What happens if you try to compile and run this program?

What happens if you try to compile and run this program?

#include <stdio.h>

int main (int argc, char *argv[]) {

int i =2, j = 1;

if(i / j)

j += j;

else

i += i;

printf("%d",i + j);

return 0;

}

Choose the right answer:
A . The program outputs 1
B . The program outputs 5
C . The program outputs 3
D . Compilation fails
E . The program outputs 4

View Answer

Answer: E

Explanation:

In the if statement, i / j is 2 / 1, which is true. Therefore, the if block is executed, and j += j; doubles the value of j (j becomes 2).

After the if-else statement, printf("%d", i + j); prints the sum of i and the updated val-ue of j (2 + 2), which is 4.