Is there an easier way than using #ifdef in C?

Question

From what I understand assert is a macro in C and supposedly if you use it at compile time but leave it disabled then there won't be overhead (which might not be correct I don't know). The problem for me is that what I'd like to do is get all the variables passed to my function and print out that output, but only if I want debugging enabled. Here is what I have so far:

int exampleFunction (int a, int b)
{
  #ifdef debugmode
  printf("a = %i, b = %i", a, b);
  #endif 
}

I'm wondering if there is any easier (and less ugly) method for doing something like this. xdebug for php has this feature and I've found is saves me an enormous amount of time when debugging so i want to do it for each function.

Thanks

Answer 1

Look at

Answer 2

try this:

#ifdef debugmode
#define DEBUG(cmd) cmd
#else
#define DEBUG(cmd)
#endif


DEBUG(printf("a = %i, b = %i", a, b));

now, if you have debugmode defined, you get your print statement. otherwise, it never shows up in the binary.

Answer 3

You can define PRINTF_IF_DEBUGGING as

#ifndef NDEBUG
#define PRINTF_IF_DEBUGGING(X) printf(X)
#else
#define PRINTF_IF_DEBUGGING(X)
#endif

This will centralize the #ifdefs in only one place.

Answer 4

if (MY_DEBUG_DEFINE) {
        do_debug_stuff();
}

Any half decent compiler would optimize the block away. Note you need to define MY_DEBUG_DEFINE as a boolean (ie 0 or not 0).

#define MY_DEBUG_DEFINE defined(NDEBUG)

If you happen to compile with maximum warning level, this trick avoids unreferenced argument.

Answer 5

Using GCC, I really enjoy to add, per file:

#if 0
#define TRACE(pattern,args...)   fprintf(stderr,"%s:%s/%u" pattern "\n",__FILE__,__FUNCTION__,__LINE__,##args)
#else
#define TRACE(dummy,args...)
#endif

and then in the code:

i++;
TRACE("i=%d",i);

i will be printed only when I activate the TRACE() macro in the top of the file. Works really great, plus it prints the source file, line and function it occurred.

Answer 6

Well, the problem with the PRINT_DEBUG type macros as given here, is that they only allow one parameter. For a proper printf() we'll need several, but C macros don't (presently) allow variable arguments.

So, to pull this off, we've got to get creative.

#ifdef debugmode
     #define PRINTF   printf
#else
     #define PRINTF    1 ? NULL : printf
#endif

Then when you write PRINTF("a = %i, b = %i", a, b);, in non-debug mode, it will be renders as (effectively):

 if (true) NULL;
 else printf("a = %i, b = %i", a, b);

The compiler is happy, but the printf is never execute, and if the compiler if bright (i.e, any modern C compiler), the code for the printf() will never be generated, as the compiler will recognize that path can never be taken.

Note, however, that the parameters will still be evaluated, so if they have any side effects (i.e, ++x or a function call), they code may be generated (but not executed)

Answer 7

Workaround to get vararg with preprocessors that don't support it

#define DEBUG

#ifdef DEBUG
#define trace(args) printf args
#else
#define trace(args)
#endif


int dostuff(int value)
{


    trace(("%d", value));

}

Answer 8

I would also print some other C preprocessor flags that can help you track problems down

printf("%s:%d {a=%i, b=%i}\n", __FILE__, __LINE__, a, b);