How are variable arguments implemented in gcc?

Question

int max(int n, ...)

I am using cdecl calling convention where the caller cleans up the variable after the callee returns.

I am interested in knowing how do the macros va_end, va_start and va_arg work?

Does the caller pass in the address of the array of arguments as the second argument to max?

Answer 1

If you look at the way the C language stores the parameters on the stack, the way the macros work should become clear:-

Higher memory address    Last parameter
                         Penultimate parameter
                         ....
                         Second parameter
Lower memory address     First parameter
       StackPointer  ->  Return address

(note, depending on the hardware the stack pointer maybe one line down and the higher and lower may be swapped)

The arguments are always stored like this¹, even without the ... parameter type.

The va_start macro just sets up a pointer to the first function parameter, e.g.:-

 void func (int a, ...)
 { 
   // va_start
   char *p = (char *) &a + sizeof a;
 }

which makes p point to the second parameter. The va_arg macro does this:-

 void func (int a, ...)
 { 
   // va_start
   char *p = (char *) &a + sizeof a;

   // va_arg
   int i1 = *((int *)p);
   p += sizeof (int);

   // va_arg
   int i2 = *((int *)p);
   p += sizeof (int);

   // va_arg
   long i2 = *((long *)p);
   p += sizeof (long);
 }

The va_end macro just sets the p value to NULL.

NOTES:

1. Optimising compilers and some RISC CPUs store parameters in registers rather than use the stack. The presence of the ... parameter would switch off this ability and for the compiler to use the stack.