如何确保输入是一个双在C编程语言

Question

我怎么确保我已经有了一个双人和不是其他什么东西？

int main() {
    int flagOk = 0;
    double number;
    while(!flagOk) {
        printf("Put in a double");
        scanf("%lf", &number);
        if(number == "%lf"); //this want make sure
        flagOk = 1;
    }
}

Answer 1

您可能需要一个代码片段更是这样的：

double number;
do {
    printf("Enter a double: ");
    scanf("%*c"); // burn stdin so as not to buffer up responses
} while (1 != scanf("%lf", &number));

然而，由于乔纳森所指出的，一些更好的行由行解析可能更好。直接从stdin扫描这种方式是不直观给用户。

Answer 2

检查返回值 scanf();它会告诉你多少转换是成功的。

在该背景下，如果转失败了，你会得到0;如果成功，你将获得1.

错误的恢复需要思考。我通常会发现更容易阅读的线数据的使用 fgets() (及 永远不获取()!)，那么处理它 sscanf().很容易放弃错误的数据，如果转换失败。

Answer 3

如果你想可以确保用户的预期值正好与返回的值过字符串 atof(str) - 其中包括指数符号 -然后下列代码的工作。

你可以得到输入的使用 fgets(str,大小，stdin), 和你甚至不需要删除结尾的新行之前通过串的分析程序。

此外，如果有一个分析错误，功能报告位置的所有罪字指通过作为附加参数。

还有很长的形式更易于阅读，并简短形式的-更容易类型。

只要形式：

/*
    Copyright (C) 2010  S. Randall Sawyer

This code is in the public domain.  It is intended to be usefully illustrative.
It is not intended to be used for any particular application. If this code is
used in any application - whether open source or proprietary, then please give
credit to the author.

*/

#include  <ctype.h>

#define FAILURE (0)
#define SUCCESS (!FAILURE)

enum {
  END_EXPRESSION        = 0x00,
  ALLOW_POINT           = 0x01,
  ALLOW_NEGATIVE        = 0x02,
  REQUIRE_DIGIT         = 0x04,
  ALLOW_EXPONENT        = 0x08,
  HAVE_EXPONENT         = 0x10,
  EXPECT_EXPRESSION     = ALLOW_POINT | ALLOW_NEGATIVE | REQUIRE_DIGIT,
  EXPECT_INTEGER        = ~ALLOW_POINT,
  EXPECT_POS_EXPRESSION = ~ALLOW_NEGATIVE,
  EXPECT_POS_INTEGER    = EXPECT_INTEGER & EXPECT_POS_EXPRESSION,
  EXPECT_EXPONENT       = EXPECT_INTEGER ^ HAVE_EXPONENT,
  EXPONENT_FLAGS        = REQUIRE_DIGIT | ALLOW_EXPONENT | HAVE_EXPONENT
} DoubleParseFlag;

int double_parse_long ( const char * str, const char ** err_ptr )
{
  int ret_val, flags;
  const char * ptr;

  flags = EXPECT_EXPRESSION;
  ptr   = str;

  do  {

    if ( isdigit ( *ptr ) ) {
      ret_val = SUCCESS;
      /*  The '+' here is the key:  It toggles 'ALLOW_EXPONENT' and
          'HAVE_EXPONENT' successively  */
      flags = ( flags + ( flags & REQUIRE_DIGIT ) ) & EXPECT_POS_EXPRESSION;
    }
    else if ( (*ptr & 0x5f) == 'E' )  {
      ret_val = ( ( flags & ( EXPONENT_FLAGS ) ) == ALLOW_EXPONENT );
      flags = EXPECT_EXPONENT;
    }
    else if ( *ptr == '.' ) {
      ret_val = ( flags & ALLOW_POINT );
      flags = ( flags & EXPECT_POS_INTEGER );
    }
    else if ( *ptr == '-' ) {
      ret_val = ( flags & ALLOW_NEGATIVE );
      flags = ( flags & EXPECT_POS_EXPRESSION );
    }
    else if ( iscntrl ( *ptr ) )  {
      ret_val = !( flags & REQUIRE_DIGIT );
      flags = END_EXPRESSION;
    }
    else  {
      ret_val = FAILURE;
      flags = END_EXPRESSION;
    }

    ptr++;

  } while (ret_val && flags);

  if (err_ptr != NULL)
    *err_ptr = ptr - 1;

  return ret_val;
}

短形式：

/*
    Copyright (C) 2010  S. Randall Sawyer

This code is in the public domain.  It is intended to be usefully illustrative.
It is not intended to be used for any particular application. If this code is
used in any application - whether open source or proprietary, then please give
credit to the author.

*/

#include  <ctype.h>

int double_parse_short ( const char * str, const char ** err_ptr )
{
  int ret_val, flags;
  const char * ptr;

  flags = 0x07;
  ptr   = str;

  do  {

    ret_val = ( iscntrl ( *ptr ) ) ? !(flags & 0x04) : (
              ( *ptr == '.' ) ? flags & 0x01 : (
              ( *ptr == '-' ) ? flags & 0x02 : (
              ( (*ptr & 0x5f) == 'E' ) ? ((flags & 0x1c) == 0x08) : (
              ( isdigit ( *ptr ) ) ? 1 : 0 ) ) ) );

    flags =   ( isdigit ( *ptr ) ) ?
                (flags + (flags & 0x04)) & 0x1d : (
              ( (*ptr & 0x5f) == 'E' ) ? 0x0e : (
              ( *ptr == '-' ) ? flags & 0x1d : (
              ( *ptr == '.' ) ? flags & 0x1c : 0 ) ) );

    ptr++;

  } while (ret_val && flags);

  if (err_ptr != NULL)
    *err_ptr = ptr - 1;

  return ret_val;
}

我希望这可以帮助!