__va_sizeof__ Keyword

According to the ANSI-C specification, you must promote character arguments in open parameter lists to int. The use of char in the va_arg macro to access this parameter may not work as per the ANSI-C specification, as shown in the following listing.

Listing: Inappropriate use of char with the va_arg Macro

int f(int n, ...) {
  int res;

  va_list l= va_start(n, int);

  res= va_arg(l, char); /* should be va_arg(l, int) */

  va_end(l);

  return res;

}

void main(void) {

  char c=2;

  int res=f(1,c);

}

With the __va_sizeof__ operator, the f function in the va_arg macro returns 2.

A safe implementation of the f function is to use va_arg(l, int) instead of va_arg(l, char).

The __va_sizeof__ unary operator, which is used exactly as the sizeof keyword, returns the size of its argument after promotion as in an open parameter list, as shown in the following listing.

Listing: __va_sizeof__ Examples

__va_sizeof__(char) == sizeof (int)
__va_sizeof__(float) == sizeof (double)

struct A { char a; };

__va_sizeof__(struct A) >= 1 (1 if the target needs no padding bytes)

Note: In ANSI-C, it is impossible to distinguish a 1-byte structure without alignment or padding from a character variable in a va_arg macro. They need a different space on the open parameter calls stack for some processors.