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digitalmars.D.learn - check instance of nested variadic template

reply Gianni Pisetta <pisetta.gianni alice.it> writes:
Hi all,
I am having issues finding a solution for this, i want to check 
if an alias is an istance of a variadic template nested in 
another variadic template.
Here is an example:

template A(As...) {
     template B(Bs...) {
     }
}

alias BI = A!(1,2).B!(3,4,5);

Now i want to test if BI is an istance of A.B, but i can't do 
something like this:

static if ( is( BI == A!As.B!Bs, As..., Bs... ) ) {
     // Do Something
}

there is some sort of workaround?

Thanks,
Gianni
Nov 04 2016
parent reply Basile B. <b2.temp gmx.com> writes:
On Friday, 4 November 2016 at 15:50:36 UTC, Gianni Pisetta wrote:

 Hi all,
 I am having issues finding a solution for this, i want to check 
 if an alias is an istance of a variadic template nested in 
 another variadic template.
 [...]
 there is some sort of workaround?

 Thanks,
 Gianni


Hello, I'm not sure that's exactly what you want but check this:

============================
template A(As...) {
     template B(Bs...) {
     }
}

alias BI = A!(1,2).B!(3,4,5);

import std.traits;

template NestedTemplateArgsOf(alias T)
{
     alias NestedTemplateArgsOf = TemplateArgsOf!(__traits(parent, 
T));
}

alias Bs = TemplateArgsOf!BI;
alias As = NestedTemplateArgsOf!BI;


static if (is(typeof(BI) == typeof(A!As.B!Bs)))
{
     pragma(msg, "for the win");
}
============================

The missing key was NestedTemplateArgsOf. With it you can solve 
the problem.
Nov 04 2016
next sibling parent Basile B. <b2.temp gmx.com> writes:
On Friday, 4 November 2016 at 17:37:10 UTC, Basile B. wrote:

 On Friday, 4 November 2016 at 15:50:36 UTC, Gianni Pisetta 
 wrote:

 Hi all,
 I am having issues finding a solution for this, i want to 
 check if an alias is an istance of a variadic template nested 
 in another variadic template.
 [...]
 there is some sort of workaround?

 Thanks,
 Gianni


 Hello, I'm not sure that's exactly what you want but check this:

 ============================
 [...]
 ============================


A better generic solution:

https://github.com/BBasile/iz/blob/1a6452b376a3a1977c8287f9ef294d2dde76952d/import/iz/types.d#L398
Nov 05 2016
prev sibling parent reply Gianni Pisetta <pisetta.gianni alice.it> writes:
On Friday, 4 November 2016 at 17:37:10 UTC, Basile B. wrote:

 Hello, I'm not sure that's exactly what you want but check this:

 ============================
 template A(As...) {
     template B(Bs...) {
     }
 }

 alias BI = A!(1,2).B!(3,4,5);

 import std.traits;

 template NestedTemplateArgsOf(alias T)
 {
     alias NestedTemplateArgsOf = 
 TemplateArgsOf!(__traits(parent, T));
 }

 alias Bs = TemplateArgsOf!BI;
 alias As = NestedTemplateArgsOf!BI;


 static if (is(typeof(BI) == typeof(A!As.B!Bs)))
 {
     pragma(msg, "for the win");
 }
 ============================

 The missing key was NestedTemplateArgsOf. With it you can solve 
 the problem.


Well, kind of. But i think i can make it with what i got from 
your example, so thanks.
Another thing that I encountered and while messing with your code 
is that __traits( parent, T ) does not work as expected when you 
have structs instead of template. I think because something like

struct A(As...) {}

is downgraded to

template A(As...) {
   struct A {}
}

when i use __traits( parent, A!(1,2) ) i get in return A!(1,2), 
looping around the same symbol.
When you compile this

struct A(As...) {}

import std.conv;

pragma( msg, "The parent symbol is the same? " ~ to!string( 
__traits( isSame, A!(1,2), __traits( parent, A!(1,2) ) ) ) );

void main() {}

you get a really interesting result:

The parent symbol is the same? true

Gianni Pisetta
Nov 05 2016
parent reply Gianni Pisetta <pisetta.gianni alice.it> writes:
When i have time i will test it with ldc and see if the result is 
the same, then it will probably be a front-end bug and i will 
report it as an issue.
Nov 05 2016
next sibling parent reply Lodovico Giaretta <lodovico giaretart.net> writes:
On Saturday, 5 November 2016 at 13:34:51 UTC, Gianni Pisetta 
wrote:

 When i have time i will test it with ldc and see if the result 
 is the same, then it will probably be a front-end bug and i 
 will report it as an issue.


I think it has already been reported.

https://issues.dlang.org/show_bug.cgi?id=13372
Nov 05 2016
parent Basile B. <b2.temp gmx.com> writes:
On Saturday, 5 November 2016 at 13:43:34 UTC, Lodovico Giaretta 
wrote:

 On Saturday, 5 November 2016 at 13:34:51 UTC, Gianni Pisetta 
 wrote:

 When i have time i will test it with ldc and see if the result 
 is the same, then it will probably be a front-end bug and i 
 will report it as an issue.


 I think it has already been reported.

 https://issues.dlang.org/show_bug.cgi?id=13372


Yes. Though the report is more generalized now. It applies to any 
eponymous type template.
Nov 05 2016
prev sibling parent reply Basile B. <b2.temp gmx.com> writes:
On Saturday, 5 November 2016 at 13:34:51 UTC, Gianni Pisetta 
wrote:

 When i have time i will test it with ldc and see if the result 
 is the same, then it will probably be a front-end bug and i 
 will report it as an issue.


Indeed, I've been fighting against that since a few minutes. We 
cant select the Base of an eponymous template.


template isEponymousTemplate(T)
{
     static if (is(T == class) || is(T == interface) || is(T == 
struct) || is(T == union))
     {
         enum p = __traits(parent, T).stringof == T.stringof;
         enum isEponymousTemplate = p && isTemplateInstance!T;
     }
     else
         enum isEponymousTemplate = false;
}

template TemplateBase(T : Base!Args, alias Base, Args...)
if (is(T == class) || is(T == interface) || is(T == struct) || 
is(T == union))
{
     alias TemplateBase = Base;
}
Nov 05 2016
parent Basile B. <b2.temp gmx.com> writes:
On Saturday, 5 November 2016 at 14:37:53 UTC, Basile B. wrote:

 On Saturday, 5 November 2016 at 13:34:51 UTC, Gianni Pisetta 
 wrote:

 [...]


 Indeed, I've been fighting against that since a few minutes. We 
 cant select the Base of an eponymous template.


 template isEponymousTemplate(T)
 {
     static if (is(T == class) || is(T == interface) || is(T == 
 struct) || is(T == union))
     {
         enum p = __traits(parent, T).stringof == T.stringof;
         enum isEponymousTemplate = p && isTemplateInstance!T;
     }
     else
         enum isEponymousTemplate = false;
 }

 template TemplateBase(T : Base!Args, alias Base, Args...)
 if (is(T == class) || is(T == interface) || is(T == struct) || 
 is(T == union))
 {
     alias TemplateBase = Base;
 }


To be more accurate the template base can be selected for

     template A(T) {class A{}}

but not for

     class A(T){}
Nov 05 2016