函数后浮点值变化 Return

Question

我发现我的浮点值在返回后发生了变化。我在返回之前输出浮点数的十六进制值，然后在返回之后输出。我希望十六进制输出是相同的，但它们已经改变了。您能否解释为什么会发生这种情况以及如何防止这种情况发生？

谢谢！

  #include <stdio.h>
  typedef int             uint32;
  typedef unsigned char   uint8;
  static const float CONST_FLOAT = 150.36e6;

  uint32 foo( uint8 param ) {
     return 1;
  }

  float get_float( uint8 param )
  {
       float fVal = 0.0;

       fVal = (foo( param )*1.0) / ( 60.0 * CONST_FLOAT );

       printf("fVal = 0x%X\n", fVal);
       printf("fVal = %f\n", fVal);

       return fVal;
  }

  int main(void) {
     float fTest;

     fTest = get_float(1);

     printf("fTest = 0x%X\n",fTest);
     printf("fTest = %f\n",fTest);

     return;
  }

输出： fVal = 0x6A96C728 fVal = 0.000000

fTest = 0x9AAF9000 fTest = 0.000000

我希望以十六进制格式显示的 fVal 与十六进制格式的 fTest 相同，但结果却不同，这是为什么？

谢谢！

Answer 1

您不能打印带有 "%x" 说明符的 float 值。这是未定义的行为，您的程序显然会相应地表现。

这是来自 C11 草稿 1570，

7.21.6.1 The fprintf function

The conversion specifiers and their meanings are:

o,u,x,X

The unsigned int argument is converted to unsigned octal (o), unsigned decimal (u), or unsigned hexadecimal notation (x or X) in the style dddd; the letters abcdef are used for x conversion and the letters ABCDEF for X conversion. The precision specifies the minimum number of digits to appear; if the value being converted can be represented in fewer digits, it is expanded| with leading zeros. The default precision is 1. The result of converting a zero value with a precision of zero is no characters. ?

If a conversion specification is invalid, the behavior is undefined. 282) If any argument is not the correct type for the corresponding conversion specification, the behavior is undefined.

函数后浮点值变化 Return

Floating Point Value Changes After Function Return

c

floating-point