDecimal = True Me._decimalIndex = Me._stringOfNumbers.Length ElseIf theDigitToAppend = NEGATIVE_NUMBER_STRING_REPRESENTATION Then Me._isNegative = Not Me._isNegative End If End Sub Public Shared Function ConvertDoubleToNumber(ByVal numberThatIsADouble As Double) As Number Dim numberResult As New Number For Each character As Char In numberThatIsADouble.ToString.ToCharArray numberResult.UpdateNumber(character) Next Return numberResult End Function End Class Public MustInherit Class Operation Protected _firstnumber As New Number Protected _secondnumber As New Number Public Property FirstNumber() As Number Get Return _firstnumber End Get Set(ByVal value As Number) _firstnumber = value End Set End Property Public Property SecondNumber() As Number Get Return _secondnumber End Get Set(ByVal value As Number) _secondnumber = value End Set End Property End Class Public Interface IOperatable Function PerformOperation(ByVal number1 As Double, ByVal number2 As Number) As Double End Interface Public Class Addition Inherits Operation Implements IOperatable Public Const ADDITION_STRING_REPRESENTATION As String = "+" Public Sub New() End Sub Public Function PerformOperation(ByVal number1 As Double, ByVal number2 As Number) As Double Implements IOperatable.PerformOperation Dim result As Double = 0 result = number1 + number2.ActualNumber Return result End Function End Class Public Class Multiplication Inherits Operation Implements IOperatable Public Const MULTIPLICATION_STRING_REPRESENTATION As String = "*" Public Sub New() End Sub Public Function PerformOperation(ByVal number1 As Double, ByVal number2 As Number) As Double Implements IOperatable.PerformOperation Dim result As Double = 0 result = number1 * number2.ActualNumber Return result End Function End Class Public Class Division Inherits Operation Implements IOperatable Public Const DIVISION_STRING_REPRESENTATION As String = "/" Public Const DIVIDE_BY_ZERO_ERROR_MESSAGE As String = "I took a lot of time to write this program. Please don't be a child and try to defile it by dividing by zero. Nobody thinks you are funny." Public Sub New() End Sub Public Function PerformOperation(ByVal number1 As Double, ByVal number2 As Number) As Double Implements IOperatable.PerformOperation If Not number2.ActualNumber = 0 Then Dim result As Double = 0 result = number1 / number2.ActualNumber Return result Else Dim divideByZeroException As New Exception(DIVIDE_BY_ZERO_ERROR_MESSAGE) Throw divideByZeroException End If End Function End Class Public Class Parenthetical Public Const LEFT_PARENTHESIS_STRING_REPRESENTATION As String = "(" Public Const RIGHT_PARENTHESIS_STRING_REPRESENTATION As String = ")" Private _allNumbers As New List(Of Number) Public Property AllNumbers() As List(Of Number) Get Return _allNumbers End Get Set(ByVal value As List(Of Number)) _allNumbers = value End Set End Property Private _allOperators As New List(Of IOperatable) Public Property AllOperators() As List(Of IOperatable) Get Return _allOperators End Get Set(ByVal value As List(Of IOperatable)) _allOperators = value End Set End Property Public Sub New() End Sub Public Function EvaluateParentheticalStatement() As Number Dim result As Double = 0 Dim operationIndex As Integer = 0 For Each numberOnWhichToPerformOperations As Number In AllNumbers result = AllOperators(operationIndex).PerformOperation(result, numberOnWhichToPerformOperations) operationIndex = operationIndex + 1 Next Dim numberToReturn As New Number numberToReturn = Number.ConvertDoubleToNumber(result) Return numberToReturn End Function End Class End Class

的Haskell

字符的号：182

没有尝试在聪明，只是一些压缩：4行，312个字节

import Data.Char;import Text.ParserCombinators.Parsec q=either(error.show)id.runParser t id"".filter(' '/=);t=do s<-getState;a<-fmap read(many1$oneOf".-"<|>digit)<|>between(char '('>>setState id)(char ')'>>setState s)t option(s a)$choice(zipWith(\c o->char c>>return(o$s a))"+-*/"[(+),(-),(*),(/)])>>=setState>>t

现在，真正进入高尔夫球精神，3行和182个字节：

q=snd.(`e`id).filter(' '/=) e s c|[(f,h)]<-readsPrec 0 s=g h(c f);e('(':s)c=g h(c f)where(')':h,f)=e s id g('+':h)=e h.(+);g('-':h)=e h.(-);g('*':h)=e h.(*);g('/':h)=e h.(/);g h=(,)h

分解：

-- Strip spaces from the input, evaluate with empty accumulator, -- and output the second field of the result. q :: String -> Double q = snd . flip eval id . filter (not . isSpace) -- eval takes a string and an accumulator, and returns -- the final value and what’s left unused from the string. eval :: (Fractional a, Read a) => String -> (a -> a) -> (String, a) -- If the beginning of the string parses as a number, add it to the accumulator, -- then try to read an operator and further. eval str accum | [(num, rest)] <- readsPrec 0 str = oper rest (accum num) -- If the string starts parentheses, evaluate the inside with a fresh -- accumulator, and continue after the closing paren. eval ('(':str) accum = oper rest (accum num) where (')':rest, num) = eval str id -- oper takes a string and current value, and tries to read an operator -- to apply to the value. If there is none, it’s okay. oper :: (Fractional a, Read a) => String -> a -> (String, a) -- Handle operations by giving eval a pre-seeded accumulator. oper ('+':str) num = eval str (num +) oper ('-':str) num = eval str (num -) oper ('*':str) num = eval str (num *) oper ('/':str) num = eval str (num /) -- If there’s no operation parsable, just return. oper str num = (str, num)

的Python

字符的号：237

全混淆功能：

from operator import* def e(s,l=[]): if s:l+=list(s.replace(' ','')+')') a=0;o=add;d=dict(zip(')*+-/',(0,mul,o,sub,div)));p=l.pop while o: c=p(0) if c=='(':c=e(0) while l[0]not in d:c+=p(0) a=o(a,float(c));o=d[p(0)] return a

清除/半混淆功能：

import operator def calc(source, stack=[]): if source: stack += list(source.replace(' ', '') + ')') answer = 0 ops = { ')': 0, '*': operator.mul, '+': operator.add, '-': operator.sub, '/': operator.div, } op = operator.add while op: cur = stack.pop(0) if cur == '(': cur = calc(0) while stack[0] not in ops: cur += stack.pop(0) answer = op(answer, float(cur)) op = ops[stack.pop(0)] return answer

