I'd like to know if someone knows the way a compiler would interpret the following code:
#include <iostream>
using namespace std;
int main() {
cout << (true && true || false && false) << endl; // true
}
Is this true because && has a higher precedence than || or because || is a short-circuit operator (in other words, does a short circuit operator disregard all subsequent expressions, or just the next expression)?
about the true && true || infiniteLoop() && infiniteLoop() example, neither of the infinite loop calls are being evaluated because of the two characteristics combined: && has precedence over ||, and || short-circuits when the left side is true.
if && and || had the same precedence, the evaluation would have to go like this:
((( true && true ) || infiniteLoop ) && infiniteLoop )
(( true || infiniteLoop ) && infiniteLoop )
=> first call to infiniteLoop is short-circuited
(true && infiniteLoop) => second call to infiniteLoop would have to be evaluated
but because of &&'s precedence, the evaluation actually goes
(( true && true ) || ( infiniteLoop && infiniteLoop ))
( true || ( infiniteLoop && infiniteLoop ))
=> the entire ( infiniteLoop && infiniteLoop ) expression is short circuited
( true )
This is a sequential illustration:
(true && true || false && false)
= ((true && true) || (false && false)) // because && is higher precedence than ||,
// like 1 + 2 * 3 = 7 (* higher than +)
= ((true) || (false))
= true
but also note that if it is
(true || ( ... ))
then the right hand side is not evaluated, so any function there is not called, and the expression will just return true.