DEBUG statement question

djgerbavore · 01-25-2005, 10:25 PM

when looking at some c code. I found the following statement alot of the time and want to know what it does,

Code:

DEBUG('t', "a message");

is this a function that puts your program in debug mode, and does this help you out in big programing projects?

thanks,
djgerbavor3

jtshaw · 01-25-2005, 10:34 PM

I'm guessing that is a macro defined somewhere in the project. As far as I know there is no DEBUG function in any of the stdlib's.

What I usually do is have a macro function called PRINTDB that is just a wrapper for printf basically. However, I set it up so it only gets defined so you see output if the symbol DEBUG is defined. So you can determine at compile time whether or not the debug messages will get printed.

Matir · 01-26-2005, 09:33 AM

Wow, i've been stupid. I'm not sure I should publically admit this, but I've been stupid enough to do:

Code:

#ifdef DEBUG
fprintf(stderr,"DEBUG: MESSAGE");
#endif
...

I guess the following would make more sense:

Code:

#define DBGMSG(x) fprintf(stderr,"DEBUG: %s\n",(x))

and then use that. Or perhaps I can find some way to emulate printf with variable length arguments... as in

Code:

DBGMSG("ival: %d xval: %d\n",1,2);

giving me, on stderr:

Code:

DEBUG: ival: 1 xval: 2

jtshaw · 01-26-2005, 09:37 AM

I use the C99 standard... but this is how I do it:

Code:

#define DEBUGMSG(format, ...) fprintf (stderr, format, __VA_ARGS__)