Question

Consider the following code. Assume that the array begins at memory address Ox200. What is printed by the following code. Assume sizeof(int) is 4 bytes and sizeof(char) is 1 byte. Do not include the Ox or leading zeros in your answer. Enter -1 if the answer is indeterminate, or -2 if the code generates a compile error, or -3 if the code generates a runtime error. #include <stdio.h> int main() { int a[] {1,2,3}; int *iptr a; printf("%p\n", iptr+1); return 0; }

Answer 1

iptr is the address of array a
So, iptr is 200

iptr + 1 
= 200 + 1 * sizeof(int)
= 200 + 1*4
= 200 + 4
= 204

So, output of the code is
204

Answer

204

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