C Programming - Variable Number of Arguments - Discussion

Wants to know more. Read this code.

#include<stdio.h>
//Header file contains Macros
#include<stdarg.h>
//Compute avg of given no's
double avg(int num, ...)
{
int i;
double sum=0.0;
//Create va_list variable
va_list valist;
//Initializa va_list with num
va_start(valist,num);
//Run untill all variables are visted once
for(i=1;i<=num;++i)
{
//va_arg will pick up variable from argument list and each
//time move 4 bytes ahead (because we are using here integer
//second argument in va_arg.
sum+=va_arg(valist,int);
}
return sum/num;
}
int main()
{
printf("\nAVG:%e",avg(5, 1,2,3,4));
return 0;
}

@Niyati.

while(n-- > 0) means u understand so n is indicate total number of argument.

va_start is indicate first argument. va_args(p,int) return first argument value and va_args(p,int) second argument value.

so that print first function and second function and not print first value.

Print first value by simply write printf("%d", n); of the above example.

I hope you understand.

In both dumplist 2, 3 initial values represent no of arguments taken remaining are values stored so n->0 indicates for 1st while (n->0) indicates n is equal to 0 for both times of dumplist.

It is checked but initial value of n=2 no of arguments taken by dumplist accordingly n=3 for 2nd one n--is done according to printf function respective values included initially is printed.

In the list, the first parameter always represents the number of arguments to be passed.

So the 2 in the first function represents that there are 2 arguments.

And 3 in second function represents there are 3 arguments.

So first it prints 4 and 8.

Next, it will print 6, 9, 7. Excluding 4 and 8.
4 and 8 are actually to define a number of parameters. And they are not get printed.

First dumplist function is called with elements 2, 4 and 8. When va_start is called, it points to element 2.

When it sees va_arg, it points to the next element 4.

Since va_arg is within while loop, it prints 4 8 for the first function and 6 9 7 for the second function.

So the answer is option C.

@Raj... When it sees va_arg, it points to the next element 4.

Since va_arg is within while loop, it prints 4 8 for the first function and 6 9 7 for the second function.What you are saying here is incorrect.va_arg returns the element it is pointing to and the move to next element in list

Here first variable passed in both calls will be considered as the starting point to work with variable list.

In first part,

va start : 2.

So n would be, 2 so loop runs for 4 and 7.

Whereas in second case, va start:3 so 6 9 and 7 is printed on the screen.

Can anyone please explain this in detail.

@Raj well, ques2 also talks about the same concept, but here, va_arg doesn't seem to point to the next argument.

while(n-- >0) ...loop until the value of n is greater than 0. I think here n represents the no of arguments passed excluding n.

what does while loop condition indicates??

i.e. while(n-->0)

what this condition indicates????