What is output of following C – Code?
#include <stdio.h> #include <stdarg.h> int fun( int n, ...)
{ int i, j = 1, val = 0;
va_list p;
va_start (p, n);
for (; j < n; ++j)
{
i = va_arg (p, int );
val += i;
}
va_end (p);
return val;
} int main()
{ printf ( "%d\n" , fun(4, 1, 2, 3));
return 0;
} |
(A) 3
(B) 5
(C) 6
(D) 10
Answer: (C)
Explanation: The function receives variable number of arguments as there are three dots after first argument. The first argument is count of all arguments including first. The function mainly returns sum of all remaining arguments. See https://www.geeksforgeeks.org/how-to-count-variable-numbers-of-arguments-in-c for details.
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