C Programming - Variable Number of Arguments - Discussion
Discussion Forum : Variable Number of Arguments - Find Output of Program (Q.No. 3)
3.
What will be the output of the program?
#include<stdio.h>
#include<stdarg.h>
void dumplist(int, ...);
int main()
{
dumplist(2, 4, 8);
dumplist(3, 6, 9, 7);
return 0;
}
void dumplist(int n, ...)
{
va_list p; int i;
va_start(p, n);
while(n-->0)
{
i = va_arg(p, int);
printf("%d", i);
}
va_end(p);
printf("\n");
}
Discussion:
20 comments Page 2 of 2.
Harini said:
1 decade ago
Actually n-->0 means that,
n-- > 0 i.e. n is first decremented and is compared with 0.
n-- > 0 i.e. n is first decremented and is compared with 0.
Niyati said:
1 decade ago
Please can anyone clearly explain this problem.I am not able to understand this.please..
Priya said:
1 decade ago
I can't understand, please anyone explain the above program clearly.
Kalpana said:
1 decade ago
I need a clear explanation about this program.
Pooja said:
2 decades ago
Please describe how to solve this problem.
Mahi said:
8 years ago
Decrement value of n.
Untill (n--) to 0.
Untill (n--) to 0.
Akshaya Shelke said:
8 years ago
Thanks for your explanation @Vvvvvvv.
K.karthik raja said:
1 decade ago
Please clearly explain this problem?
Teju said:
9 years ago
while(n-->0) Explain this line.
Sandipan said:
9 years ago
What is va_end?
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