How this logical operator works?

Question

#include<stdio.h>
int main()
{
    int a=-10,b=3,c=0,d;
    d= a++||++b &&c++;
    printf("%d %d %d %d ",a,b,c,d);
}

How above expression is evaluates. I have read preceedence but still i am getting confused. Kindly give me the right solution of this in step by step.

[Precedence](https://en.cppreference.com/w/c/language/operator_precedence) makes it `(a++) || ( (++b) && (c++) )`. The real question is what does pre-increment (`++x`) and post-increment (`x++`) return... — ikegami, Nov 23 '21 at 04:15

score 0 · Answer 1 · answered Nov 23 '21 at 04:34

0

In case of || and && operators the order of evaluation is indeed defined as left-to-right.So, operator precedence rules tell you that the grouping should be

(a++) || ((++b) && (c++))

Now, order-of-evaluation rules tell you that first we evaluate a++, then (if necessary) we evaluate ++b, then (if necessary) we evaluate c++, then we evaluate && and finally we evaluate ||.

answered Nov 23 '21 at 04:34

vegan_meat

878
4
10

The '(if necessary)' comments sorta mean that the 'then we evaluate `&&` and finally we evaluate `||`' parts are misleading (at best) IMO. If the `a++` evaluates to non-zero (true), nothing is done with `b` or `c` and the `&&` is never evaluated. – Jonathan Leffler Nov 23 '21 at 05:23

machfour · Answer 2 · 2021-11-23T23:41:07.200

I have a feeling this is a homework question, so I'll try to give extra explanation. There are actually a lot of concepts going on here!

Here are the main concepts:

Pre- and post- increment and decrement. ++a increments a before the value is used in an expression, while a++ increments a after the value is used in the expression.
Operator precedence. Specifically, && has higher precedence than ||, which means that the assignment of d should be should be read as d = (a++) || (++b && c++);
Expression evaluation order. The link shows a whole list of evaluation rules for C. In particular, item 2 says (paraphrasing) that for operators || and &&, the entire left side is always fully evaluated before any evaluation of the right-hand side begins. This is important because of...
Short-circuit boolean evaluation, which says that, for operators && and ||, if the value of the left-hand term is enough to determine the result of the expression, then the right-hand term is not evaluated at all.
Behaviour of boolean operators in C. There is no inbuilt boolean type in C, and operators && and || work on integer values. In short, an argument is 'true' if it is nonzero, and false if it equals zero. The return value is 1 for true and 0 for false.

Putting this all together, here is what happens:

After the first line, a is -10, b is 3, c is 0 and d is unset.
Next, the variable d needs to be assigned. To determine the value assigned to d, the left hand term of (a++) || (++b && c++) is evaluated, which is a++. The value USED in the expression is 10, however the value of a after this expression is -9 due to the post-increment.
For the purposes of the boolean operator, the value 10 is true, and therefore value of the || expression is 1. Because of short-circuit evaluation, this means that ++b && c++ is not evaluated at all, so the increments do not happen. Thus we have d = 1.
At the end, the values are: a = -9, b = 3, c = 0, d = 1.

So the program prints out -9 3 0 1.

How this logical operator works?

2 Answers2