But... I don't want my delegates to be lazy - breeding advice

[Sean Kelly]

Walter Bright wrote:
> Ok, I've thought about it a bit:
> 
>     void foo(int x);    // same as before 0.165
>     void foo(int delegate() x)    // same as before 0.165
> 
> and now:
> 
>     void foo(lazy int x);
> 
> In other words, 'lazy' is now a parameter storage class. This means that:
> 
>     void foo(int x);
>     void foo(lazy int x);
> 
> cannot be distinguished based on overloading, but:
> 
>     void foo(lazy int x);
>     void foo(int delegate() x);
> 
> can be. The implicit conversion of a value to a delegate returning that 
> value would be removed. The conversion happens always (not implicitly) 
> if the parameter storage class is 'lazy'.

Fantastic!  This will even allow lazy values to be pre-constructed, 
which is a nice perk.


Sean