lazy variables cannot be lvalues - why?

[Jonathan M Davis]

On Monday, November 01, 2010 08:57:09 Adam Cigánek wrote:
> Hello,
> 
> why is the following code illegal?
> 
> 
>   import std.stdio;
> 
>   void delegate() fun;
> 
>   void capture(lazy void f) {
>     fun = &f;
>   }
> 
>   void main() {
>     capture(writeln("hello"));
>     fun();
>   }
> 
> 
> It says "Error: lazy variables cannot be lvalues", pointing to the
> "fun = &f" line.
> 
> It can be worked around by rewriting it like this:
> 
> void capture(lazy void f) {
>   fun = delegate void() { f(); };
> }
> 
> So it's not big deal, just a minor inconvenience. But still, why is it
> illegal? According to the docs
> (http://www.digitalmars.com/d/2.0/lazy-evaluation.html), lazy
> expressions are implicitly converted to delegates, so it seems to me
> that it should work.
> 
> adam.

1. I 'm stunned that the compiler doesn't complain about you declaring f as 
void. It strikes me as a bug with lazy. You can't declare variables of type 
void. It makes no sense.

2. A lazy parameter is for all intents an purposes the exact same thing as a 
non-lazy parameter except that it's not actually calculated until the function 
is called. The fact that a delegate to make it lazy is used is an implementation 
detail. Use typeof on it, and you'll notice that its type is the same as if it 
weren't lazy, not a delegate.

- Jonathan M Davis