Question
#include<stdio.h>
int main()
{
int i=-3, j=2, k=0, m;
m = ++i && ++j || ++k;
printf("%d, %d, %d, %d\n", i, j, k, m);
return 0;
}
What will be the output of the program?
#include<stdio.h>
int main()
{
int i=-3, j=2, k=0, m;
m = ++i && ++j || ++k;
printf("%d, %d, %d, %d\n", i, j, k, m);
return 0;
}
Answer: Option C
Was this answer helpful ?
Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and
variable i, j, k are initialized to -3, 2, 0 respectively.
Step 2: m = ++i && ++j || ++k;
becomes m = (-2 && 3) || ++k;
becomes m = TRUE || ++k;.
(++k) is not executed because (-2 && 3) alone return TRUE.
Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.
Step 3: printf("%d, %d, %d, %d`setminus`n", i, j, k, m); In the previous step the value of i,j are increemented by '1'(one).
Hence the output is "-2, 3, 0, 1".
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution