Thread: weird problem in constant definition

Hello all!

My first post here.

I am investigating constants and I am seeing weird effect in defining constants.

Here is the problem.

Code:

#include <stdio.h>
#define CONST 5/9

int main()
{
	double test_const = 5.0/9.0;
	printf("test_const: %f\n",test_const);
	printf("     CONST: %f\n",CONST);	
	return 0;	
}

gives

Code:

test_const: 0.555556
     CONST: 0.555555

First question, why the rounding difference, no rounding at all in constant definition?
However, if I comment out first print function I get even weirder result:

Code:

#include <stdio.h>
#define CONST 5/9

int main()
{
	double test_const = 5.0/9.0;
	//printf("test_const: %f\n",test_const);
	printf("     CONST: %f\n",CONST);	
	return 0;	
}

Code:

CONST: 0.000000

How does the commenting out of one simple print function result such effect?

However, If I use correct type casting in constant definition

Code:

#define CONST (float)5/9

No such problems appears.

However, in linux I get for out commented print function random outputs:

Code:

[root@xxx test]# ./a.out
     CONST: -0.510853
[root@xxx test]# ./a.out
     CONST: -0.030291
[root@xxx test]# ./a.out
     CONST: -0.023501
[root@xxx test]# ./a.out
     CONST: -0.051702

Witch gives me good reason to think uninitialized memory area. Again, correct casting solves problem.

But the main question is: how does commenting out first print function affect the output?

On windows I used

Code:

C:\test>gcc --version
gcc (GCC) 3.4.2 (mingw-special)
Copyright (C) 2004 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

and linux (debian)

Code:

[root@xxx test]# gcc --version
gcc (GCC) 4.1.2 20080704 (Red Hat 4.1.2-44)
Copyright (C) 2006 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

5/9 should always give you 0. Perhaps what is going on in the first instance is that you are passing only an int (0 is an int, and so is 5/9) but printf is expecting a double, so it needs to get four bytes of information from somewhere, so it's probably reading the old information from the previous call that is still left on the stack.

Yeah, that is it!

If I used this:

Code:

#include <stdio.h>
#define CONST 5/9

int main()
{
	double test_const = 5.0/9.0;
	printf("test_const: %f\n",test_const);
	printf("     CONST: %f\n",CONST);
	printf("      NULL: %f\n",0);
	printf("  HARDCODE: %f\n",0.007);
	printf("     CONST: %f\n",CONST);
	printf("      TEXT: %s\n","AAaaAAaa");
	printf("     CONST: %f\n",CONST);
	printf("  HARDCODE: %f\n",20007.1);
	printf("     CONST: %f\n",CONST);
	
		

	
	printf("%x\n",test_const);
	printf("%x\n",CONST);

	return 0;	
}

i get

Code:

test_const: 0.555556
     CONST: 0.555555
      NULL: 0.555555
  HARDCODE: 0.007000
     CONST: 0.007000
      TEXT: AAaaAAaa
     CONST: 0.007000
  HARDCODE: 20007.100000
     CONST: 20007.093750
71c71c72
0

So obviousely printf gets trash as you mentioned. Thank you for replay, it cleared my mind as fresh water

Obviousely printf gets CONST (replaced by 0 in pre compile) witch is hard coded and taking float data from code previousely used... debug would probably clear it up correctly.