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.)
Originally, in C (on which C++ is based) there was no Boolean type. Instead, the value "true" was assigned to any non-zero value and the value "false" was assigned to anything which evaluates to zero. This behavior still exists in C++. So for an int x, the expressions !x means "x not true", which is "x not non-zero", i.e. it's true if x is zero.
The test for int is true for non-zero values and false for zero values, so not is just true for zero values and false for non-zero values.
The build-in ! operator converts its argument to bool. The standard specifies that there exists a conversion from any arithmetic type(int, char,.... float, double...) to bool. If the source value is 0 the result is true, otherwise it is false
