What is the output of the below program?
#include <stdio.h>
int arr_1[5] = {1,2,3};
int arr_2[3] = {1,2,3};
void main()
{
int *ptr_1 = (&arr_1+1);
printf("%d\n %d\n", *(arr_1+1), *(ptr_1-1));
int *ptr_2 = (&arr_2+1);
printf("%d\n %d\n", *(arr_2+1), *(ptr_2-1));
}
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In the above program two integer arrays arr_1 and arr_2 are declared.
arr_1 is partially initialized
arr_2 is completely initialized
Consider the case of ptr_1,
pointer ptr_1 is assigned (&arr_1+1) which means it shall be assigned arraysize+1, if each integer is considered, 4 bytes and array address starts at X, then ptr_1 could be X+4*5 = X+20.
Value at arr_1[X] = 1
Value at arr_1[X+4] = 2
Value at arr_1[X+8] = 3
Value at arr_1[X+12] = 0
Value at arr_1[X+16] = 0
Values at arr_1[X+12] and arr_1[X+16] are zero as they are not initialized in the array.
In the print function, *(arr_1+1) is printed first which meant, value that is incremented by 1 from the base value of array. which is arr_1[1] which is 2.
Also, in the print function, *(ptr_1-1) is printed which meant ptr_1 value decremented by 1 which as considered above shall be X+16 which meant it prints 0.
So, for the first printf output is 2,0
Consider the case of ptr_2,
pointer ptr_2 is assigned (&arr_2+1) which means it shall be assigned arraysize+1, if each integer is considered, 4 bytes and array address starts at X, then ptr_1 could be be X+4*3 = X+12.
Value at arr_1[X] = 1
Value at arr_1[X+4] = 2
Value at arr_1[X+8] = 3
In the print function, *(arr_1+1) is printed first which meant, value that is incremented by 1 from the base value of array. which is arr_1[1] which is 2.
Also, in the print function, *(ptr_1-1) is printed which meant ptr_1 value decremented by 1 which as considered above shall be X+8 which meant it prints 3.
So, for the first printf output is 2,3
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