Defining a variable arguments function in a preprocessor #define - c

Question

I am trying to define a debug_log function in a preprocessor #define in order for this function to exist only in Debug mode.
The thing is that I wish to use a variable_argument function :
#ifdef DEBUG
#define DEBUG_ENABLED 1
#else
#define DEBUG_ENABLED 0
#endif
#define debug_log(msg, ...) \
do { \
if (DEBUG_ENABLED) { \
char str[300]; \
int length = -1; \
va_list argList; \
va_start( argList, msg ); \
length = vsnprintf(str, sizeof(str), msg, argList); \
va_end( argList ); \
if (length > 0) \
{ \
fprintf(stderr, "%s, %d ",__func__, __LINE__); \
fprintf(stderr, "%s", str); \
fprintf(stderr,"\n"); \
} \
} \
} while (0) \
The compiler is returning :
error: ‘va_start’ used in function with fixed args
[build] 20 | va_start( argList, msg ); \
Thank you for your help ;)

Answer 1

You're not actually defining a function here. You're defining a block of code that expects to be part of a variadic function. You're most likely not in a variadic function when you call this macro, hence the error.
Instead, define an actual function inside of an #if block, along with a dummy function-like macro in the #else block that does nothing.
#ifdef DEBUG
#define debug_log(...) debug_log_impl(__func__, __LINE__, __VA_ARGS__)
void debug_log_impl(const char *func, int line, const char *msg, ...)
{
char str[300];
int length = -1;
va_list argList;
va_start( argList, msg );
length = vsnprintf(str, sizeof(str), msg, argList);
va_end( argList );
if (length > 0)
{
fprintf(stderr, "%s, %d ", func, line);
fprintf(stderr, "%s", str);
fprintf(stderr,"\n");
}
}
#else
#define debug_log(...) (void)0
#fi
There was a question in the comments regarding why (void)0 should be used in the #else case instead of an empty expression. Suppose you were to use debug_log as the left operand of the comma operator:
while (debug_log("iterating, x=%d",x), x>0)
With the definition above, if DEBUG is not defined this line expands to:
while ((void)0, x>0)
If it were an empty expression, it would expand to:
while (, x>0)
Which is a syntax error.

Answer 2

You are mixing up variable arguments for macros with variable argument access for functions.
The ... in the macro parameter list represents the variable arguments of the macro. The identifier __VA_ARGS__ in the macro replacement text expands to the variable arguments.
The ... in a function parameter list represents the variable arguments of the function. On object of type va_list can be used to access these variable arguments using the va_start, va_arg and va_end macros (and perhaps the va_copy macro if needed). Those are defined by #include <stdarg.h>.
Your debug_log macro is not a function, so it is not a function with variable arguments, so it cannot use va_start etc. to access those arguments. What it can do is pass the macros variable arguments as a whole to something else. For your example, snprintf would be a good choice as a replacement for the vsnprintf you orignally used:
#define debug_log(msg, ...) \
do { \
if (DEBUG_ENABLED) { \
char str[300]; \
int length = -1; \
length = snprintf(str, sizeof(str), msg, __VA_ARGS__); \
if (length > 0) \
{ \
fprintf(stderr, "%s, %d ",__func__, __LINE__); \
fprintf(stderr, "%s", str); \
fprintf(stderr,"\n"); \
} \
} \
} while (0)
The above form requires at least two arguments, so you couldn't use it to print a simple debug message such as "got here". As a workaround, the macro can be defined with no fixed parameters, omitting the msg parameter:
#define debug_log(...) \
do { \
if (DEBUG_ENABLED) { \
char str[300]; \
int length = -1; \
length = snprintf(str, sizeof(str), __VA_ARGS__); \
if (length > 0) \
{ \
fprintf(stderr, "%s, %d ",__func__, __LINE__); \
fprintf(stderr, "%s", str); \
fprintf(stderr,"\n"); \
} \
} \
} while (0)
That would allow you to attempt to call the macro with no parameters at all, such as debug_log();, but that will result in a compiler error at the call to snprintf.