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

Question

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.

Eolanda

Answer 1

Yes, the && operator in C++ uses short-circuit evaluation so that if bool1 evaluates to false it doesn't bother evaluating bool2.

"Short-circuit evaluation" is the fancy term that you want to Google and look for in indexes.

The same happens with the || operator, if bool1 evaluates to true then the whole expression will evaluate to true, without evaluating bool2.

In case you want to evaluate all expressions anyway you can use the & and | operators.

Bushelman

What about short circuit evaluation with or || – Beata

The || operator uses short-circuit evaluation too. The difference is that given an expression bool1 || bool2, when bool1 evaluates to false then bool2 will be evaluated too. – Intendance

> "in case you want to evaluate all expressions anyway you can use the & and | operators." Replacing logical && and ||, with bitwise operators & and | is not safe unless you're strictly using bool expressions. Since the logical operators can act on other types, you are not guaranteed to get reliable results. For example: 4 && 2 evaluates to true, while 4 & 2 evaluates to 0. 4 || 2 evaluates to true, while 4 | 2 evaluates to 6. If you want to force both expressions to be evaluated and do a logical comparison, explicitly evaluate, then compare. – Rustic

It's probably clearer to follow Money47's advice too. If I saw if (foo(x) | bar(y), my first thought wouldn't be "oh, foo and bar return bools but the author wanted to make sure they were both called. It would be "okay, foo and bar seem to return a word of bit-flags and we're testing to see if any of the flags are set after calling both of them; this looks a little hairy". – Merras

Answer 2

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.

Tightfisted

Can you be certain pointer will be evaluated before pointer->someMethod()? – Deckert

Yes, as the operands of && are evaluated from left to right. – Marconigraph

Answer 3

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

Dysgenic

But short-circuiting is by no means limited to native types. Overloading a conversion to bool (or safe bool) instead of operator && makes this work just fine. – Gennie

Interesting. I didn't know that. Do you know the section/text from the standard that defines this? – Astatic

That is correct, reference in this SO. I'll try to rephrase my answer. – Dysgenic

Answer 4

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

Fleenor

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.

Gummous

Answer 6

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.

Bitternut

Answer 7

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,

Morentz