Discussion

What will be the output of the program if the array begins at 65472 and each integer occupies 2 bytes?

#include <stdio.h>

int main()
{
    int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0};
    printf("%u, %u\n", a+1, &a+1);
    return 0;
}

Answer: B

Correct answer: 65480, 65496

a is a 3×4 array of 2-byte ints. a alone decays to a pointer to its first row, so adding 1 skips one row of 4 ints:

\[ 65472 + 4 \times 2 = 65480 \]

&a is a pointer to the whole array, so adding 1 skips all 12 ints:

\[ 65472 + 12 \times 2 = 65496 \]

Same address, different type - and the type is what decides how far +1 moves.

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