Re: Checking template parameter types of class
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
Re: Checking template parameter types of class
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.
Re: Checking template parameter types of class
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
Re: Checking template parameter types of class
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
Re: Checking template parameter types of class
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
Re: Checking template parameter types of class
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.
Checking template parameter types of class
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.
