Debug function like printf()

Hi, I need some help!

I'd like to do a function for debug my code (at compile time) that follows this scheme:

#ifdef DEBUG
my_printf( <list arguments> ) printf( <list arguments> )
#else
my_printf( <list arguments> ) NOTHING
#end

In other words if DEBUG is defined the code is replace with the printf() function else do nothing.

How can I write this macro for my purpose??

Thanks a lots.

#ifdef DEBUG
#define my_printf printf
#else
#define my_printf
#end

Thanks but with this define I get a lot of:

Code:

---

cripto.c: In function ‘print_error’:
cripto.c:90: warning: left-hand operand of comma expression has no effect
cripto.c:90: warning: left-hand operand of comma expression has no effect
cripto.c:90: warning: statement with no effect

---

where the function is present and DEBUG not defined.

If you have C99, you can use variadic macro.
#define debug_print(...) fprintf(stderr,__VA_ARGS)
You might also want to read:
Question 10.26

Thanks, I used this:

Code:

---

#ifdef DEBUG_CRIPTO
        #define lib_debug(...) fprintf(stderr, __VA_ARGS__)
#else
        #define lib_debug(...)
#endif

---

and it works fine.

Hrm...
How about:

#ifdef DEBUG
#define my_printf printf
#else
#define my_printf //
#end
(Assuming you have C99.)

For your method, I think you'd have to do something like this:

Quote:

---

Originally Posted by Elysia

#ifdef DEBUG
#define my_printf( x ) printf( x )
#else
#define my_printf
#end

---

Then I think you need double parentheses when calling it: my_printf(( "%d %s", 5, "foo" ));

Something like that anyways.

It is #endif not #end

A small addition to Elysia's idea:

Code:

---

#ifdef DEBUG
#define my_printf printf
#else
#define my_printf ;//
#end

---

Since you might have

Code:

---

if (something)
  my_printf(...);

---

Of course what Mordor actually did would be the best for C99.

testing cpjust idea

Code:

---

#include <stdlib.h>
#include <stdio.h>

#define DEBUG
#ifdef DEBUG
#define my_printf(x) printf(x)
#else
#define my_printf
#endif

int main()
{
  printf(("%s", "10"));
}

---

this works. This doesn't compile

Code:

---

#include <stdlib.h>
#include <stdio.h>

#define DEBUG
#ifdef DEBUG
#define my_printf(x) printf(x)
#else
#define my_printf
#endif

int main()
{
  printf(("%d", 10));
}

---

Both give a warning (in GCC). That the left hand operant on the comma has no effect. The error the later gets is that
"passing arg 1 of 'printf' makes pointer from integer without a cast"

What I think happens is that this

Code:

---

("%s", "hey") --> ("%s" "hey") --> ("%shey")

---

so the first one will run. The second I have no idea. Or something equally weird happens.

Conclusion: be careful when being tricky with macros

What maybe you could do is change it to

Code:

---

#define my_printf(x, y) printf(x, y)

---

and always use printf with two inputs and combine multiple of them for more inputs