Correct Answer - Option 4 : 24
code:
#include<stdio.h>
int main(){
int arr[4][5];
int i, j;
for(i =0; i<4; i++){
for (j =0; j<5; j++){
arr [i][j] = 10 +i + j;
}
}
print("%d", *(arr[1] + 9));
return 0;
}
- C doesn't check array boundaries. A segmentation fault will only occur if you try to dereference a pointer to memory that your program doesn't have permission to access. Simply going past the end of an array is unlikely to cause that behavior. Undefined behavior is just that - undefined. It may appear to work just fine, but you shouldn't be relying on its safety.
- Your program causes undefined behavior by accessing memory past the end of the array. In this case, it looks like one of your str[i] = c writes overwrite the value in i.
Explanation:
*(arr[1] + 9) can be written as arr[1][9].
as C doesn't follow bound check and follow the row major ordering
arr[1][5] = arr[2][0] // arr[1][4] will be first row of array and then arr[2][0] will be second row of array
arr[1][6] = arr[2][1]
arr[1][7] = arr[2][2]
arr[1][8] = arr[2][3]
arr[1][9] = arr[2][4]
arr[2][4] = 10*i + j = 10*2+4 = 24
Option 4 is the answer.
