I noticed a difference between how static and lazy variables evaluate closures
and thought that it was a bug:
https://bugs.swift.org/browse/SR-1178
but apparently it’s not.
The difference can be illustrated in a small example. The following code will
evaluate the closure for the static variable, even when *setting* the variable,
but won’t evaluate the closure for the lazy variable. Thus, it prints “static”,
but not “lazy”.
class Foo {
static var bar: String = {
print("static")
return "Default"
}()
lazy var baz: String = {
print("lazy")
return "Lazy"
}()
}
Foo.bar = "Set"
let foo = Foo()
foo.baz = “Set"
I would have thought that neither case should evaluate the closure when setting
the variable, since the result from the closure is never used in that case. I
don’t feel that strongly about if the closure is evaluated or not. But I would
like both types (static and lazy) to behave the same.
- David
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