CODEWITHSUNDEEP
c
ravivasani75
ravivasani75

Reputation: 123

How does __builtin_va_arg() works even after preprocessing?

Here is source code:

#include <stdarg.h> 
void foo(char *fmt, ...)
{
    va_list ap;
    va_start(ap, fmt);
    int a, b, d, e;
    double c;
    a = va_arg(ap, int);
    va_end(ap);
}

And here is the code after preprocessing(ignoring #include and preprocessor comments(#)):

void foo(char *fmt, ...)
{
     va_list ap;
     __builtin_va_start(ap, fmt);
     int a, b, d, e;
     double c;
     a = __builtin_va_arg(ap, int); // int as an argument after processing!
     __builtin_va_end(ap);
}

so, how does __builtin_va_arg(ap, int) works ? When this macro expanded ?

Upvotes: 1

Views: 3310

Answers (1)

rici
rici

Reputation: 241691

These are not macros. They are built-in primitives implemented by the compiler itself. (That is, the particular compiler you are using.)

The actual syntax and semantics of these built-ins is deliberately not documented. You must not use them directly in a program, as indicated by the fact that their names start with two underscores. (Only documented symbols starting with two underscores may be used in a program.) When the compiler encounters them after preprocessing, it will emit the appropriate code (which likely cannot be written in C).

Upvotes: 3

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