的Fortran 77（gfortran方言，现在用G77支持）

<强>的字符数： 2059

<强>模糊版本：

function e(c) character*99 c character b real f(24) integer i(24) nf=0 ni=0 20 nf=kf(0.0,nf,f) ni=ki(43,ni,i) 30 if (isp(c).eq.1) goto 20 h=fr(c) 31 g=fp(nf,f) j=ip(ni,i) select case(j) case (40) goto 20 case (42) d=g*h case (43) d=g+h case (45) d=g-h case (47) d=g/h end select 50 nf=kf(d,nf,f) 60 j=nop(c) goto (20, 70, 75, 75, 60, 75, 60, 75) (j-39) 65 e=fp(nf,f) return 70 h=fp(nf,f) goto 31 75 ni=ki(j,ni,i) goto 30 end function kf(v,n,f) real f(24) kf=n+1 f(n+1)=v return end function ki(j,n,i) integer i(24) ki=n+1 i(n+1)=j return end function fp(n,f) real f(24) fp=f(n) n=n-1 return end function ip(n,i) integer i(24) ip=i(n) n=n-1 return end function nop(s) character*99 s l=1 do while(s(l:l).eq." ".and.l.lt.99) l=l+1 enddo nop=ichar(s(l:l)) s(l:l)=" " return end function isp(s) character*99 s isp=0 l=1 do while(s(l:l).eq." ".and.l.lt.99) l=l+1 enddo isp=41-ichar(s(l:l)) if (isp.eq.1) s(l:l)=" " return end function fr(s) character*99 s m=1 n=1 i=1 do while(i.le.99) j=ichar(s(i:i)) if (j.eq.32) goto 90 if (j.ge.48.and.j.lt.58) goto 89 if (j.eq.43.or.j.eq.45) goto (89,80) m if (j.eq.46) goto (83,80) n 80 exit 83 n=2 89 m=2 90 i=i+1 enddo read(s(1:i-1),*) fr do 91 j=1,i-1 s(j:j)=" " 91 continue return end

清除版本：（3340个字符支架）

program infixeval character*99 c do while (.true.) do 10 i=1,99 c(i:i)=" " 10 continue read(*,"(A99)") c f=e(c) write(*,*)f enddo end function e(c) character*99 c character b real f(24) ! value stack integer i(24) ! operator stack nf=0 ! number of items on the value stack ni=0 ! number of items on the operator stack 20 nf=pushf(0.0,nf,f) ni=pushi(43,ni,i) ! ichar(+) = 43 D write (*,*) "'",c,"'" 30 if (isp(c).eq.1) goto 20 h=fr(c) D write (*,*) "'",c,"'" 31 g=fpop(nf,f) j=ipop(ni,i) D write(*,*) "Opperate ",g," ",char(j)," ",h select case(j) case (40) goto 20 case (42) ! "*" d=g*h case (43) ! "+" d=g+h case (45) ! "-" d=g-h case (47) ! "*" d=g/h end select 50 nf=pushf(d,nf,f) 60 j=nop(c) D write(*,*) "Got op: ", char(j) goto (20, 70, 75, 75, 60, 75, 60, 75) (j-39) 65 e=fpop(nf,f) return 70 h=fpop(nf,f) ! Encountered a "(" goto 31 75 ni=pushi(j,ni,i) goto 30 end c push onto a real stack c OB as kf function pushf(v,n,f) real f(24) pushf=n+1 f(n+1)=v D write(*,*) "Push ", v return end c push onto a integer stack c OB as ki function pushi(j,n,i) integer i(24) pushi=n+1 i(n+1)=j D write(*,*) "Push ", char(j) return end c pop from real stack c OB as fp function fpop(n,f) real f(24) fpop=f(n) n=n-1 D write (*,*) "Pop ", fpop return end c pop from integer stack c OB as ip function ipop(n,i) integer i(24) ipop=i(n) n=n-1 D write (*,*) "Pop ", char(ipop) return end c Next OPerator: returns the next nonws character, and removes it c from the string function nop(s) character*99 s l=1 do while(s(l:l).eq." ".and.l.lt.99) l=l+1 enddo nop=ichar(s(l:l)) s(l:l)=" " return end c IS an open Paren: return 1 if the next non-ws character is "(" c (also overwrite it with a space. Otherwise return not 1 function isp(s) character*99 s isp=0 l=1 do while(s(l:l).eq." ".and.l.lt.99) l=l+1 enddo isp=41-ichar(s(l:l)) if (isp.eq.1) s(l:l)=" " return end c Float Read: return the next real number in the string and removes the c character function fr(s) character*99 s m=1 ! No sign (Minus or plus) so far n=1 ! No decimal so far i=1 do while(i.le.99) j=ichar(s(i:i)) if (j.eq.32) goto 90 ! skip spaces if (j.ge.48.and.j.lt.58) goto 89 if (j.eq.43.or.j.eq.45) goto (89,80) m if (j.eq.46) goto (83,80) n c not part of a number 80 exit 83 n=2 89 m=2 90 i=i+1 enddo read(s(1:i-1),*) fr do 91 j=1,i-1 s(j:j)=" " 91 continue return end

备注：此编辑后的版本，而不是我第一次尝试更邪恶。同样的算法，但现在gotos的一个可怕的混乱在线。我抛弃了协程，但现在使用的是一对夫妇计算分支机构的口味。所有的错误检查和报告已被删除，但这个版本会悄悄地从输入意外的字符一些类恢复。此版本还编译与G77。

主要限制仍然FORTRAN的刚性格式化，长和无处不在的关键字，以及简单的原语。

C99

数字：239 (但是参看下面 209)

压缩的功能：

#define S while(*e==32)++e #define F float F strtof();char*e;F v();F g(){S;return*e++-40?strtof(e-1,&e):v();}F v(){F b,a=g();for(;;){S;F o=*e++;if(!o|o==41)return a;b=g();a=o==43?a+b:o==45?a-b:o==42?a*b:a/b;}}F f(char*x){e=x;return v();}

解压功能：

float strtof(); char* e; float v(); float g() { while (*e == ' ') ++e; return *e++ != '(' ? strtof(e-1, &e) : v(); } float v() { float b, a = g(); for (;;) { while (*e == ' ') ++e; float op = *e++; if (op == 0 || op == ')') return a; b = g(); a = op == '+' ? a + b : op == '-' ? a - b : op == '*' ? a * b : a / b; } } float eval(char* x) { e = x; return v(); }

功能不再参赛者。

编辑从克里斯*卢茨:我讨厌到践踏上另一个人的代码，但是这里是一个 209-字的版本：

#define S for(;*e==32;e++) #define X (*e++-40?strtof(e-1,&e):v()) float strtof();char*e;float v(){float o,a=X;for(;;){S;o=*e++;if(!o|o==41)return a;S;a=o-43?o-45?o-42?a/X:a*X:a-X:a+X;}} #define f(x) (e=x,v())

可读(嗯，不是真的非常可读的，但解压):

float strtof(); char *e; float v() { float o, a = *e++ != '(' ? strtof(e - 1, &e) : v(); for(;;) { for(; *e == ' '; e++); o = *e++; if(o == 0 || o==')') return a; for(; *e == ' '; e++); // I have no idea how to properly indent nested conditionals // and this is far too long to fit on one line. a = o != '+' ? o != '-' ? o != '*' ? a / (*e++ != '(' ? strtof(e - 1, &e) : v()) : a * (*e++ != '(' ? strtof(e - 1, &e) : v()) : a - (*e++ != '(' ? strtof(e - 1, &e) : v()) : a + (*e++ != '(' ? strtof(e - 1, &e) : v()); } } #define f(x) (e = x, v())

是啊， f() 是一个宏观的，不一功能，但它的工作。可读的版本具有一些逻辑改写，但不重新排序(喜欢 o != '+' 而不是的 o - '+')，但是否只是一个缩进(和预处理过的)版本的另一个。我一直试图简化 if(!o|o==41)return a; 部分成 for() 环，但是它从来没有使它更短。我仍然相信这是可以做到的，但我做高尔夫球。如果我工作这个问题了，它将在语言必须不被命名为.

