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What will be the output of following program code?#include <stdio.h>int main(void){ char p; char buf[10] = {1, 2, 3, 4, 5, 6, 9, 8}; p = (buf + 1)[5]; printf("%d", p); return 0;}
- A5
- B6
- C9
- DError
- ENone of the above
Correct answer: C
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CCommunity
9
x[i] is equivalent to *(x + i),
so (buf + 1)[5] is *(buf + 1 + 5), i.e. buf[6].
x[i] is equivalent to *(x + i),
so (buf + 1)[5] is *(buf + 1 + 5), i.e. buf[6].