On executing this piece of C command, the output of num is 7. I was expecting it to be 6, can anyone explain why and how it turns out to be 7?
#include <stdio.h>
int main() {
int a[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 };
int i = 0, num = 0;
num = a[++i + a[++i]] + a[++i];
printf("%d ", i);
printf("%d ", num);
return 0;
}
i is incremented 3 times within the same expression in num = a[++i + a[++i]] + a[++i];
This has undefined behavior in C. It is a moot point trying to explain why it outputs 7 or 6 or anything whatsoever. Undefined behavior can be anything.
llvm on OS X gives me 6 and a warning:
wi.c:5:8: warning: multiple unsequenced modifications to 'i' [-Wunsequenced]
which suggests that we are looking at an undefined behavior. The exact nature of what it undefined here is not clear to me thus far, but feels a bit irrelevant.
This is a bit tricky, the expression : a[++i+a[++i]], involves the increment of the variable i two times, and comes out to be a[i + 2 + a[i + 2]], which is a[0 + 2 + a[2]] = a[4] = 4, and the second operand, a[++i] becomes a[3], which is equal to 3, hence, the final answer is 7. In other words, this is undefined behaviour.
i = 0;
num = a[ ++i + a[++i]] + a[++i]
Will evaluate to
num = a[1+ a[2]] + a[3]
num = a[1 + 2] + a[3]
num = a[3] + a[3]
num = 3 + 3
num = 6
