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; } ?->(Show Answer!)

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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; } ?->(Show Answer!)

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1. 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; }

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By: guest on 01 Jun 2017 06.00 pm

Step 1: int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0}; The array a[3][4] is declared as an integer array having the 3 rows and 4 colums dimensions. Step 2: printf("%u, %u\n", a+1, &a+1); The base address(also the address of the first element) of array is 65472. For a two-dimensional array like a reference to array has type "pointer to array of 4 ints". Therefore, a+1 is pointing to the memory location of first element of the second row in array a. Hence 65472 + (4 ints * 2 bytes) = 65480 Then, &a has type "pointer to array of 3 arrays of 4 ints", totally 12 ints. Therefore, &a+1 denotes "12 ints * 2 bytes * 1 = 24 bytes". Hence, begining address 65472 + 24 = 65496. So, &a+1 = 65496 Hence the output of the program is 65480, 65496

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