Common Lisp的

（SBCL），点击字符的号：251

(defun g(e)(if(numberp e)e(let((m (g (pop e)))(o(loop for x in e by #'cddr collect x))(n(loop for x in (cdr e)by #'cddr collect (g x))))(mapcar(lambda(x y)(setf m(apply x(list m y))))o n)m)))(defun w(e)(g(read-from-string(concatenate'string"("e")"))))

正确版本（387个字符）：

(defun wrapper (exp) (golf-eval (read-from-string (concatenate 'string "(" exp ")")))) (defun golf-eval (exp) (if (numberp exp) exp (let ((mem (golf-eval (pop exp))) (op-list (loop for x in exp by #'cddr collect x)) (num-list (loop for x in (cdr exp) by #'cddr collect (golf-eval x)))) (mapcar (lambda (x y) (setf mem (apply x (list mem y)))) op-list num-list) mem)))

输入是形式w()，它接受一个字符串参数。它使用的把戏，NUMS /操作数和运算都在图样n 0：N 0：N ...和递归地评估所有操作数，并因此获得筑巢很便宜。 ;）

的JavaScript（不兼容IE）

字符的编号：260分之268

全混淆功能：

function e(x){x=x.replace(/ /g,'')+')' function P(n){return x[0]=='('?(x=x.substr(1),E()):(n=/^[-+]?[\d.]+/(x)[0],x=x.substr(n.length),+n)}function E(a,o,b){a=P() for(;;){o=x[0] x=x.substr(1) if(o==')')return a b=P() a=o=='+'?a+b:o=='-'?a-b:o=='*'?a*b:a/b}}return E()}

或，在JavaScript 1.8（火狐3+），则可以通过使用表达式封闭保存几个字符：

e=function(x,P,E)(x=x.replace(/ /g,'')+')',P=function(n)(x[0]=='('?(x=x.substr(1),E()):(n=/^[-+]?[\d.]+/(x)[0],x=x.substr(n.length),+n)),E=function(a,o,b){a=P() for(;;){o=x[0] x=x.substr(1) if(o==')')return a b=P() a=o=='+'?a+b:o=='-'?a-b:o=='*'?a*b:a/b}},E())

清除/半混淆功能：

function evaluate(x) { x = x.replace(/ /g, "") + ")"; function primary() { if (x[0] == '(') { x = x.substr(1); return expression(); } var n = /^[-+]?\d*\.?\d*/.exec(x)[0]; x = x.substr(n.length); return +n; } function expression() { var a = primary(); for (;;) { var operator = x[0]; x = x.substr(1); if (operator == ')') { return a; } var b = primary(); a = (operator == '+') ? a + b : (operator == '-') ? a - b : (operator == '*') ? a * b : a / b; } } return expression(); }

无论版本将在IE工作，因为它们使用阵列式的下标上的字符串。如果您更换x[0] x.charAt(0)两次出现的，第一个应该在任何地方工作。

我通过转动变量成函数参数和if语句与条件运算替换另一个切出由于第一版本的一些更多的字符。

C＃与 Regex的爱

<强>的字符数： 384

<强>全混淆：

float E(string i){i=i.Replace(" ","");Regex b=new Regex(@"$(?>[^()]+|\((?<D>)|$(?<-D>))*(?(D)(?!))\)");i=b.Replace(i,m=>Eval(m.Value.Substring(1,m.Length-2)).ToString());float r=0;foreach(Match m in Regex.Matches(i,@"(?<=^|\D)-?[\d.]+")){float f=float.Parse(m.Value);if(m.Index==0)r=f;else{char o=i[m.Index-1];if(o=='+')r+=f;if(o=='-')r-=f;if(o=='*')r*=f;if(o=='/')r/=f;}}return r;}

不会混淆的：

private static float Eval(string input) { input = input.Replace(" ", ""); Regex balancedMatcher = new Regex(@"$ (?> [^()]+ | \( (?<Depth>) | $ (?<-Depth>) )* (?(Depth)(?!)) \)", RegexOptions.IgnorePatternWhitespace); input = balancedMatcher.Replace(input, m => Eval(m.Value.Substring(1, m.Length - 2)).ToString()); float result = 0; foreach (Match m in Regex.Matches(input, @"(?<=^|\D)-?[\d.]+")) { float floatVal = float.Parse(m.Value); if (m.Index == 0) { result = floatVal; } else { char op = input[m.Index - 1]; if (op == '+') result += floatVal; if (op == '-') result -= floatVal; if (op == '*') result *= floatVal; if (op == '/') result /= floatVal; } } return result; }
利用.NET的正则表达式的均衡组特征。

PHP

字符的号：284

混淆：

function f($m){return c($m[1]);}function g($n,$m){$o=$m[0];$m[0]=' ';return$o=='+'?$n+$m:($o=='-'?$n-$m:($o=='*'?$n*$m:$n/$m));}function c($s){while($s!=($t=preg_replace_callback('/$([^()]*)$/',f,$s)))$s=$t;preg_match_all('![-+/*].*?[\d.]+!',"+$s",$m);return array_reduce($m[0],g);}

可读：

function callback1($m) {return c($m[1]);} function callback2($n,$m) { $o=$m[0]; $m[0]=' '; return $o=='+' ? $n+$m : ($o=='-' ? $n-$m : ($o=='*' ? $n*$m : $n/$m)); } function c($s){ while ($s != ($t = preg_replace_callback('/$([^()]*)$/','callback1',$s))) $s=$t; preg_match_all('![-+/*].*?[\d.]+!', "+$s", $m); return array_reduce($m[0], 'callback2'); } $str = ' 2.45/8.5 * -9.27 + ( 5 * 0.0023 ) '; var_dump(c($str)); # float(-2.66044117647)

