What does this code mean?

Question

#include <iostream>
using namespace std;

int main() {
int x=5;
x=!(x>8);
cout<<x;
return 0;
}
why output is 1?

Answer 1

hi my friend ! attention : x>8 returns false ( that is 0 )
then NOT of false becomes true : !(false)= true
so integer value of (true) is 1 and value of x becomes 1. (x=1)
bye

Answer 2

دمت گرم داداش

Answer 3

output is 1 because x>8 is false and ! operater is logical not this to 1 x=not false that is x=true

Answer 4

x>8 is a logical operation, but since x is an integer therefore logical results are displayed in form of 0 (means FALSE). Then the !0 becomes 1(means False). therefore output is 1.

Answer 5

In x>8 if x=5 then 5>8.it becomes false. But not of false is true. When it is true it returns output 1.

Answer 6

Included iostream;
namespace std loaded;

x = !(5>8); !(false) -- true, int cannot be transformed into bool */
1