C Programming - Expressions - Discussion
Discussion Forum : Expressions - Find Output of Program (Q.No. 3)
3.
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
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.
Hence the output is "-2, 2, 0, 1".
Discussion:
41 comments Page 5 of 5.
Vivek said:
1 decade ago
{
int a=10,b=2,c;
a=!(c=c==c)&&++b;
printf("%d",b);
}
Why value of b is 2? Not 3 but if we replace && by || b is 3?
int a=10,b=2,c;
a=!(c=c==c)&&++b;
printf("%d",b);
}
Why value of b is 2? Not 3 but if we replace && by || b is 3?
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