Convert char to int in C#

Question

I have a char in c#:

char foo = '2';

Now I want to get the 2 into an int. I find that Convert.ToInt32 returns the actual decimal value of the char and not the number 2. The following will work:

int bar = Convert.ToInt32(new string(foo, 1));

int.parse only works on strings as well.

Is there no native function in C# to go from a char to int without making it a string? I know this is trivial but it just seems odd that there's nothing native to directly make the conversion.

Answer 1

Interesting answers but the docs say differently:

Use the GetNumericValue methods to convert a Char object that represents a number to a numeric value type. Use Parse and TryParse to convert a character in a string into a Char object. Use ToString to convert a Char object to a String object.

http://msdn.microsoft.com/en-us/library/system.char.aspx

Answer 2

This will convert it to an int:

char foo = '2';
int bar = foo - '0';

This works because each character is internally represented by a number. The characters '0' to '9' are represented by consecutive numbers, so finding the difference between the characters '0' and '2' results in the number 2.

Answer 3

Has anyone considered using int.Parse() and int.TryParse() like this

int bar = int.Parse(foo.ToString());

Even better like this

int bar;
if (!int.TryParse(foo.ToString(), out bar))
{
    //Do something to correct the problem
}

It's a lot safer and less error prone

Answer 4

char c = '1';
int i = (int)(c-'0');

and you can create a static method out of it:

static int ToInt(this char c)
{
    return (int)(c - '0');
}

Answer 5

Try This

char x = '9'; // '9' = ASCII 57

int b = x - '0'; //That is '9' - '0' = 57 - 48 = 9

Answer 6

By default you use UNICODE so I suggest using faulty's method

int bar = int.Parse(foo.ToString());

Even though the numeric values under are the same for digits and basic Latin chars.

Answer 7

This converts to an integer and handles unicode

CharUnicodeInfo.GetDecimalDigitValue('2')

You can read more here.

Answer 8

The real way is:

int theNameOfYourInt = (int).Char.GetNumericValue(theNameOfYourChar);

"theNameOfYourInt" - the int you want your char to be transformed to.

"theNameOfYourChar" - The Char you want to be used so it will be transformed into an int.

Leave everything else be.

Answer 9

Principle:

char foo = '2';
int bar = foo & 15;

The binary of the ASCII charecters 0-9 is:

0   -   0011 0000
1   -   0011 0001
2   -   0011 0010
3   -   0011 0011
4   -   0011 0100
5   -   0011 0101
6   -   0011 0110
7   -   0011 0111
8   -   0011 1000
9   -   0011 1001

and if you take in each one of them the first 4 LSB (using bitwise AND with 8'b00001111 that equals to 15) you get the actual number (0000 = 0,0001=1,0010=2,... )

Usage:

public static int CharToInt(char c)
{
    return 0b0000_1111 & (byte) c;
}

Answer 10

Comparison of some of the methods based on the result when the character is not an ASCII digit:

char c = '\n';                              
Debug.Print($"{c & 15}");                   // 10
Debug.Print($"{c ^ 48}");                   // 58
Debug.Print($"{c - 48}");                   // -38
Debug.Print($"{(uint)c - 48}");             // 4294967258
Debug.Print($"{char.GetNumericValue(c)}");  // -1 

Answer 11

I'm using Compact Framework 3.5, and not has a "char.Parse" method. I think is not bad to use the Convert class. (See CLR via C#, Jeffrey Richter)

char letterA = Convert.ToChar(65);
Console.WriteLine(letterA);
letterA = 'あ';
ushort valueA = Convert.ToUInt16(letterA);
Console.WriteLine(valueA);
char japaneseA = Convert.ToChar(valueA);
Console.WriteLine(japaneseA);

Works with ASCII char or Unicode char

Answer 12

I am agree with @Chad Grant

Also right if you convert to string then you can use that value as numeric as said in the question

int bar = Convert.ToInt32(new string(foo, 1)); // => gives bar=2

I tried to create a more simple and understandable example

char v = '1';
int vv = (int)char.GetNumericValue(v); 

char.GetNumericValue(v) returns as double and converts to (int)

More Advenced usage as an array

int[] values = "41234".ToArray().Select(c=> (int)char.GetNumericValue(c)).ToArray();

Answer 13

I've seen many answers but they seem confusing to me. Can't we just simply use Type Casting.

For ex:-

int s;
char i= '2';
s = (int) i;

Answer 14

This worked for me:

int bar = int.Parse("" + foo);