If an expression evaluates multiple &&
operators, and 开发者_如何学Cdoes not evaluate any operators of lower precedence (eg. ||
, ?:
), will the expression evaluate to 0 as soon as one of the &&
s return 0, or will it finish evaluating the remaining &&
s?
For example,
q=0; w=1; e=1; r=1;
if(q && w && r && e) {}
Will this if()
evaluate to false as soon as q && w
evaluates to 0 (since the remaining &&
must all evaluate to 0 regardless of the right hand operators)?
Yes, evaluation will terminate early ("short circuit") as soon as a false expression is encountered. There is one exception to this: if the &&
operator has been overloaded for the relevant types of arguments. This is strongly discouraged and extremely rare, but could happen.
Yes, it does do short-circuit evaluation, for built-in types. Custom types where you've overloaded &&
or ||
won't have short-circuiting performed, which can cause subtle bugs.
Others have already stated the answer (i.e. "yes").
I will answer to add an example of one particularly idiomatic usage of this:
if ((p != NULL) && (*p == 42))
{
/* Do something */
}
This would have to be written in a much clumsier manner if short-circuiting didn't happen.
Note that you can also use this in a Perl-esque manner, e.g.:
someCondition && doSomething();
so doSomething()
only gets called if someCondition
is true
. But this only compiles if doSomething()
returns a type that can be converted to bool
, and it's not considered idiomatic C++. (Note also that this technique doesn't compile in C.)
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