== Auszug aus useo (u...@start.bg)'s Artikel > == Auszug aus Steven Schveighoffer (schvei...@yahoo.com)'s Artikel > > On Thu, 10 Feb 2011 09:09:03 -0500, useo <u...@start.bg> wrote: > > > I implemented all I wanted and it works perfectly ;). > > > > > > But I'm using the "if (is(T == delegate) || is(T == function))"- > > > statement in another class/template. > > > > > > class Example(T) if (is(T == delegate) || is(T == function)) { > > > ... > > > } > > > > > > Now, when I declare Example!(void function()) myVar; I always get: > > > > > > Error: template instance Example!(void function()) does not match > > > template declaration Example(T) if (is(T == delegate) || is(T == > > > function)) > > > Error: Example!(void function()) is used as a type > > > > > > When I declare myVar as Example!(function), I get some other > errors. > > > > > > What's wrong with my code? > > Please post a full example that creates the error. From that error > > message, it appears that your code is correct, but I'd have to play > with a > > real example to be sure. > > -Steve > I created a complete, new file with the following code: > module example; > void main(string[] args) { > Example!(void function()) myVar; > } > class Example(T) if (is(T == delegate) || is(T == function)) { > } > And what I get is: > example.d(4): Error: template instance Example!(void function()) does > not match template declaration Example(T) if (is(T == delegate) || is > (T == function)) > example.d(4): Error: Example!(void function()) is used as a type > I'm using the current stable version 2.051.
I noticed, that this error only occurs when I use function as template-type. When I use Example!(void delegate()) myVar; it compiles without any error. With function - Example!(void function()) myVar; I get the error(s) above.