应与任何有效的输入（包括负数和任意空白）工作

SQL（SQL Server 2008的）
字符的
编号：4202

全混淆功能：

WITH Input(id,str)AS(SELECT 1,'1 + 3 / -8'UNION ALL SELECT 2,'2*3*4*5+99'UNION ALL SELECT 3,'4 * (9 - 4)/ (2 * 6 - 2)+ 8'UNION ALL SELECT 4,'1 + ((123 * 3 - 69)/ 100)'UNION ALL SELECT 5,'2.45/8.5*9.27+(5*0.0023)'),Separators(i,ch,str_src,priority)AS(SELECT 1,'-',1,1UNION ALL SELECT 2,'+',1,1UNION ALL SELECT 3,'*',1,1UNION ALL SELECT 4,'/',1,1UNION ALL SELECT 5,'(',0,0UNION ALL SELECT 6,')',0,0),SeparatorsStrSrc(str,i)AS(SELECT CAST('['AS varchar(max)),0UNION ALL SELECT str+ch,SSS.i+1FROM SeparatorsStrSrc SSS INNER JOIN Separators S ON SSS.i=S.i-1WHERE str_src<>0),SeparatorsStr(str)AS(SELECT str+']'FROM SeparatorsStrSrc WHERE i=(SELECT COUNT(*)FROM Separators WHERE str_src<>0)),ExprElementsSrc(id,i,tmp,ele,pre_ch,input_str)AS(SELECT id,1,CAST(LEFT(str,1)AS varchar(max)),CAST(''AS varchar(max)),CAST(' 'AS char(1)),SUBSTRING(str,2,LEN(str))FROM Input UNION ALL SELECT id,CASE ele WHEN''THEN i ELSE i+1 END,CAST(CASE WHEN LEFT(input_str,1)=' 'THEN''WHEN tmp='-'THEN CASE WHEN pre_ch LIKE(SELECT str FROM SeparatorsStr)THEN tmp+LEFT(input_str,1)ELSE LEFT(input_str,1)END WHEN LEFT(input_str,1)IN(SELECT ch FROM Separators)OR tmp IN(SELECT ch FROM Separators)THEN LEFT(input_str,1)ELSE tmp+LEFT(input_str,1)END AS varchar(max)),CAST(CASE WHEN LEFT(input_str,1)=' 'THEN tmp WHEN LEFT(input_str,1)='-'THEN CASE WHEN tmp IN(SELECT ch FROM Separators)THEN tmp ELSE''END WHEN LEFT(input_str,1)IN(SELECT ch FROM Separators)OR tmp IN(SELECT ch FROM Separators)THEN CASE WHEN tmp='-'AND pre_ch LIKE(SELECT str FROM SeparatorsStr)THEN''ELSE tmp END ELSE''END AS varchar(max)),CAST(LEFT(ele,1)AS char(1)),SUBSTRING(input_str,2,LEN(input_str))FROM ExprElementsSrc WHERE input_str<>''OR tmp<>''),ExprElements(id,i,ele)AS(SELECT id,i,ele FROM ExprElementsSrc WHERE ele<>''),Scanner(id,i,val)AS(SELECT id,i,CAST(ele AS varchar(max))FROM ExprElements WHERE ele<>''UNION ALL SELECT id,MAX(i)+1,NULL FROM ExprElements GROUP BY id),Operator(op,priority)AS(SELECT ch,priority FROM Separators WHERE priority<>0),Calc(id,c,i,pop_count,s0,s1,s2,stack,status)AS(SELECT Scanner.id,1,1,0,CAST(scanner.val AS varchar(max)),CAST(NULL AS varchar(max)),CAST(NULL AS varchar(max)),CAST(''AS varchar(max)),CAST('init'AS varchar(max))FROM Scanner WHERE Scanner.i=1UNION ALL SELECT Calc.id,Calc.c+1,Calc.i,3,NULL,NULL,NULL,CASE Calc.s1 WHEN'+'THEN CAST(CAST(Calc.s2 AS real)+CAST(Calc.s0 AS real)AS varchar(max))WHEN'-'THEN CAST(CAST(Calc.s2 AS real)-CAST(Calc.s0 AS real)AS varchar(max))WHEN'*'THEN CAST(CAST(Calc.s2 AS real)*CAST(Calc.s0 AS real)AS varchar(max))WHEN'/'THEN CAST(CAST(Calc.s2 AS real)/CAST(Calc.s0 AS real)AS varchar(max))ELSE NULL END+' '+stack,CAST('calc '+Calc.s1 AS varchar(max))FROM Calc INNER JOIN Scanner NextVal ON Calc.id=NextVal.id AND Calc.i+1=NextVal.i WHERE Calc.pop_count=0AND ISNUMERIC(Calc.s2)=1AND Calc.s1 IN(SELECT op FROM Operator)AND ISNUMERIC(Calc.s0)=1AND(SELECT priority FROM Operator WHERE op=Calc.s1)>=COALESCE((SELECT priority FROM Operator WHERE op=NextVal.val),0)UNION ALL SELECT Calc.id,Calc.c+1,Calc.i,3,NULL,NULL,NULL,s1+' '+stack,CAST('paren'AS varchar(max))FROM Calc WHERE pop_count=0AND s2='('AND ISNUMERIC(s1)=1AND s0=')'UNION ALL SELECT Calc.id,Calc.c+1,Calc.i,Calc.pop_count-1,s1,s2,CASE WHEN LEN(stack)>0THEN SUBSTRING(stack,1,CHARINDEX(' ',stack)-1)ELSE NULL END,CASE WHEN LEN(stack)>0THEN SUBSTRING(stack,CHARINDEX(' ',stack)+1,LEN(stack))ELSE''END,CAST('pop'AS varchar(max))FROM Calc WHERE Calc.pop_count>0UNION ALL SELECT Calc.id,Calc.c+1,Calc.i+1,Calc.pop_count,CAST(NextVal.val AS varchar(max)),s0,s1,coalesce(s2,'')+' '+stack,cast('read'as varchar(max))FROM Calc INNER JOIN Scanner NextVal ON Calc.id=NextVal.id AND Calc.i+1=NextVal.i WHERE NextVal.val IS NOT NULL AND Calc.pop_count=0AND((Calc.s0 IS NULL OR calc.s1 IS NULL OR calc.s2 IS NULL)OR NOT(ISNUMERIC(Calc.s2)=1AND Calc.s1 IN(SELECT op FROM Operator)AND ISNUMERIC(calc.s0)=1AND (SELECT priority FROM Operator WHERE op=Calc.s1)>=COALESCE((SELECT priority FROM Operator WHERE op=NextVal.val),0))AND NOT(s2='('AND ISNUMERIC(s1)=1AND s0=')')))SELECT Calc.id,Input.str,Calc.s0 AS result FROM Calc INNER JOIN Input ON Calc.id=Input.id WHERE Calc.c=(SELECT MAX(c)FROM Calc calc2 WHERE Calc.id=Calc2.id)ORDER BY id

