What's the meaning of !int_variable--?

Question

I can't understand what the following code is doing on s:

 if(!s--)

s is an int

1) that's not a function. 2) use divide and conquer. there are only three things there: `if`. `!` and `--`. — Karoly Horvath, Dec 05 '14 at 10:40
At first, compiler is checking if s is equal to zero to enter into the if block. Then s is decremented by 1. — Mukit09, Dec 05 '14 at 10:45

score 3 · Answer 1 · answered Dec 05 '14 at 10:40

Actually, it's misleading.

You are testing is s is different from 0 (with if (!s)). And then, afterward, whatever the result is, you're decreasing it.

So, it's two different operations. It could be written this way :

if (!s)
{
     s--;
     //...
}
else
{
     s--;
}

Natasha Dutta · Accepted Answer · 2014-12-05T11:03:21.653

2

! is called negation operator. It is a logical operator.

if(!s--)

The order in which it executes

check the value of s is 0 or not , if s is 0, if condition is success [thanks to the ! operator], otherwise, failure.
After that, decrement s by one unit.
Based on the evaluation of if condition, continue the execution [code under if condition, or next block of code].

edited Dec 05 '14 at 11:03

answered Dec 05 '14 at 10:39

Natasha Dutta

2 Answers2