CODEWITHSUNDEEP
c
kusur
kusur

Reputation: 618

In C: passing and printing variables arguments of a function

I am trying to write a program which uses function with variable number of arguments. An additional task is to print all the arguments of every function call separately. The code is as follows:-

#include<stdio.h>
#include<stdarg.h>
#include<string.h>
int mul(int x,...);
int main()
{
int a=1,b=2,c=3,d=4,x;
printf("The product of %d is :%d\n",a,mul(1,a));
printf("The product of %d, %d is :%d\n",a,b,mul(2,a,b));
printf("The product of %d, %d, %d is :%d\n",a,b,c,mul(3,a,b,c));
printf("The product of %d, %d, %d, %d is :%d\n",a,b,c,d,mul(4,a,b,c,d));
return 0;
}
int mul(int x,...)
{
    int i,prod=1;
    va_list arglist;
    va_start(arglist, x);
    for(i=0;i<x;i++)
    {   
        prod*=va_arg(arglist,int);
    }
    printf("\n");
    for(i=0;i<x;i++)
    {   
        printf("The argument is %d\n",va_arg(arglist,int));
    }
    va_end(arglist);
    return prod;
}

The output of this program is as follows:-

Output of the above program

The other piece of code is:- 

#include<stdio.h>
#include<stdarg.h>
#include<string.h>
int mul(int x,...);
int main()
{
int a=1,b=2,c=3,d=4,x;
printf("The product of %d is :%d\n",a,mul(1,a));
printf("The product of %d, %d is :%d\n",a,b,mul(2,a,b));
printf("The product of %d, %d, %d is :%d\n",a,b,c,mul(3,a,b,c));
printf("The product of %d, %d, %d, %d is :%d\n",a,b,c,d,mul(4,a,b,c,d));
return 0;
}
int mul(int x,...)
{
    int i,prod=1;
    va_list arglist;
    va_start(arglist, x);
    for(i=0;i<x;i++)
    {   
        prod*=va_arg(arglist,int);
    }
    printf("\n");
    va_end(arglist);
    va_start(arglist,x);
    for(i=0;i<x;i++)
    {   
        printf("The argument is %d\n",va_arg(arglist,int));
    }
    va_end(arglist);
    return prod;
}

The output is as follows:-

second output

Why is this difference? Any help?

Upvotes: 1

Views: 2201

Answers (2)

woryzower
woryzower

Reputation: 976

va_arg(va_list ap, type) retrieves next argument in the argument list.So in the first code you are consuming the arguments after one loop . Instead of 2nd code you can use the following code which prints argument and maintains multiplication in a single loop

int mul(int x,...)

{
    int i,m,prod=1;
    enter code here
    va_list arglist;
    enter code here
    va_start(arglist, x);
    for(i=0;i<x;i++)
    {   
        m=va_arg(arglist,int);
        prod*=m
        printf("The argument is %d\n",m);
    }
    printf("\n");

    return prod;
}

Upvotes: 0

Mark Byers
Mark Byers

Reputation: 838156

In the first example you are missing two lines:

va_end(arglist);
va_start(arglist,x);

This means that after doing the multiplication you are reading past the end of the parameters. The values that are displayed are whatever happened to be on the stack.

Upvotes: 2

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