清除/半混淆功能：

WITH Input(id, str) AS ( SELECT 1, '1 + 3 / -8' UNION ALL SELECT 2, '2*3*4*5+99' UNION ALL SELECT 3, '4 * (9 - 4) / (2 * 6 - 2) + 8' UNION ALL SELECT 4, '1 + ((123 * 3 - 69) / 100)' UNION ALL SELECT 5, '2.45/8.5*9.27+(5*0.0023)' ) , Separators(i, ch, str_src, priority) AS ( SELECT 1, '-', 1, 1 UNION ALL SELECT 2, '+', 1, 1 UNION ALL SELECT 3, '*', 1, 1 UNION ALL SELECT 4, '/', 1, 1 UNION ALL SELECT 5, '(', 0, 0 UNION ALL SELECT 6, ')', 0, 0 ) , SeparatorsStrSrc(str, i) AS ( SELECT CAST('[' AS varchar(max)), 0 UNION ALL SELECT str + ch , SSS.i + 1 FROM SeparatorsStrSrc SSS INNER JOIN Separators S ON SSS.i = S.i - 1 WHERE str_src <> 0 ) , SeparatorsStr(str) AS ( SELECT str + ']' FROM SeparatorsStrSrc WHERE i = (SELECT COUNT(*) FROM Separators WHERE str_src <> 0) ) , ExprElementsSrc(id, i, tmp, ele, pre_ch, input_str) AS ( SELECT id , 1 , CAST(LEFT(str, 1) AS varchar(max)) , CAST('' AS varchar(max)) , CAST(' ' AS char(1)) , SUBSTRING(str, 2, LEN(str)) FROM Input UNION ALL SELECT id , CASE ele WHEN '' THEN i ELSE i + 1 END , CAST( CASE WHEN LEFT(input_str, 1) = ' ' THEN '' WHEN tmp = '-' THEN CASE WHEN pre_ch LIKE (SELECT str FROM SeparatorsStr) THEN tmp + LEFT(input_str, 1) ELSE LEFT(input_str, 1) END WHEN LEFT(input_str, 1) IN (SELECT ch FROM Separators) OR tmp IN (SELECT ch FROM Separators) THEN LEFT(input_str, 1) ELSE tmp + LEFT(input_str, 1) END AS varchar(max)) , CAST( CASE WHEN LEFT(input_str, 1) = ' ' THEN tmp WHEN LEFT(input_str, 1) = '-' THEN CASE WHEN tmp IN (SELECT ch FROM Separators) THEN tmp ELSE '' END WHEN LEFT(input_str, 1) IN (SELECT ch FROM Separators) OR tmp IN (SELECT ch FROM Separators) THEN CASE WHEN tmp = '-' AND pre_ch LIKE (SELECT str FROM SeparatorsStr) THEN '' ELSE tmp END ELSE '' END AS varchar(max)) , CAST(LEFT(ele, 1) AS char(1)) , SUBSTRING(input_str, 2, LEN(input_str)) FROM ExprElementsSrc WHERE input_str <> '' OR tmp <> '' ) , ExprElements(id, i, ele) AS ( SELECT id , i , ele FROM ExprElementsSrc WHERE ele <> '' ) , Scanner(id, i, val) AS ( SELECT id , i , CAST(ele AS varchar(max)) FROM ExprElements WHERE ele <> '' UNION ALL SELECT id , MAX(i) + 1 , NULL FROM ExprElements GROUP BY id ) , Operator(op, priority) AS ( SELECT ch , priority FROM Separators WHERE priority <> 0 ) , Calc(id, c, i, pop_count, s0, s1, s2, stack, status) AS ( SELECT Scanner.id , 1 , 1 , 0 , CAST(scanner.val AS varchar(max)) , CAST(NULL AS varchar(max)) , CAST(NULL AS varchar(max)) , CAST('' AS varchar(max)) , CAST('init' AS varchar(max)) FROM Scanner WHERE Scanner.i = 1 UNION ALL SELECT Calc.id , Calc.c + 1 , Calc.i , 3 , NULL , NULL , NULL , CASE Calc.s1 WHEN '+' THEN CAST(CAST(Calc.s2 AS real) + CAST(Calc.s0 AS real) AS varchar(max)) WHEN '-' THEN CAST(CAST(Calc.s2 AS real) - CAST(Calc.s0 AS real) AS varchar(max)) WHEN '*' THEN CAST(CAST(Calc.s2 AS real) * CAST(Calc.s0 AS real) AS varchar(max)) WHEN '/' THEN CAST(CAST(Calc.s2 AS real) / CAST(Calc.s0 AS real) AS varchar(max)) ELSE NULL END + ' ' + stack , CAST('calc ' + Calc.s1 AS varchar(max)) FROM Calc INNER JOIN Scanner NextVal ON Calc.id = NextVal.id AND Calc.i + 1 = NextVal.i WHERE Calc.pop_count = 0 AND ISNUMERIC(Calc.s2) = 1 AND Calc.s1 IN (SELECT op FROM Operator) AND ISNUMERIC(Calc.s0) = 1 AND (SELECT priority FROM Operator WHERE op = Calc.s1) >= COALESCE((SELECT priority FROM Operator WHERE op = NextVal.val), 0) UNION ALL SELECT Calc.id , Calc.c + 1 , Calc.i , 3 , NULL , NULL , NULL , s1 + ' ' + stack , CAST('paren' AS varchar(max)) FROM Calc WHERE pop_count = 0 AND s2 = '(' AND ISNUMERIC(s1) = 1 AND s0 = ')' UNION ALL SELECT Calc.id , Calc.c + 1 , Calc.i , Calc.pop_count - 1 , s1 , s2 , CASE WHEN LEN(stack) > 0 THEN SUBSTRING(stack, 1, CHARINDEX(' ', stack) - 1) ELSE NULL END , CASE WHEN LEN(stack) > 0 THEN SUBSTRING(stack, CHARINDEX(' ', stack) + 1, LEN(stack)) ELSE '' END , CAST('pop' AS varchar(max)) FROM Calc WHERE Calc.pop_count > 0 UNION ALL SELECT Calc.id , Calc.c + 1 , Calc.i + 1 , Calc.pop_count , CAST(NextVal.val AS varchar(max)) , s0 , s1 , coalesce(s2, '') + ' ' + stack , cast('read' as varchar(max)) FROM Calc INNER JOIN Scanner NextVal ON Calc.id = NextVal.id AND Calc.i + 1 = NextVal.i WHERE NextVal.val IS NOT NULL AND Calc.pop_count = 0 AND ( (Calc.s0 IS NULL or calc.s1 is null or calc.s2 is null) OR NOT( ISNUMERIC(Calc.s2) = 1 AND Calc.s1 IN (SELECT op FROM Operator) AND ISNUMERIC(calc.s0) = 1 AND (SELECT priority FROM Operator WHERE op = Calc.s1) >= COALESCE((SELECT priority FROM Operator WHERE op = NextVal.val), 0) ) AND NOT(s2 = '(' AND ISNUMERIC(s1) = 1 AND s0 = ')') ) ) SELECT Calc.id , Input.str , Calc.s0 AS result FROM Calc INNER JOIN Input ON Calc.id = Input.id WHERE Calc.c = (SELECT MAX(c) FROM Calc calc2 WHERE Calc.id = Calc2.id) ORDER BY id

这不是最短的。但我认为这是SQL非常灵活。这很容易增加新的运营商。这很容易改变运营商的优先级。

F＃

字符的号：327

OP一直在寻找的F＃版本，在这儿呢。因为我上滥用的 REF 的这里保存字符可以做到的要好很多。它处理大多数的东西，如<强> - （1.0），第3 - -3 并甚至<强> 0 - 0.5 等

let g s= let c=ref[for x in System.Text.RegularExpressions.Regex.Matches(s,"[0-9.]+|[^\s]")->x.Value] let rec e v=if (!c).IsEmpty then v else let h=(!c).Head c:=(!c).Tail match h with|"("->e(e 0.0)|")"->v|"+"->e(v+(e 0.0))|"-"->e(v-(e 0.0))|"/"->e(v/(e 0.0))|"*"->e(v*(e 0.0))|x->float x e(e 0.0)

