using macros for multilevel debug

**zolfaghar** · 06-21-2016

Hi
I am following a text, and in it, the author suggests implementing multi level debug messages with a macro definition so you can execute the program as follows:

Code:

a.out -d3

And it would provide a more verbose level of debug as opposed to level 1. He says I can put the following code in a header file

Code:

#ifdef DEBON
#define DEBUG(level,fmt, ...)        if (Debug >= level) fprintf (stderr, fmt, __VA_ARGS__)
#else
#    define DEBUG(level, fmt, ...)
#endif

I just put the code in my sample program as follows:

Code:

#include <stdio.h>
#include <stdlib.h>
// int Debug;
//#define DEBUG(fmt, ...)  fprintf (stderr, fmt, __VA_ARGS__ )
#ifdef DEBON
#define DEBUG(level,fmt, ...)        if (Debug >= level) fprintf (stderr, fmt, __VA_ARGS__)
#else
#    define DEBUG(level, fmt, ...)
#endif
int process ( int i1, int i2)
{
  int val;
  DEBUG (1, "process (%i, %i)\n", i1, i2);
  val = i1 * i2;
  DEBUG (3, "return %i\n", val);
  return val;
}
int main ( int argc, char *argv[])
{
int arg1 = 0, arg2 = 0;
if (argc > 1)
arg1 = atoi (argv[1]);
if ( argc == 3)
  arg2 = atoi (argv[2]);
DEBUG (1, "processed %i arguments\n", argc -1 );
DEBUG (3, "arg1 = %i, arg2 = %i\n" , arg1, arg2);
printf ("%d\n", process (arg1, arg2));
return 0;
}

But I get the following error messages:

Code:

$ gcc -g -D DEBON 173.c -o 173
173.c: In function 'process':
173.c:6:39: error: expected expression before 'if'
 #define DEBUG(level,fmt, ...)        (if (Debug >= level) fprintf (stderr, fmt, __VA_ARGS__))
                                       ^
173.c:14:3: note: in expansion of macro 'DEBUG'
   DEBUG (1, "process (%i, %i)\n", i1, i2);
   ^
173.c:6:39: error: expected expression before 'if'
 #define DEBUG(level,fmt, ...)        (if (Debug >= level) fprintf (stderr, fmt, __VA_ARGS__))
                                       ^
173.c:16:3: note: in expansion of macro 'DEBUG'
   DEBUG (3, "return %i\n", val);
   ^
173.c: In function 'main':
173.c:6:39: error: expected expression before 'if'
 #define DEBUG(level,fmt, ...)        (if (Debug >= level) fprintf (stderr, fmt, __VA_ARGS__))
                                       ^
173.c:27:1: note: in expansion of macro 'DEBUG'
 DEBUG (1, "processed %i arguments\n", argc -1 );
 ^
173.c:6:39: error: expected expression before 'if'
 #define DEBUG(level,fmt, ...)        (if (Debug >= level) fprintf (stderr, fmt, __VA_ARGS__))
                                       ^
173.c:28:1: note: in expansion of macro 'DEBUG'
 DEBUG (3, "arg1 = %i, arg2 = %i\n" , arg1, arg2);
 ^

I have seen some similar posts about this. One of them was Good practice of writing debug print statements
But I can not quite follow the advice. What am I doing wrong?

**whiteflags** · 06-21-2016

I probably shouldn't show you this, but this is how your macros expanded....

Code:

gcc -g -D DEBON -E debug.c
< whiteflags snipped unhelpful preprocessor code >

if (Debug >= 1) fprintf ((&_iob[2]), "processed %i arguments\n", argc -1);
if (Debug >= 3) fprintf ((&_iob[2]), "arg1 = %i, arg2 = %i\n", arg1, arg2);

< code continues >

Now, other than you seeing the guts of stdout, we can also see that the code depends on Debug, a variable you commented out.

The C compiler removes comments before it expands macros.

**laserlight** · 06-22-2016

By the way, your way (or the author's way) of implementing the macro can be problematic. Consider:

Code:

if (whatever)
    foo();
    DEBUG(1, "test #%d", 1);
else
    bar();

At a glance, it looks like there should be a compile error since the else does not match the if as braces were not used to combine the two statements into a single block of code forming the body of the if statement. However, the code may well compile because if DEBON is defined, after preprocessing it would look something like:

Code:

if (whatever)
    foo();
    if (Debug >= level)
        fprintf(stderr, "test #%d", 1);
else
    bar();

or with braces:

Code:

if (whatever)
{
    foo();
}
if (Debug >= level)
{
    fprintf(stderr, "test #%d", 1);
}
else
{
    bar();
}

One way to fix this is to enclose the implementation of the function-style macro in a do while (0) block, e.g.,

Code:

#define DEBUG(level, fmt, ...) do {\
    if (Debug >= level)\
        fprintf(stderr, fmt, __VA_ARGS__);\
} while (0)

Personally, instead of both an int global variable named Debug and a macro named DEBON, I wonder it would be simpler to have a macro named DEBUG_LEVEL that is left undefined if these debugging outputs are to be disabled, and defined to an integer if they are to be enabled. You would then also define named constants (maybe using an enum) for your debugging levels instead of using magic numbers like 1 and 3 (ideas on such names are all over various programming languages, libraries, frameworks, etc, that can be found on the Internet, though the more commonly used term is "logging" whereas debugging is more of an activity that might make use of logs).

**whiteflags** · 06-22-2016

You could also give up and use a debugger. I find RMS's gdb Tutorial to be a gentle introduction to them, since even if you don't use GDB, the features that GDB provides and the way the tutorial exposes them to the reader echo real world uses of other debuggers that other IDEs come bundled with.

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