C语言 将浮点数转换为字符*

Converting float to char*

提问by boom

How can I convert a floatvalue to char*in Clanguage?

我如何能转换float值char*的C语言？

回答by Delan Azabani

char buffer[64];
int ret = snprintf(buffer, sizeof buffer, "%f", myFloat);

if (ret < 0) {
    return EXIT_FAILURE;
}
if (ret >= sizeof buffer) {
    /* Result was truncated - resize the buffer and retry.
}

That will store the string representation of myFloatin myCharPointer. Make sure that the string is large enough to hold it, though.

这将存储myFloatin的字符串表示形式myCharPointer。但是，请确保绳子足够大以容纳它。

snprintfis a better option than sprintfas it guarantees it will never write past the size of the buffer you supply in argument 2.

snprintf是一个更好的选择，sprintf因为它保证它永远不会超过您在参数 2 中提供的缓冲区的大小。

回答by aJ.

char array[10];
sprintf(array, "%f", 3.123);

sprintf: (from MSDN)

sprintf：（来自 MSDN）

回答by baligena

In Arduino:

在阿杜诺：

//temporarily holds data from vals
char charVal[10];                

//4 is mininum width, 3 is precision; float value is copied onto buff
dtostrf(123.234, 4, 3, charVal);

monitor.print("charVal: ");
monitor.println(charVal);

回答by chux - Reinstate Monica

Long after accept answer.

很久以后接受答案。

Use sprintf(), or related functions, as many others have answers suggested, but use a better format specifier.

使用sprintf()或相关函数，正如许多其他人建议的答案一样，但使用更好的格式说明符。

Using "%.*e", code solves various issues:

使用"%.*e", 代码解决了各种问题：

• The maximum buffer size needed is far more reasonable, like 18. sprintf(buf, "%f", FLT_MAX);could need 47+. sprintf(buf, "%f", DBL_MAX);may need 317+

• Using ".*"allows code to define the number of decimal places needed to distinguish a string version of float xand it next highest float. For deatils, see Printf width specifier to maintain precision of floating-point value

• Using "%e"allows code to distinguish small floats from each other rather than all printing "0.000000"which is the result when |x| < 0.0000005.

  #define FLT_STRING_SIZE (1+1+1+(FLT_DECIMAL_DIG-1)+1+1+ 4   +1)
                       //  - d .  dddddddd           e - dddd 
  #define FLT_STRING_SIZE (1+1+1+(FLT_DECIMAL_DIG-1)+1+1+ 4   +1)
                       //  - d .  dddddddd           e - dddd 
  typedef union{
      float a;
      char b[4];
  } my_union_t;
  
  
  char buf[FLT_STRING_SIZE];
  sprintf(buf, "%.*e", FLT_DECIMAL_DIG-1, some_float);
  
  
  char buf[FLT_STRING_SIZE];
  sprintf(buf, "%.*e", FLT_DECIMAL_DIG-1, some_float);
  

• 所需的最大缓冲区大小要合理得多，例如 18。sprintf(buf, "%f", FLT_MAX);可能需要 47+。sprintf(buf, "%f", DBL_MAX);可能需要 317+

• 使用".*"允许代码定义区分字符串版本float x和下一个最高版本所需的小数位数float。有关详细信息，请参阅Printf 宽度说明符以保持浮点值的精度

• Using"%e"允许代码将小floats 彼此区分开来，而不是所有打印"0.000000"，这是|x| < 0.0000005.

  char* str=NULL;
  int len = asprintf(&str, "%g", float_var);
  if (len == -1)
    fprintf(stderr, "Error converting float: %m\n");
  else
    printf("float is %s\n", str);
  free(str);
  

Ideas:
IMO, better to use 2x buffer size for scratch pads like buf[FLT_STRING_SIZE*2].
For added robustness, use snprintf().

想法：
IMO，最好将 2x 缓冲区大小用于buf[FLT_STRING_SIZE*2].
为了增加健壮性，请使用snprintf().

回答by Peter

char array[10];
snprintf(array, sizeof(array), "%f", 3.333333);

You can access to float data value byte by byte and send it through 8-bit output buffer (e.g. USART) without casting.

您可以逐字节访问浮点数据值，并通过 8 位输出缓冲区（例如 USART）将其发送，而无需进行强制转换。

回答by pixelbeat

##代码##

回答by pcent

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