Why does the value of c not change?

Question

#include <stdio.h>
void main()
{
    int a = 5, b = -7, c = 0, d;
    d = ++a && ++b || ++c;
    printf("\n %d,%d,%d,%d", a,  b, c, d);
}

In the above code, the output is 6,-6,0,1. Why is the value of c 0 and the value of d 1? how did d get the value as 1?

Answer 1

That is because || first checks the left part and if it is true it return true without evaluating the right part.

In C any non zero is treated as True and zero as False

int a = 5, b = -7, c = 0, d;
    d = ++a && ++b || ++c;

Here ++a and ++b are non zero and both are treated as True so ++a&&++b becomes True and the expression stops evaluating over there.

Answer 2

There are already good answers that explain why c is 0 after the line

d = ++a && ++b || ++c;

is executed.

The line is equivalent to:

d = (++a && ++b || ++c);

That explains why d is 1 .

Answer 3

int a = 5, b = -7, c = 0, d;
d = ++a && ++b || ++c;

let's analyze the statement part by part

++a : a=6

++a && ++b: b becomes -6 ,and then it does : 6 && -6 ,which is equal to 1

now there is a || (or symbol) ,but this cannot affect the value of d since

1||"value2" =1 , so the compiler does not evaluate "++c" .

so c remains 0 and d becomes 6 && -6 =1

Answer 4

&& has higher precedence than || so first ++a&&++b is evaluated first and it becomes true. The expression becomes true || ++c true || ++c and || will evaluate from left to right since || encounters true and it returns 1 to d without evaluating ++c. So c value will not be incremented

Answer 5

An evaluation tree is created which does the evaluation. The || part is parent and the two parts on its side are its parents. When it evaluates one child, if it gets true, it doesn't evaluate the other child.