• Eyyub (25/25) Jul 20 2012 Hello,
• =?UTF-8?B?QWxpIMOHZWhyZWxp?= (14/37) Jul 20 2012 The confusion is due to the two different meanings of the ':' operator.
• Artur Skawina (7/42) Jul 21 2012 No; it would be considered. The problem is likely related to "Function t...
• Eyyub (4/4) Jul 21 2012 Okay !

"Eyyub" <eyyub.pangearaion gmail.com> writes:

Hello,

I have a question about the semantic of parameter 
specialization(TemplateTypeParameterSpecialization)
In this following code, there are 2 forms of the same function 
'add' :
<code>

T add(T, U : T) (T a, U b)   //doesn't work
{
     return a + b;
}

T add(T, U) (T a, U b) if(is(U : T)) //works
{
     return a + b;
}


void main()
{
     assert(add(2, cast(short)2) == 4);
}
</code>
So, I infer that, in this case, 
TemplateTypeParameterSpecialization and TypeSpecialization(of 
IsExpression) aren't semantically equal ? What are differences 
between this 2 forms ?

Eyyub,

(I hope that I have an english a bit compréhensible)

=?UTF-8?B?QWxpIMOHZWhyZWxp?= <acehreli yahoo.com> writes:

On 07/20/2012 06:47 PM, Eyyub wrote:
 Hello,

 I have a question about the semantic of parameter
 specialization(TemplateTypeParameterSpecialization)
 In this following code, there are 2 forms of the same function 'add' :
 <code>

 T add(T, U : T) (T a, U b) //doesn't work
 {
 return a + b;
 }

 T add(T, U) (T a, U b) if(is(U : T)) //works
 {
 return a + b;
 }


 void main()
 {
 assert(add(2, cast(short)2) == 4);
 }
 </code>
 So, I infer that, in this case, TemplateTypeParameterSpecialization and
 TypeSpecialization(of IsExpression) aren't semantically equal ? What are
 differences between this 2 forms ?

 Eyyub,

 (I hope that I have an english a bit compréhensible)

The confusion is due to the two different meanings of the ':' operator.

1) In the template parameter list, ':' specializes U. "U : T" means: 
"this is a specialization where U is T".

   http://dlang.org/template.html

(Search for the "Specialization" section.)

When you use the template, T is int and U is short; so that 
specialization is not considered.

2) In an is expression, ':' means "implicitly convertible to".

   http://dlang.org/expression.html#IsExpression

So "is(U : T)" matches because "short is implicitly convertible to int."

Ali

-- 
D Programming Language Tutorial: http://ddili.org/ders/d.en/index.html

Artur Skawina <art.08.09 gmail.com> writes:

On 07/21/12 08:29, Ali Çehreli wrote:
 On 07/20/2012 06:47 PM, Eyyub wrote:
 I have a question about the semantic of parameter
 specialization(TemplateTypeParameterSpecialization)
 In this following code, there are 2 forms of the same function 'add' :
 <code>

 T add(T, U : T) (T a, U b) //doesn't work
 {
 return a + b;
 }

 T add(T, U) (T a, U b) if(is(U : T)) //works
 {
 return a + b;
 }


 void main()
 {
 assert(add(2, cast(short)2) == 4);
 }
 </code>
 So, I infer that, in this case, TemplateTypeParameterSpecialization and
 TypeSpecialization(of IsExpression) aren't semantically equal ? What are
 differences between this 2 forms ?

 
 The confusion is due to the two different meanings of the ':' operator.
 
 1) In the template parameter list, ':' specializes U. "U : T" means: "this is
a specialization where U is T".
 
   http://dlang.org/template.html
 
 (Search for the "Specialization" section.)
 
 When you use the template, T is int and U is short; so that specialization is
not considered.

No; it would be considered. The problem is likely related to "Function template
type parameters that are to be implicitly deduced may not have specializations".

Note that

   T add(T, U : int) (T a, U b) {...}

already works, and "add(T, U : T)" could probably be made to work too.

artur

"Eyyub" <eyyub.pangearaion gmail.com> writes:

Okay !

The spec' should be more explicit, I'm sure that I'm not the 
first who asked this question !

Thanks Guys,