Floating Point Value Changes After Function Return

Question

I'm finding my float value has changed after being returned. I output the hex value of the float before being returned and then after it gets returned. I expect the HEX output to be identical but they have changed. Can you please explain why this is happening and how do I prevent this from happening?

Thanks!

  #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;
  }

OUTPUT: fVal = 0x6A96C728 fVal = 0.000000

fTest = 0x9AAF9000 fTest = 0.000000

I expect the fVal displayed in Hex format to be the same as fTest in Hex but they turned out different why is that?

Thanks!

Iharob Al Asimi · Accepted Answer · 2015-04-29T21:08:33.837

3

You cannot print a float value with the "%x" specifier. It's undefined behavior, and your program is obviously behaving accordingly.

This is from the C11 draft 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.

edited Apr 29 '15 at 21:08

answered Apr 29 '15 at 20:47

Iharob Al Asimi

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How do I dump the hex representation of the floating point value? ie. what it looks like in memory. – user1144251 Apr 29 '15 at 20:52
2

@user1144251 Try `printf("fTest = %a\n",fTest);` to print in "hexadecimal floating-point". – chux - Reinstate Monica Apr 29 '15 at 21:02
1

@chux I have never used the `a` specifier, but your are absolutely right, just read it in the draft pdf. – Iharob Al Asimi Apr 29 '15 at 21:03
@chux interesting! `fTest` is a rather small number and the `%a` format reveals more than the unqualified `%f`does. – Weather Vane Apr 29 '15 at 21:30
@Weather Vane for more detail on revealing more: http://stackoverflow.com/q/16839658/2410359 – chux - Reinstate Monica Apr 29 '15 at 21:35

Floating Point Value Changes After Function Return

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