C Programming - Variable Number of Arguments - Discussion

In 1 place in random test you say that va_end should be present for every va_start. But here it compiles with no error. How?

@irshad::thanks for your explination.

We can also use va_arg(ptr, char);

It depends on the variable arguments ur passing to the function

Ex: fun("IndiaBIX", 'I','N','D','I','A' );

What does int in va_arg(ptr, int) indicates? and is there any case where we can use char over there?

#include<stdio.h>
#include<stdarg.h>
void irshad(char *arr, ...)
{
int num;
va_list ptr; /* create a new list ptr */
va_start(ptr,arr); /* attach ptr with arr */
printf("\n num=%d",arr); /* it will print 1st argument */
num = va_arg(ptr,int); /* fetch 2nd argument */
printf("\n num=%d",num);
num = va_arg(ptr,int); /* fetch 3rd argument */
printf("\n num=%d",num);
}
int main()
{

printf("\n variable argument:");
irshad(10,20,30,40,50,0);
return 0;
}


/* output */

variable argument:
num=10
num=20
num=30


Please let me know if any problem to understand this one.
irshadalam.mca[at]gmail.com

I didn't get it ! Can anyone expalin in little more detail ?

Sir please explane how it comes.

Yes I also thinks what shiv z saying, dats correct.

I am aggry with shiv.

va_list ptr is an argument pointer, used to traverse through the variable arguments passed in the function.

va_start points 'ptr' to the first argument, in this case 'IndiaBix'

Every call to va_arg moves the ptr to next variable argument.

Hence after 2 calls to va_arg, the pointer ends up at '7' and num is '4'.