Is there any syntax for something like that:

class Resource(T) if( is(T: FileResource) ){
}


I tried it as above, but it is not accepted. Maybe I am following a wrong syntax.

I tried

class Resource(T: FileResource){
}

But it is not accepted as well.

On Monday, May 25, 2015 03:19:29 tcak via Digitalmars-d-learn wrote:
> Is there any syntax for something like that:
>
> class Resource(T) if( is(T: FileResource) ){
> }
>
>
> I tried it as above, but it is not accepted. Maybe I am following a wrong syntax.
>
> I tried
>
> class Resource(T: FileResource){
> }
>
> But it is not accepted as well.

What about it isn't accepted? This code compiles just fine


class FileResource
{
}

class SubFileResource : FileResource
{
}

class Resource(T)
    if(is(T : FileResource))
{
}

void main()
{
    Resource!SubFileResource foo;
}

- Jonathan M Davis

On Monday, 25 May 2015 at 03:35:22 UTC, Jonathan M Davis wrote:
> On Monday, May 25, 2015 03:19:29 tcak via Digitalmars-d-learn wrote:
>> Is there any syntax for something like that:
>>
>> class Resource(T) if( is(T: FileResource) ){
>> }
>>
>>
>> I tried it as above, but it is not accepted. Maybe I am following
>> a wrong syntax.
>>
>> I tried
>>
>> class Resource(T: FileResource){
>> }
>>
>> But it is not accepted as well.
>
> What about it isn't accepted? This code compiles just fine
>
>
> class FileResource
> {
> }
>
> class SubFileResource : FileResource
> {
> }
>
> class Resource(T)
>     if(is(T : FileResource))
> {
> }
>
> void main()
> {
>     Resource!SubFileResource foo;
> }
>
> - Jonathan M Davis

Well, if I do not check the line number of error, this happens. It was giving error on the line of creating a new instance.

Line 243: auto fileResourceList = new shared FileResourceList( 2 );

main.d(243): Error: class main.FileResourceList(T) if (is(T : FileResource)) is used as a type

On Monday, May 25, 2015 03:42:22 tcak via Digitalmars-d-learn wrote:
> On Monday, 25 May 2015 at 03:35:22 UTC, Jonathan M Davis wrote:
> > On Monday, May 25, 2015 03:19:29 tcak via Digitalmars-d-learn wrote:
> >> Is there any syntax for something like that:
> >>
> >> class Resource(T) if( is(T: FileResource) ){
> >> }
> >>
> >>
> >> I tried it as above, but it is not accepted. Maybe I am
> >> following
> >> a wrong syntax.
> >>
> >> I tried
> >>
> >> class Resource(T: FileResource){
> >> }
> >>
> >> But it is not accepted as well.
> >
> > What about it isn't accepted? This code compiles just fine
> >
> >
> > class FileResource
> > {
> > }
> >
> > class SubFileResource : FileResource
> > {
> > }
> >
> > class Resource(T)
> >     if(is(T : FileResource))
> > {
> > }
> >
> > void main()
> > {
> >     Resource!SubFileResource foo;
> > }
> >
> > - Jonathan M Davis
>
> Well, if I do not check the line number of error, this happens. It was giving error on the line of creating a new instance.
>
> Line 243: auto fileResourceList = new shared FileResourceList( 2
> );
>
> main.d(243): Error: class main.FileResourceList(T) if (is(T :
> FileResource)) is used as a type

shared(FileResourceList) is not implicitly convertible to FileResource. It may be implicitly convertible to shared(FileResource), but not FileResource). Code that uses shared usually has to be written specifically for shared. Also, any time that you construct a templated type, you have to provide the template argument. IFTI (implicit function template instantiation) doesn't work with constructors. To make that work, you need a wrapper function - e.g.

auto resource(T)(T t)
{
    return new Resource!T(t);
}

So, I suspect that you're either running into problems, because you're using shared, and Resource's template constraint requires non-shared, and/or because you aren't explicitly instantiating Resource with a template argument.

- Jonathan M Davis

On Monday, 25 May 2015 at 04:07:06 UTC, Jonathan M Davis wrote:
> On Monday, May 25, 2015 03:42:22 tcak via Digitalmars-d-learn wrote:
>>
>> Well, if I do not check the line number of error, this happens.
>> It was giving error on the line of creating a new instance.
>>
>> Line 243: auto fileResourceList = new shared FileResourceList( 2
>> );
>>
>> main.d(243): Error: class main.FileResourceList(T) if (is(T :
>> FileResource)) is used as a type
>
> shared(FileResourceList) is not implicitly convertible to FileResource. It
> may be implicitly convertible to shared(FileResource), but not
> FileResource). Code that uses shared usually has to be written specifically
> for shared. Also, any time that you construct a templated type, you have to
> provide the template argument. IFTI (implicit function template
> instantiation) doesn't work with constructors. To make that work, you need a
> wrapper function - e.g.
>
> auto resource(T)(T t)
> {
>     return new Resource!T(t);
> }
>
> So, I suspect that you're either running into problems, because you're using
> shared, and Resource's template constraint requires non-shared, and/or
> because you aren't explicitly instantiating Resource with a template
> argument.
>
> - Jonathan M Davis

Nah. Its all about giving the T value on object creation.

auto fileResourceList = new shared FileResourceList!FileResource( 2 );

Then it works. The error message is not indicating directly this, though logically it is still correct.

On 05/24/2015 09:14 PM, tcak wrote:

>>> Line 243: auto fileResourceList = new shared FileResourceList( 2
>>> );
>>>
>>> main.d(243): Error: class main.FileResourceList(T) if (is(T :
>>> FileResource)) is used as a type

struct and class templates do not have automatic type deduction; function templates do. The solution is to provide a convenience function template alongside the struct or class.

> auto fileResourceList = new shared FileResourceList!FileResource( 2 );
>
> Then it works. The error message is not indicating directly this, though
> logically it is still correct.

'if (is(T : FileResource))' part means "if T can implicitly be converted to FileResource". You still provide T so that it gets checked.

Ali

On Monday, 25 May 2015 at 04:15:00 UTC, tcak wrote:

>>> main.d(243): Error: class main.FileResourceList(T) if (is(T :
>>> FileResource)) is used as a type

> The error message is not indicating directly this, though logically it is still correct.

Compiler means "template is used as a type" which is an error since templates are not types. I guess adding the word "template" to that message would make it more clear.