How does the following code work?
#define ENABLE_DEBUG 1
#if ENABLE_DEBUG
#define LOG_MSG printf
#else
#define LOG_MSG(...)
#endif
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2 Answers
Depending on the value of ENABLE_DEBUG, LOG_MSG is either defined to be an alias for printf() or it is defined as a no-op macro. It is implied that you can change the value to 0 to disable debugging. This is a common technique for making it easy to switch between debugging builds which display lots of output and release builds which are quiet.
#define LOG_MSG printf
This makes it an alias for printf().
#define LOG_MSG(...) /* empty */
And this defines it as an empty macro. Notice that here it has a set of parentheses, which means the macro takes parameters. It has nothing afterwards which means it expands to absolutely nothing. And the ... indicates that this macro can take a varying number of arguments. This syntax is a C99 extension so it may not be available on older C compilers.
LOG_MSG("file not found\n");
The result is that a LOG_MSG() call will either print a message or do nothing depending on whether logging is enabled.
// If ENABLE_DEBUG is non-zero, a debugging printout:
printf("file not found\n");
// If ENABLE_DEBUG is zero, an empty statement:
;
For what it's worth, whoever authored this macro could've done a better job by replacing the first definition with one using the ... syntax (which he/she is clearly familiar with), printing to stderr instead of stdout:
#define LOG_MSG(...) fprintf(stderr, __VA_ARGS__)
Comments
This uses the preprocessor to change code before compilation.
If ENABLE_DEBUG is defined as 1, whenever the preprocessor sees
LOG_MSG("something happened");
It will replace it with
printf("something happened");
If it is defined as 0, or not defined it will replace it with nothing (as the other answer that has just been published says).
