Bug 6207 – Mixin template evaluated to string can convert to string mixin expression implicitly

If mixin template instantiation makes manifest constant string and it appears in expression context, the compiler converts it implicitly to string mixin expression. This enhancement feature does not conflict with any existing features. Because normal instantiating with mixin template is illegal. Sample code: ---- mixin template expand(string code) { static if (code.length >= 2 && code[0..2] == "$x") { enum expand = `x` ~ code[2..$]; pragma(msg, expand); } else enum expand = code; } void main() { int x = 1; int y = expand!q{$x+2}; // Rhs is implicitly converted to mixin(expand!(q{$x+2})) assert(y == 3); } ----

We can improve the string lambda features in std.algorithm! mixin template map(string pred) { enum map = `map!((a){ return `~pred~`; })`; } template map(alias pred) { auto map(E)(E[] r) { E[] result; foreach (e; r) result ~= pred(e); return result; } } void main() { int b = 10; auto r = map!q{ a * b }([1,2,3]); // --> mixin(`map!((a){ return ` ~ q{ a * b } ~ `; })`)([1,2,3]) // --> map!((a){ return a * b ; })([1,2,3]); assert(r == [10,20,30]); }

See the pull request for more discussion.

What if we add some special syntax for mixin template instantiation. I mean something like this: int y = expand#"$x+2"; // instead of mixin(expand!"$x+2") What do you think?

(In reply to Walter Bright from comment #3) > See the pull request for more discussion. I believe such a change today would require a DIP, so I'm going to close this.