Ĵ
字符的
数：208

的杰夫·莫泽的评论后，我意识到，我已经完全忘记了这门语言。我不是专家，但我第一次尝试去比较好。

e=:>@{:@f@;: f=:''&(4 :0) 'y x'=.x g y while.($y)*-.')'={.>{.y do.'y x'=.(x,>(-.'/'={.>{.y){('%';y))g}.y end.y;x ) g=:4 :0 z=.>{.y if.z='('do.'y z'=.f}.y else.if.z='-'do.z=.'_',>{.}.y end.end.(}.y);":".x,z )

这有点恼人的，不必x/y和-z映射到J的x%y和_z。如果没有，也许这代码的50％可能会消失。

的Python（没有导入任何东西）

字符的号：222

我偷了Dave的答案花样繁多，但我设法刮掉了一些更多的字符。

def e(s,l=0,n=0,f='+'): if s:l=[c for c in s+')'if' '!=c] while f!=')': p=l.pop;m=p(0) if m=='(':m=e(0,l) while l[0]not in'+-*/)':m+=p(0) m=float(m);n={'+':n+m,'-':n-m,'*':n*m,'/':n/(m or 1)}[f];f=p(0) return n

评论版：

def evaluate(stringexpr, listexpr=0, n=0, f_operation='+'): # start out as taking 0 + the expression... (or could use 1 * ;) # We'll prefer to keep the expression as a list of characters, # so we can use .pop(0) to eat up the expression as we go. if stringexpr: listexpr = [c for c in stringexpr+')' if c!=' '] # use ')' as sentinel to return the answer while f_operation != ')': m_next = listexpr.pop(0) if m_next == '(': # lists are passed by reference, so this call will eat the (parexp) m_next = evaluate(None, listexpr) else: # rebuild any upcoming numeric chars into a string while listexpr[0] not in '+-*/)': m_next += listexpr.pop(0) # Update n as the current answer. But never divide by 0. m = float(m_next) n = {'+':n+m, '-':n-m, '*':n*m, '/':n/(m or 1)}[f_operation] # prepare the next operation (known to be one of '+-*/)') f_operation = listexpr.pop(0) return n

C＃

字符的号：403

因此，这里是我的解决方案...我还在等待有人在C＃中可以击败它张贴之一。（马克Gravell接近，经过一番摆弄多还可以做的比我好。）

全混淆功能：

float e(string x){float v=0;if(float.TryParse(x,out v))return v;x+=';';int t=0; char o,s='?',p='+';float n=0;int l=0;for(int i=0;i<x.Length;i++){o=s;if( x[i]!=' '){s=x[i];if(char.IsDigit(x[i])|s=='.'|(s=='-'&o!='1'))s='1';if(s==')') l--;if(s!=o&l==0){if(o=='1'|o==')'){n=e(x.Substring(t,i-t));if(p=='+')v+=n; if(p=='-')v-=n;if(p=='*')v*=n;if(p=='/')v/=n;p=x[i];}t=i;if(s=='(')t++;} if(s=='(')l++;}}return v;}

半混淆功能：

public static float Eval(string expr) { float val = 0; if (float.TryParse(expr, out val)) return val; expr += ';'; int tokenStart = 0; char oldState, state = '?', op = '+'; float num = 0; int level = 0; for (int i = 0; i < expr.Length; i++) { oldState = state; if (expr[i] != ' ') { state = expr[i]; if (char.IsDigit(expr[i]) || state == '.' || (state == '-' && oldState != '1')) state = '1'; if (state == ')') level--; if (state != oldState && level == 0) { if (oldState == '1' || oldState == ')') { num = Eval(expr.Substring(tokenStart, i - tokenStart)); if (op == '+') val += num; if (op == '-') val -= num; if (op == '*') val *= num; if (op == '/') val /= num; op = expr[i]; } tokenStart = i; if (state == '(') tokenStart++; } if (state == '(') level++; } } return val; }

没有什么太聪明会在这里，它似乎竟被。功能并但是具有作为优势重入（即线程安全的）。

我也相当满意，字符的数量，因为它是写在C＃（有效1.0，2.0，和3.0我相信）。

下面来另一个问题：

壳脚本（使用SED + AWK）

字符的号：295

混淆：

e(){ a="$1";while echo "$a"|grep -q $;do eval "`echo "$a"|sed 's/\(.*$($[^()]*$)$.*$/a="\1\`e \"\2\"\`\3"/'`";done; echo "$a"|sed 's/$[-+*/]$ *$-\?$ */ \1 \2/g'|awk '{t=$1;for(i=2;i<NF;i+=2){j=$(i+1);if($i=="+") t+=j; else if($i=="-") t-=j; else if($i=="*") t*=j; else t/=j}print t}';}

可读

e () { a="$1" # Recursively process bracket-expressions while echo "$a"|grep -q $; do eval "`echo "$a"| sed 's/\(.*$($[^()]*$)$.*$/a="\1\`e \"\2\"\`\3"/'`" done # Compute expression without brackets echo "$a"| sed 's/$[-+*/]$ *$-\?$ */ \1 \2/g'| awk '{ t=$1; for(i=2;i<NF;i+=2){ j=$(i+1); if($i=="+") t+=j; else if($i=="-") t-=j; else if($i=="*") t*=j; else t/=j } print t }' }

测试：

str=' 2.45 / 8.5 * 9.27 + ( 5 * 0.0023 ) ' echo "$str"|bc -l e "$str"

结果：

2.68344117647058823526 2.68344

MATLAB（v7.8.0）

字符的号：239

混淆功能：

function [v,s]=m(s),r=1;while s,s=regexp(s,'( ?)(?(1)-?)[\.\d]+|\S','match');c=s{end};s=[s{1:end-1}];if any(c>47),v=str2num(c);elseif c>41,[l,s]=m(s);v=[l/v l*v l+v l-v];v=v(c=='/*+-');if r,break;end;r=1;elseif c<41,break;end;r=r&c~=41;end

清除（ER）功能：

function [value,str] = math(str) returnNow = 1; while str, str = regexp(str,'( ?)(?(1)-?)[\.\d]+|\S','match'); current = str{end}; str = [str{1:end-1}]; if any(current > 47), value = str2num(current); elseif current > 41, [leftValue,str] = math(str); value = [leftValue/value leftValue*value ... leftValue+value leftValue-value]; value = value(current == '/*+-'); if returnNow, break; end; returnNow = 1; elseif current < 41, break; end; returnNow = returnNow & (c ~= 41); end

测试：

>> [math('1 + 3 / -8'); ... math('2*3*4*5+99'); ... math('4 * (9 - 4) / (2 * 6 - 2) + 8'); ... math('1 + ((123 * 3 - 69) / 100)'); ... math('2.45/8.5*9.27+(5*0.0023)')] ans = -0.5000 219.0000 10.0000 4.0000 2.6834

