Strange question, but someone showed me this, I was wondering can you use the not ! operator for int in C++? (its strange to me).
#include <iostream>
using namespace std;
int main()
{
int a=5, b=4, c=4, d;
d = !( a > b && b <= c) || a > c && !b;
cout << d;
system ("pause");
return 0;
}
Yes. For integral types, !
returns true
if the operand is zero, and false
otherwise.
So !b
here just means b == 0
.
This is a particular case where a value is converted to a bool
. The !b
can be viewed as !((bool)b)
so the question is what is the "truthness" of b
. In C++, arithmetic types, pointer types and enum can be converted to bool
. When the value is 0 or null, the result is false
, otherwise it is true
(C++ §4.1.2).
Of course custom classes can even overload the operator!
or operator
<types can be convert to bool> to allow the !b
for their classes. For instance, std::stream
has overloaded the operator!
and operator void*
for checking the failbit, so that idioms like
while (std::cin >> x) { // <-- conversion to bool needed here
...
can be used.
(But your code !( a > b && b <= c) || a > c && !b
is just cryptic.)