Question

I have tried this:

#define format(f, ...) \
                int size = strlen(f) + (sizeof((int[]){__VA_ARGS__})/sizeof(int)) + 1); \
                char *buf = malloc(size); \
                snprintf(buf, size, f, __VA_ARGS__); \
                buf

But it returns a lot of syntactic errors. How do I do this properly?

Answer 1

C macros are not functions but 1:1 substitutions. So if you want to use your macro like this:

mystring = format("%d", 5);

You get this:

mystring = int size = strlen(f) + (sizeof((int[]){5})/sizeof(int)) + 1); \
            char *buf = malloc(size); \
            snprintf(buf, size, f, 5); \
            buf;

Which does not make any sense. In your case you are better off defining an inline function which should not be any worse in terms of performance on a decent compiler.

If it really has to be a macro and you are on GCC, you can use the compound statement to achieve your goal. It allows you to do this: mystring = ({ statement1, statement2, ..., statementN}) which will execute all your statements in a local scope and then assign statementN to mystring. However it will make your code non-portable and be a hell to debug.

So here you go, but please don’t use this in real applications:

#define format(f, ...) \
    ({ int size = snprintf(NULL, 0, f, __VA_ARGS__) + 1;\
    char * buf = malloc(size);\
    snprintf(buf, size, f, __VA_ARGS__); buf; })

I’m serious. Don’t use this. Use an inline function. You can also have variadic arguments in normal functions, using va_arg and va_start:

inline char * format(f, ...) {
    va_list args;
    va_start(args, f);
    int size = vsnprintf(NULL, 0, f, args) + 1;
    char * buf = malloc(size);
    vsnprintf(buf, size, f, args);
    return buf;
}