梗概：的正则表达式和递归的混合物。几乎是最好的，我已经能够到目前为止做的，没有欺骗和使用EVAL。

红宝石

字符数：170

混淆：

def s(x) while x.sub!(/$([^\($]*?)\)/){s($1)} x.gsub!('--','') end while x.sub!(/(-?[\d.]+)[ ]*([+\-*\/])[ ]*(-?[\d.]+)/){$1.to_f.send($2,$3.to_f)} end x.strip.to_f end

可读：

def s(x) while x.sub!(/$([^\($]*?)\)/){s($1)} x.gsub!('--','') end while x.sub!(/(-?[\d.]+)[ ]*([+\-*\/])[ ]*(-?[\d.]+)/){$1.to_f.send($2,$3.to_f)} end x.strip.to_f end [ ['1 + 3 / -8', -0.5], ['2*3*4*5+99', 219], ['4 * (9 - 4) / (2 * 6 - 2) + 8', 10], ['1 + ((123 * 3 - 69) / 100)', 4], ['2.45/8.5*9.27+(5*0.0023)',2.68344117647059], ['(3+7) - (5+2)', 3] ].each do |pair| a,b = s(String.new(pair[0])),pair[1] print pair[0].ljust(25), ' = ', b, ' (', a==b, ')' puts end

有没有真正的混淆这一个，我决定后新鲜，因为它是从我第一次完全不同。我应该从一开始就看到了这一点。这个过程是一个非常简单的排除过程：发现并解决了对最高括号（嵌在最内层）成数，直到没有更多的发现，然后解决所有现存数量和经营到结果。而且，在解析括号中的语句我有它去除所有双破折号（Float.to_f不知道该怎么跟他们做）。

因此，它仅仅通过处理顺序支持括号内正数和负数（3，3，＆-3），甚至否定的子表达式。唯一的更短的实施是Perl（W / O EVAL）之一。

编辑：我还在追逐Perl的，但是这是第二个最小的答案现在。我更改第二正则表达式，并通过改变字符串的处理是破坏性的（取代了旧的字符串）缩小它。这消除了需要复制的字符串，我发现仅仅是一个新的字符串指针。和从重新命名功能，以取值解决保存几个字符。

Python和正则表达式

字符的号：283

全混淆功能：

import re from operator import* def c(e): O=dict(zip("+-/*()",(add,sub,truediv,mul))) a=[add,0];s=a for v,o in re.findall("(-?[.\d]+)|([+-/*()])",e): if v:s=[float(v)]+s elif o=="(":s=a+s elif o!=")":s=[O[o]]+s if v or o==")":s[:3]=[s[1](s[2],s[0])] return s[0]

不混乱：

import re from operator import * def compute(s): operators = dict(zip("+-/*()", (add, sub, truediv, mul))) stack = [add, 0] for val, op in re.findall("(-?[.\d]+)|([+-/*()])", s): if val: stack = [float(val)] + stack elif op == "(": stack = [add, 0] + stack elif op != ")": stack = [operators[op]] + stack if val or op == ")": stack[:3] = [stack[1](stack[2], stack[0])] return stack[0]

我想看看我的驾驶室使用正则表达式击败其他Python的解决方案。

不能。

我使用的正则表达式创建双（VAL，OP），其中在每对仅一个项目是有效的列表。的代码的其余部分是相当标准的基于堆栈的解析器与使用Python列表赋值语法的计算结果代替顶部3细胞在堆的一个巧妙的方法。使用负数这项工作只需要两个附加的字符（ - ？在正则表达式）。

的Python

字符的号：382

又一Python的溶液，大量使用正则表达式替换。每个经过循环的最简单的表达式被计算运行并且结果放回字符串。

这是是非模糊代码，除非你考虑正则表达式来进行模糊处理。

import re from operator import * operators = dict(zip("+-/*", (add, sub, truediv, mul))) def compute(s): def repl(m): v1, op, v2 = m.groups() return str(operators[op](float(v1), float(v2))) while not re.match("^\d+\.\d+$", s): s = re.sub("([.\d]+)\s*([+-/*])\s*([.\d]+)", repl, s) s = re.sub("$([.\d]+)$", r"\1", s) return s

有了这样的想法，就像我在转向，不能让他走，直到我把它写下来，并使其正常工作。

C＃

<强>的字符数：396 （更新）

（但没有你添加“/ -8”的考验，我不倾向于解决它......

static float Eval(string s){int i,j;s=s.Trim();while((i=s.IndexOf(')'))>=0){j=s.LastIndexOf('(',i,i);s=s.Substring(0,j++)+Eval(s.Substring(j,i-j))+s.Substring(i+1);}if((i=s.LastIndexOfAny("+-*/".ToCharArray()))<0) return float.Parse(s);var r=float.Parse(s.Substring(i+1));var l=i>0?Eval(s.Substring(0,i)):(float?)null;return s[i]=='+'?(l??0)+r:(s[i]=='-'?(l??0)-r:(s[i]=='/'?(l??1)/r:(l??1)*r));}

自：

static float Eval(string s) { int i, j; s = s.Trim(); while ((i = s.IndexOf(')')) >= 0) { j = s.LastIndexOf('(', i, i); s = s.Substring(0, j++) + Eval(s.Substring(j, i - j)) + s.Substring(i + 1); } if ((i = s.LastIndexOfAny("+-*/".ToCharArray())) < 0) return float.Parse(s); var r = float.Parse(s.Substring(i + 1)); var l = i > 0 ? Eval(s.Substring(0, i)) : (float?)null; return s[i] == '+' ? (l ?? 0) + r : (s[i] == '-' ? (l ?? 0) - r : (s[i] == '/' ? (l ?? 1) / r : (l ?? 1) * r)); }

的Python

字符的号：235

全混淆功能：

def g(a): i=len(a) while i: try:m=g(a[i+1:]);n=g(a[:i]);a=str({'+':n+m,'-':n-m,'*':n*m,'/':n/(m or 1)}[a[i]]) except:i-=1;j=a.rfind('(')+1 if j:k=a.find(')',j);a=a[:j-1]+str(g(a[j:k]))+a[k+1:] return float(a.replace('--',''))

半混淆：

def g(a): i=len(a); # do the math while i: try: # recursively evaluate left and right m=g(a[i+1:]) n=g(a[:i]) # try to do the math assuming that a[i] is an operator a=str({'+':n+m,'-':n-m,'*':n*m,'/':n/(m or 1)}[a[i]]) except: # failure -> next try i-=1 j=a.rfind('(')+1 # replace brackets in parallel (this part is executed first) if j: k=a.find(')',j) a=a[:j-1]+str(g(a[j:k]))+a[k+1:] return float(a.replace('--',''))

FWIW，第n + 1的Python溶液。在公然滥用尝试 - 除了我使用的是试错法。它应该处理所有情况下，此包括像-(8)，--8和g('-(1 - 3)')东西。这是重入的。而无需为--情况下，其许多实现方式中不支持的支持，这是在217个字符（参见前面的修改）。

感谢在周日和周一另一个30分钟一个有趣的时刻。由于 krubo 他漂亮的字典。

红宝石

字符的数：<德尔> 217 179

~~这是最短的红宝石解决方案到现在（一个主要基于正则表达式产生不正确的答案时，字符串中包含括号的几组）~~ - 不再是真实的。基于正则表达式和替代解决方案更短。这一类是基于蓄电池的堆叠并解析整个表达式从左到右。这是重入，不修改输入字符串。它可以被指责打破不使用eval的规则，因为它调用Float的方法具有相同的名称作为自己的数学助记符（+， - ，/，*）

