digitalmars.D.learn - "This for <member> needs to be type <class>, not type <otherclass>"
- Simen Kjaeraas (18/18) May 02 2008 While playing around with templates and alias parameters, I got the die ...
- Jarrett Billingsley (10/25) May 02 2008 If you make this a template:
- Simen Kjaeraas (4/38) May 02 2008 Now why didn't I think of that... I have to say the mixin syntax is ugli...
While playing around with templates and alias parameters, I got the die of trying this: struct foo(alias T) { typeof(T) opAddAssign(typeof(T) rhs) { return T += rhs; } } class bar { private: int _baz; public: foo!(_baz) baz; } When compiling this, I get the message "Error: this for_baz needs to be type bar not type foo!(_baz)*" Is there a workaround to this, or even a plan to make such things possible in the future? -- Simen
May 02 2008
"Simen Kjaeraas" <simen.kjaras gmail.com> wrote in message news:fvfg2g$1nnp$1 digitalmars.com...While playing around with templates and alias parameters, I got the die of trying this: struct foo(alias T) { typeof(T) opAddAssign(typeof(T) rhs) { return T += rhs; } }If you make this a template: template foo(alias T) // rest is the sameclass bar { private: int _baz; public: foo!(_baz) baz;And now change this to a mixin: mixin foo!(_baz) baz; It works. Of course, this has the effect of making foo (or rather an instance of foo) no longer a type. But maybe you don't need it to be.
May 02 2008
Jarrett Billingsley Wrote:"Simen Kjaeraas" <simen.kjaras gmail.com> wrote in message news:fvfg2g$1nnp$1 digitalmars.com...Now why didn't I think of that... I have to say the mixin syntax is uglier, but it gets the job done. There are times when I wish I could declare a function, template or whatever to be a mixin type, so any instantiation of it would be a mixin. If that were to be implemented, I wouldn't have the ugly 'mixin' keyword cluttering my code. It would make code cleaner, but mayhaps more hard to read. -- SimenWhile playing around with templates and alias parameters, I got the die of trying this: struct foo(alias T) { typeof(T) opAddAssign(typeof(T) rhs) { return T += rhs; } }If you make this a template: template foo(alias T) // rest is the sameclass bar { private: int _baz; public: foo!(_baz) baz;And now change this to a mixin: mixin foo!(_baz) baz; It works. Of course, this has the effect of making foo (or rather an instance of foo) no longer a type. But maybe you don't need it to be.
May 02 2008