Strange evaluation of boolean expression

Question

why is the value of the following expression false?

bool a = false;
bool b= true;
std::cout<< a || !b && !a || b;

and why does the value changes when adding parenthesis

bool a = false;
bool b= true;

std::cout<< (a || !b && !a || b);

Shouldn't the parenthesis be putted like this:

a || (!b && !a) || b

, and the result be false or false or true equal true?

Read about [*operator precedence*](http://en.cppreference.com/w/cpp/language/operator_precedence) in [any good beginners book](https://stackoverflow.com/a/388282/440558). And remember that `< — Some programmer dude, Jan 21 '18 at 15:49
See the [compiler warnings](http://coliru.stacked-crooked.com/a/32322c6b4b38148e) you'll get with that code. — , Jan 21 '18 at 15:50
&& has an higher precedence than ||, so the value should be true. — Pablito Escobar, Jan 21 '18 at 15:54
Seems like you either didn't read or didn't understand the comments. `< — JJJ, Jan 21 '18 at 15:58
http://en.cppreference.com/w/cpp/language/operator_precedence — Jesper Juhl, Jan 21 '18 at 16:04

Daniel Langr · Answer 1 · 2018-01-21T16:35:24.227

As already mentioned in comments, in the first case, the expression is due to operator precedence evaluated as

(std::cout << a) || !b && !a || b;

Result of std::cout << a is a reference to the std::cout object itself, which is in C++03 convertible to bool through operator void*() inherited from std::basic_ios. In C++11, there is operator bool() instead, which allows so-called contextual conversion.

The rest is therefore just a boolean expression and its result is discarded.

Strange evaluation of boolean expression

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