Thread: confusion in variable-length arguments

Code:

#define _INTSIZEOF(n)   ( (sizeof(n) + sizeof(int) - 1) & ~(sizeof(int) - 1) )
#define va_start(ap,v)  ( ap = (va_list)&v + _INTSIZEOF(v) )

What type is of v? Not a pointer, I think, otherwise, there is no need to get the address using &v, v itself will suffice.

Firstly, looking at stdarg.h doesn't matter, it's upto the implementation. Even if there is a way to determine the amount of arguments on your implementation (and there likely isn't), it's not part of the C standard, so you can't rely on it for portable C programs.

Looking at the code you pasted, what is happening here:

Code:

#define va_start(ap,v)  ( ap = (va_list)&v + _INTSIZEOF(v) )

Is that ap is being initialised to the address of v (your pointer, fmt) plus the size of the pointer, so that it initialises ap to point to the first variable argument on the stack. Looking at INTSIZEOF(n): ( (sizeof(n) + sizeof(int) - 1) & ~(sizeof(int) - 1) )

The &~(sizeof(int)-1) trick is to round to the nearest int, because a stack entry will also be int padded. Example, if sizeof int is 4, 4-1 is 3, ~3 is 11111100, so an & effectively zero's the two least significant bits, and because it already adds sizeof(int)-1, this will always be a round up. Example, if sizeof fmt is 6, it will round to 8, if sizeof fmt is 4, it will round to 4, if sizeof fmt is 8, it will round to 8 (assuming 4 byte ints). The reason this is done is that the last fixed parameter can be any type, not necessarily a char *

So, it is merely calculating where the first variable argument is located, it doesn't work out how many there are.

Hope I've explained clearly enough.

Originally Posted by cwr

Example, if sizeof fmt is 6, it will round to 8, if sizeof fmt is 4, it will round to 4, if sizeof fmt is 8, it will round to 8 (assuming 4 byte ints). The reason this is done is that the last fixed parameter can be any type, not necessarily a char *

Maybe I didn't make myself clear. You said "if sizeof fmt is 6", OK, there must be a way to count it so we can know about it. And what I doubt is the implementation in the STDARG.H in whihc it count the size by _sizeof_ a char pointer, which, as we know, will always get 4, the size of storage of a pointer. Say we passed some argument to miniprintf like this:

Code:

miniprintf("%c = %d\n", c, c);

Here fmt is a pointer to the address that stores the string "%c = %d\n". Obviously the size of which is not 4. But in the implement, sizeof(fmt) will only get 4.DO YOU NOT THINK SO?

Code:

char array[100], *pointer = array;

printf("array = %i\npointer = %i\n", sizeof(array), sizeof(pointer));

/*
prints (on a 32-bit computer):

array = 100
pointer = 4
*/