混淆代码（旧版本，下面微调）：

def f(p);a,o=[0],['+'] p.sub(/-/,'+-').scan(/(?:(-?\d+(?:\.\d+)?)|(.))\s*/).each{|n| q,w=n;case w;when'(';a<<0;o<<'+';when')';q=a.pop;else;o<<w end if q.nil?;a[-1]=a[-1].method(o.pop).call(q.to_f) if !q.nil?};a[0];end

更多混淆代码：

def f(p);a,o=[0],[:+] p.scan(/(?:(-?\d+(?:\.\d+)?)|(.))\s*/).each{|n|q,w=n;case w when'(';a<<0;o<<:+;when')';q=a.pop;else;o<<w;end if !q a<<a.pop.send(o.pop,q.to_f)if q};a[0];end

清洁代码：

def f(p) accumulators, operands = [0], ['+'] p.gsub(/-/,'+-').scan(/(?:(-?\d+(?:\.\d+)?)|(.))\s*/).each do |n| number, operand = n case operand when '(' accumulators << 0 operands << '+' when ')' number = accumulators.pop operands.pop else operands[-1] = operand end if number.nil? accumulators[-1] = accumulators.last.method(operands[-1]).call(number.to_f) unless number.nil? end accumulators.first end

红宝石1.8.7

字符的号：620

不要试图把它容易对我的实现，这是我第一次在我的生活写了一个表达式解析器！我保证，这是不是最好的。

<强>混淆：

def solve_expression(e) t,r,s,c,n=e.chars.to_a,[],'','','' while(c=t.shift) n=t[0] if (s+c).match(/^(-?)[.\d]+$/) || (!n.nil? && n.match(/\d/) && c=='-') s+=c elsif (c=='-' && n=='(') || c=='(' m,o,x=c=='-',1,'' while(c=t.shift) o+=1 if c=='(' o-=1 if c==')' x+=c unless c==')' && o==0 break if o==0 end r.push(m ? -solve_expression(x) : solve_expression(x)) s='' elsif c.match(/[+\-\/*]/) r.push(c) and s='' else r.push(s) if !s.empty? s='' end end r.push(s) unless s.empty? i=1 a=r[0].to_f while i<r.count b,c=r[i..i+1] c=c.to_f case b when '+': a=a+c when '-': a=a-c when '*': a=a*c when '/': a=a/c end i+=2 end a end

<强>可读：

def solve_expression(expr) chars = expr.chars.to_a # characters of the expression parts = [] # resulting parts s,c,n = '','','' # current string, character, next character while(c = chars.shift) n = chars[0] if (s + c).match(/^(-?)[.\d]+$/) || (!n.nil? && n.match(/\d/) && c == '-') # only concatenate when it is part of a valid number s += c elsif (c == '-' && n == '(') || c == '(' # begin a sub-expression negate = c == '-' open = 1 subExpr = '' while(c = chars.shift) open += 1 if c == '(' open -= 1 if c == ')' # if the number of open parenthesis equals 0, we've run to the end of the # expression. Make a new expression with the new string, and add it to the # stack. subExpr += c unless c == ')' && open == 0 break if open == 0 end parts.push(negate ? -solve_expression(subExpr) : solve_expression(subExpr)) s = '' elsif c.match(/[+\-\/*]/) parts.push(c) and s = '' else parts.push(s) if !s.empty? s = '' end end parts.push(s) unless s.empty? # expression exits 1 character too soon. # now for some solutions! i = 1 a = parts[0].to_f # left-most value is will become the result while i < parts.count b,c = parts[i..i+1] c = c.to_f case b when '+': a = a + c when '-': a = a - c when '*': a = a * c when '/': a = a / c end i += 2 end a end

红宝石1.9

（因为正则表达式）

字符的号：296

def d(s) while m = s.match(/((?<pg>$(?:\\[()]|[^()]|\g<pg>)*$))/) s.sub!(m[:pg], d(m[:pg][1,m[:pg].size-2])) end while m = s.match(/(-?\d+(\.\d+)?)\s*([*+\-\/])\s*(-?\d+(\.\d+)?)/) r=m[1].to_f.send(m[3],m[4].to_f) if %w{+ - * /}.include?m[3] s.sub!(m[0], r.to_s) end s end

编辑：包括马丁的优化
。

SNOBOL4

数字：232

a = pos(0) | '(' n = span('0123456789.') j = '!+;!-;!*;!/; output = e' d j '!' len(1) . y = " e a . q n . l '" y "' n . r = q (l " y " r) :s(p)" :s(d) k = code(j) e = input s e ' ' = :s(s) p e ('(' n . i ')') = i :s(p)f<k> end

这是一个半欺骗。它使用 code() (a变eval)de-压缩本身，但是不要评估的输入表达。

De-模糊版本，而不 code:

prefix = pos(0) | '(' num = span('0123456789.') expr = input spaces expr ' ' = '' :s(spaces) paren expr ('(' num . x ')') = x :s(paren) add expr (prefix . pfx) (num . l) '+' (num . r) = pfx (l + r) :s(paren) sub expr (prefix . pfx) (num . l) '-' (num . r) = pfx (l - r) :s(paren) mul expr (prefix . pfx) (num . l) '*' (num . r) = pfx (l * r) :s(paren) div expr (prefix . pfx) (num . l) '/' (num . r) = pfx (l / r) :s(paren) output = expr end

战略：

第一，删除所有空间(spaces)

只要有可能，删除括号内的周围一号(paren)

否则，找到一个简单的表达涉及两个数字，作为前缀 '(' 或者在开始的字符串

如果没有上述的规则适用，表达方式是完全评估。现在，如果输入的物以及形成我们应该留下一个号码。

例如：

1 + (2 * 3) + 4

1+(2*3)+4 [spaces]

1+(6)+4 [mul]

1+6+4 [paren]

7+4 [add]

11 [add]

C#

数字：355

我了 Noldorin的答案和修改，使得Noldorin99%的信贷。最好我可以做的算法被采用为408个字符。看看 Noldorin的答案为更明确的代码版本。

更改：
改变char比较，比较数字。
除去一些默认声明和联合的类型相同的声明。
重新工作的一些如果statments.

float q(string x){float v,n;if(!float.TryParse(x,out v)){x+=';';int t=0,l=0,i=0;char o,s='?',p='+';for(;i<x.Length;i++){o=s;if(x[i]!=32){s=x[i];if(char.IsDigit(x[i])|s==46|(s==45&o!=49))s='1';if(s==41)l--;if(s!=o&l==0){if(o==49|o==41){n=q(x.Substring(t,i-t));v=p==43?v+n:p==45?v-n:p==42?v*n:p==47?v/n:v;p=x[i];}t=i;if(s==40)t++;}if(s==40)l++;}}}return v;}

编辑：敲下来更多一些，从361 355，通过删除一个返回statments.

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