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June 10, 2016 Why can't I assign a mixin to an alias? | ||||
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I have the following code: private string getVariableSignalWrappersName(VarType)() { return VarType.stringof ~ "SignalWrappers"; } void addVariableListener(VarType)(int variableIndex, void delegate(int, VarType)) { alias typeSignalWrappers = mixin(getVariableSignalWrappersName!VarType); } On compilation, the following error is issued: Error: basic type expected, not mixin Why should it be like that? I believe the compiler should not impose restrictions on what mixins can or cannot do :/ |
June 10, 2016 Re: Why can't I assign a mixin to an alias? | ||||
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Posted in reply to Dechcaudron | On Friday, 10 June 2016 at 22:38:29 UTC, Dechcaudron wrote: > I have the following code: > > private string getVariableSignalWrappersName(VarType)() > { > return VarType.stringof ~ "SignalWrappers"; > } > > void addVariableListener(VarType)(int variableIndex, void delegate(int, VarType)) > { > alias typeSignalWrappers = mixin(getVariableSignalWrappersName!VarType); > } > > On compilation, the following error is issued: > Error: basic type expected, not mixin > > Why should it be like that? I believe the compiler should not impose restrictions on what mixins can or cannot do :/ I'm no expert, but this looks like a grammar issue more than anything else. You can work around it by moving the alias declaration into the mixin. mixin("alias typeSignalWrappers = " ~ getVariableSignalWrappersName!VarType ~ ";"); |
June 11, 2016 Re: Why can't I assign a mixin to an alias? | ||||
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Posted in reply to Dechcaudron | On Friday, 10 June 2016 at 22:38:29 UTC, Dechcaudron wrote:
> Error: basic type expected, not mixin
>
> Why should it be like that? I believe the compiler should not impose restrictions on what mixins can or cannot do :/
This might be a gratuitous grammar restriction. There are a few of those surrounding alias "targets". A template that simply returns its parameter might work, though, such as std.meta.Alias (alias foo = Alias!(mixin(…));).
— David
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June 11, 2016 Re: Why can't I assign a mixin to an alias? | ||||
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Posted in reply to Dechcaudron | On Friday, 10 June 2016 at 22:38:29 UTC, Dechcaudron wrote:
> I have the following code:
>
> private string getVariableSignalWrappersName(VarType)()
> {
> return VarType.stringof ~ "SignalWrappers";
> }
>
> void addVariableListener(VarType)(int variableIndex, void delegate(int, VarType))
> {
> alias typeSignalWrappers = mixin(getVariableSignalWrappersName!VarType);
> }
>
> On compilation, the following error is issued:
> Error: basic type expected, not mixin
>
> Why should it be like that? I believe the compiler should not impose restrictions on what mixins can or cannot do :/
Mixins are statements. They cannot be a part of an expression.
The other two posts have demonstrated how to get around this.
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June 11, 2016 Re: Why can't I assign a mixin to an alias? | ||||
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Posted in reply to Alex Parrill | On Saturday, 11 June 2016 at 02:33:46 UTC, Alex Parrill wrote: > Mixins are statements. No, they're not. Well, yes they are [1], but there are also mixin expressions [2]. Not to be confused with the TemplateMixin[3], which is indeed always a statement. 1: http://dlang.org/spec/grammar.html#MixinExpression 2: http://dlang.org/spec/grammar.html#MixinStatement 3: http://dlang.org/spec/grammar.html#TemplateMixin |
June 11, 2016 Re: Why can't I assign a mixin to an alias? | ||||
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Posted in reply to Adam D. Ruppe | On Saturday, 11 June 2016 at 02:46:00 UTC, Adam D. Ruppe wrote:
> On Saturday, 11 June 2016 at 02:33:46 UTC, Alex Parrill wrote:
>> Mixins are statements.
>
> No, they're not. Well, yes they are [1], but there are also mixin expressions [2]. Not to be confused with the TemplateMixin[3], which is indeed always a statement.
>
> 1: http://dlang.org/spec/grammar.html#MixinExpression
> 2: http://dlang.org/spec/grammar.html#MixinStatement
> 3: http://dlang.org/spec/grammar.html#TemplateMixin
Huh, every time I've used mixins, I've always run into the issue in the OP, so I assumed they were statements. It definitely seems like a bug then.
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June 12, 2016 Re: Why can't I assign a mixin to an alias? | ||||
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Posted in reply to David Nadlinger | On Saturday, 11 June 2016 at 01:53:10 UTC, David Nadlinger wrote:
> This might be a gratuitous grammar restriction. There are a few of those surrounding alias "targets". A template that simply returns its parameter might work, though, such as std.meta.Alias (alias foo = Alias!(mixin(…));).
It seems to be that it would be a good idea to file this in as a suggested change/fix. Do I have support?
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