Mixin Expressions, can't evalutate string variable

Tomek Sowiński just at ask.me
Thu Aug 5 12:34:44 PDT 2010

Andrej Mitrovic napisał:

> Hmm.. ok. Adding immutable helps. I'm still a bit confused tho,
>  because this will not compile:
> string input2 = "int y;";
> mixin(input2);

input2 is mutable, so theoretically there's no telling what value it holds.

> But this will compile:
> immutable string input2 = "int y;";
> mixin(input2);

And good.

> And this too will compile:
> string returnString(string input)
> {
>     return input;
> }
> mixin(returnString("int y;"));

Because the function is *nice*, compiler can evaluate the call at compile-time (see my other post). Somewhere on the D page (Lang. Ref. > 
Functions ?) there's a largish explanation what's *nice*.


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