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digitalmars.D.learn - alias and UDAs

reply Andre Pany <andre s-e-a-p.de> writes:
Hi,

in this example, both asserts fails. Is my assumption right, that 
UDA on alias have no effect? If yes, I would like to see a 
compiler warning.

But anyway, I do not understand why the second assertion fails. 
Are UDAs on arrays not allowed?

import std.traits: hasUDA;

enum Flattened;

struct Foo
{
	int bar;
}

 Flattened alias FooList = Foo[];

struct Baz
{
	FooList fooList1;
	 Flattened FooList[] fooList2;
}

void main()
{	
	Baz baz;
	static assert(hasUDA!(baz.fooList1, "Flattened")); // => false
	static assert(hasUDA!(baz.fooList2, "Flattened")); // => false
}

Kind regards
André
May 11 2017
next sibling parent reply Stanislav Blinov <stanislav.blinov gmail.com> writes:
On Thursday, 11 May 2017 at 10:39:03 UTC, Andre Pany wrote:

 Hi,

 in this example, both asserts fails. Is my assumption right, 
 that UDA on alias have no effect? If yes, I would like to see a 
 compiler warning.

 But anyway, I do not understand why the second assertion fails. 
 Are UDAs on arrays not allowed?

 import std.traits: hasUDA;

 enum Flattened;

 struct Foo
 {
 	int bar;
 }

  Flattened alias FooList = Foo[];

 struct Baz
 {
 	FooList fooList1;
 	 Flattened FooList[] fooList2;
 }

 void main()
 {	
 	Baz baz;
 	static assert(hasUDA!(baz.fooList1, "Flattened")); // => false
 	static assert(hasUDA!(baz.fooList2, "Flattened")); // => false
 }

 Kind regards
 André


It should've been

alias FooList =  Flattened Foo[];

which will generate a compile-time error (UDAs not allowed for 
alias declarations).

And then:

static assert(hasUDA!(baz.fooList2, Flattened));

No quotes, since Flattened is an enum, not a string
May 11 2017
parent Andre Pany <andre s-e-a-p.de> writes:
On Thursday, 11 May 2017 at 10:57:22 UTC, Stanislav Blinov wrote:

 On Thursday, 11 May 2017 at 10:39:03 UTC, Andre Pany wrote:

 [...]


 It should've been

 alias FooList =  Flattened Foo[];

 which will generate a compile-time error (UDAs not allowed for 
 alias declarations).

 And then:

 static assert(hasUDA!(baz.fooList2, Flattened));

 No quotes, since Flattened is an enum, not a string


Thanks for the explanation. I think I will create a bug report 
for this statement:
 Flattened alias FooList = Foo[];

The UDA has no effect as far as I understand.

Kind regards
André
May 11 2017
prev sibling parent ag0aep6g <anonymous example.com> writes:
On 05/11/2017 12:39 PM, Andre Pany wrote:

 in this example, both asserts fails. Is my assumption right, that UDA on
 alias have no effect? If yes, I would like to see a compiler warning.

 But anyway, I do not understand why the second assertion fails. Are UDAs
 on arrays not allowed?

 import std.traits: hasUDA;

 enum Flattened;

 struct Foo
 {
     int bar;
 }

  Flattened alias FooList = Foo[];

 struct Baz
 {
     FooList fooList1;
      Flattened FooList[] fooList2;
 }

 void main()
 {
     Baz baz;
     static assert(hasUDA!(baz.fooList1, "Flattened")); // => false
     static assert(hasUDA!(baz.fooList2, "Flattened")); // => false
 }


1) You have to test against `Flattened`, not `"Flattened"`. A string is 
a valid UDA, but you're not using the string on the declarations.

When you fix this, the second assert passes.

2) `Baz.fooList1` doesn't have any attributes. Attributes apply to 
declarations. If it's valid, the attribute on `FooList` applies only to 
`FooList`. It doesn't transfer to `Baz.fooList1`.

If anything, you could assert that `hasUDA!(FooList, Flattened)` holds. 
Maybe you could, if it compiled.

3) Why does `hasUDA!(FooList, Flattened)` fail to compile?

The error message reads: "template instance hasUDA!(Foo[], Flattened) 
does not match template declaration hasUDA(alias symbol, alias attribute)".

We see that `FooList` has been replaced by `Foo[]`. It's clear then why 
the instantiation fails: `Foo[]` isn't a symbol.

Unfortunately, the spec is a bit muddy on this topic. On the one hand it 
says that "AliasDeclarations create a symbol", but it also says that 
"Aliased types are semantically identical to the types they are aliased 
to" [1].

In practice, the compiler doesn't seem to create a symbol. The alias 
identifier is simply replaced with the aliased thing, and you can't use 
the alias identifier as a symbol.

That means, you might be able to add an attribute to `FooList`, but you 
can't get back to it, because whenever you use `FooList` it's always 
replaced by `Foo[]`. And `Foo[]` doesn't have the attribute, of course.

I agree that it would probably make sense to disallow putting attributes 
on aliases. You can also mark aliases `const`, `static`, `pure`, etc. 
And they all have no effect.


[1] http://dlang.org/spec/declaration.html#AliasDeclaration
May 11 2017