> On 4 Oct 2017, at 14:12, Mike Kluev <mike.kl...@gmail.com> wrote:
>
> On 4 October 2017 at 13:41, Alex Blewitt <alb...@apple.com
> <mailto:alb...@apple.com>> wrote:
>
> The difference between the & and && operators isn't to do with the implicit
> conversions; it's to do with whether both sides of the expression are
> evaluated or not.
>
> false && system('rm -rf')
>
> You really don't want to do that if both sides are executed ...
>
> actually, thanks for bringing this up as it leads up to a question:
>
> how in swift do i define my &&& operator (that i may need for whatever
> reason, e.g. logging) that will short-circuit
> the calculation of right hand side if the left hand side is false?
>
> infix operator &&&: LogicalConjunctionPrecedence
>
> func &&&(left: Bool, right: Bool) -> Bool {
> return left && right
> }
>
> as written it doesn't short-circuit. is it possible at all in swift?
When you call the function, the arguments will be evaluated prior to the
function body - so this won't work (as you correctly noted).
However, you can wrap the second argument in an @autoclosure, which means it
replaces the body of the expression with a function that evaluates the
expression automatically:
infix operator &&&: LogicalConjunctionPrecedence
func &&&(left: Bool, right: @autoclosure () -> Bool) -> Bool {
return left && right()
}
func no() -> Bool {
print("no")
return false
}
func yes() -> Bool {
print("yes")
return true
}
print(no() &&& yes())
no
false
Alex
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