Discussion
What is the output of the following pseudocode?
int arr[] = {1, 2, 3, 4, 5}
int *ptr = arr + 2
print(*ptr + *(ptr + 1))Answer: C
The output is 7.
The array holds 1, 2, 3, 4, 5 at indices 0 to 4. arr + 2 points to index 2, so *ptr is 3. ptr + 1 points to index 3, so *(ptr + 1) is 4. The sum is 3 + 4 = 7.
Pointer arithmetic moves by whole elements, not bytes, so arr + 2 is the third element. 5 is the sum of indices 1 and 2 (2 + 3), a slip of taking arr + 2 as the second element. 9 is 4 + 5, a slip of one position too far, and 3 is the first value alone.
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