无法从MSVCRT strtod / sscanf / atof函数中获取NaN

问题描述

有没有办法从Windows CRT string 到 float float 到 string 转换器在C中没有信息丢失（ strtod ， sscanf 或 atof 返回原始 float 舍入模式不会改变。

我在MinGW或Visual C ++下，因此这些调用会进入MSVC ++运行时。问题是，我不能得到它来解析任何特殊的值（如Inf或NaN ）。 Inf 是OK（在解析不适合 float 的值后返回，例如1e999）。

  / *返回浮点的最短字符串表示形式一个成功的scanf往返。 
 *保证适合13个字符（包括最后的'\0'）。 
 * / 
 char * ftoa（char * res，float f）{
 float r = 0; 
 int i，j，len，e = floor（log10（f））+ 1; 
 char fmt [8]; 
 union {float f; int32_t i; } u = {f}; 
 
 if（f> FLT_MAX）{sprintf（res，1e999）; return res; } 
 if（f  
 if（（ui& 0x7F800000）== 0x7F800000）{// NaN 
 sprintf（res，ui == 0x7FC00000？％sNaN：％sNaN％d， ui< 0？ - ：，ui& 0x7FFFFF）; 
 return res; 
} 
 
 //计算不带指数的最短字符串（123000，0.15）
 if（！f || e> -4&& e& 10）{
 for（i = 0; i  sprintf（fmt，%%。％df，i） 
 sprintf（res，fmt，f）; 
 sscanf（res，％f，& r）; if（r == f）break; 
} 
} 
 if（r == f）len = strlen（res）; 
 else len = 1e9; 
 
 if（！f）return res; //处理0和-0 
 
 //计算具有指数（123e3，15e-2）的最短字符串
 for（i = 0; i  sprintf（res，％.0fe％d，f * pow（10，-e），e）; sscanf（res，％f，& r）; if（r == f）break; 
 j = strlen（res）; if（j> = lenF）break; 
 while（res [j]！='e'）j--; 
 res [j-1]  - ; sscanf（res，％f，& r）; if（r == f）break; // try + -1 
 res [j-1] + = 2; sscanf（res，％f，& r）;如果（r == f）break; 
 e--; 
} 
 if（len  return res; 
} 
  

解决方案

如果发生溢出，则它们返回HUGE_VAL，如果不能解析输入则返回0，或者发生下溢。

Is there any way to get NaNs from the Windows CRT string to float functions?

---

Why: I'm writing an IEEE float to string converter in C with no information loss (strtod, sscanf or atof return the original float) provided the rounding mode doesn't change.

I'm under MinGW or Visual C++, so these calls go to the MSVC++ runtime. The problem is that I can't get it to parse any special values (like "Inf" or "NaN"). Inf is OK (it's returned after parsing a value that doesn't fit in a float, such as "1e999").

  /* Return the shortest string representation of a float with a successful scanf round-trip.
   * Guaranteed to fit in 13 chars (including the final '\0').
   */
  char* ftoa(char* res, float f) {
     float r = 0;
     int i, j, len, e = floor(log10(f)) + 1;
     char fmt[8];
     union { float f; int32_t i; } u = { f } ;

     if (f > FLT_MAX) { sprintf(res, "1e999"); return res; }
     if (f < -FLT_MAX) { sprintf(res, "-1e999"); return res; }

     if ((u.i & 0x7F800000) == 0x7F800000) {  // NaN
        sprintf(res, u.i == 0x7FC00000 ? "%sNaN" : "%sNaN%d", u.i<0 ? "-" : "", u.i & 0x7FFFFF);
        return res;
     }

     // compute the shortest string without exponent ("123000", "0.15")
     if (!f || e>-4 && e<10) {
        for (i=0; i<=10; i++) {
           sprintf(fmt, "%%.%df", i);
           sprintf(res, fmt, f);
           sscanf(res, "%f", &r); if (r==f) break;
        }
     }
     if (r==f) len = strlen(res);
     else len = 1e9;

     if (!f) return res;  // handle 0 and -0

     // compute the shortest string with exponent ("123e3", "15e-2")
     for (i=0; i<9; i++) {
        sprintf(res, "%.0fe%d", f * pow(10,-e), e); sscanf(res, "%f", &r); if (r==f) break;
        j = strlen(res); if (j >= lenF) break;
        while (res[j] != 'e') j--;
        res[j-1]--; sscanf(res, "%f", &r); if (r==f) break;  // try +-1
        res[j-1]+=2; sscanf(res, "%f", &r); if (r==f) break;
        e--;
     }
     if (len <= strlen(res)) sprintf(res, fmt, f);
     return res;
  }

解决方案

No. They return HUGE_VAL if overflow would occur and 0 if the input can't be parsed or underflow occurs.

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