Hi everyone,
I failed to find the reason why Swift does not allows a non-escaping parameter
to be assigned to a local variable. Here is a minimal example:
func f(_ closure: () -> Int) {
let a = closure
}
I do understand that assigning a non-escaping closure to a variable whose
lifetime exceeds that of the function would (by definition) violate the
non-escaping property. For instance, doing that is understandably illegal:
var global = { 0 }
func f(_ closure: () -> Int) {
global = closure
}
But in my first example, since `a` is stack allocated, there’s no risk that
`closure` escapes the scope of `f`.
Is there some use case I’m missing, where such assignment could be problematic?
Or is this a limitation of the compiler, which wouldn't go all the way to check
whether the lifetime of the assignee is compatible with that of the
non-escaping parameter may exceed that of the variable it is assigned to?
Thank you very much for your time and your answer.
Best,
Dimitri Racordon
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