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What will be the output of the program?

#include
 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;
 }

2, 2, 0, 1

1, 2, 1, 0

-2, 2, 0, 0

-2, 2, 0, 1

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Answer : Option D

Explanation :

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; here (++j && ++k;) this code will not get executed because ++i has non-zero value.
becomes m = -2 || ++j && ++k;
becomes m = TRUE || ++j && ++k; Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.

Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of variable 'i' only increemented by '1'(one). The variable j,k are not increemented.