1. 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; }
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By: guest on 01 Jun 2017 06.01 pm
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\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".
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\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".