Delegate parameter name shadows type name
Jonathan M Davis via Digitalmars-d-learn
digitalmars-d-learn at puremagic.com
Mon Jan 9 11:38:42 PST 2017
On Monday, January 09, 2017 11:18:02 Ali Çehreli via Digitalmars-d-learn
wrote:
> This is something that surprised me in a friend's code.
>
> (A "friend", hmmm? No, really, it wasn't me! :) )
>
> // Some type of the API
> struct MyType {
> int i;
> }
>
> // Some function of the API that takes a delegate
> void call(void delegate(MyType) dlg) {
> dlg(MyType(42));
> }
>
> void main() {
> /* The programmer simply copied the delegate definition from
> * the function and used it as-is when passing a lambda: */
> call(delegate void(MyType) {
> /* WAT? Does the following really compile? After all,
> * MyType.i is NOT a static member! */
> if (MyType.i == 42) {
> // ...
> }
> });
> }
>
> I was surprised to see it compiled and worked but of course MyType at
> the lambda definition inside main() is not a type name, rather the
> parameter name. Surprising, but I think this is according to spec.
Well, stuff inside a function is quite free to shadow stuff from outside of
it. AFAIK, the only shadowing that's prevented is declarations in a function
shadowing other declarations in a function.
- Jonathan M Davis
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