How does C++ handle &&? (Short-circuit evaluation)

Question

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• Is short-circuiting logical operators mandated? And evaluation order? 7 answers

When encountering a (bool1 && bool2), does c++ ever attempts to check bool2 if bool1 was found false or does it ignore it the way PHP does?

Sorry if it is too basic of a question, but I really could not find a mentioning of that neither in Schildt nor on the Internet.

Answer 1

That is correct (short-cicuit behavior). But beware: short-circuiting stops if the operator invoked is not the built-in operator, but a user-defined operator&& (same with operator||).

Reference in this SO

Answer 2

Short-circuit evaluation denotes the semantics of some Boolean operators in some programming languages in which the second argument is only executed or evaluated if the first argument does not suffice to determine the value of the expression: for instance, when the first argument of the AND function evaluates to false, the overall value must be false; and when the first argument of the OR function evaluates to true, the overall value must be true.

In C++, both && and || operators use short-circuit evaluation.

Answer 3

The && operator short circuits in C++ - if bool1 was false in your example, bool2 wouldn't be checked/executed.

Answer 4

C++ does use short-circuit logic, so if bool1 is false, it won't need to check bool2.

This is useful if bool2 is actually a function that returns bool, or to use a pointer:

if ( pointer && pointer->someMethod() )

without short-circuit logic, it would crash on dereferencing a NULL pointer, but with short-circuit logic, it works fine.

Answer 5

This is called short-circuit evaluation (Wikipedia)

The && operator is a short circuit operator in C++ and it will not evaluate bool2 if bool1 is false.

Answer 6

What you're referring to is short circuit evaluation. I thought that it may be compiler specific, however that article I linked to shows it as language specific, and C++ does adhere. If it is indeed compiler specific, I can't imagine a compiler that wouldn't follow it. The day to day compiler I use at the moment, VS 2008, does. Basically it will follow the operator precedence, and as soon as the condition result is guaranteed,