Delegate parameter name shadows type name

[Jacob Carlborg via Digitalmars-d-learn]

On 2017-01-09 20:18, Ali Çehreli 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.

I know this has come up before, and reported as a bug, at least once. 
Might have been me :). What's confusing is that using a type that has a 
keyword will make the parameter unnamed of the specified type, just as a 
regular function:

auto a = (int) => 3; // works, a lambda taking an int, no parameter name
auto b = (Foo) => 3; // error, cannot infer type of template lambda
alias b = (Foo) => 3; // works, since this is an alias, Foo is the 
parameter name of an unknown type

-- 
/Jacob Carlborg