• mogu (5/5) Jul 19 2016 ```
• Adam D. Ruppe (8/9) Jul 19 2016 Because S isn't a type... think of a template as being like a
• mogu (6/15) Jul 19 2016 So it's a higher kinded type aka type class in Haskell. But D's
• Lodovico Giaretta (4/23) Jul 20 2016 Note that void is a type, while S is not. So you can do:
• mogu (3/6) Jul 20 2016 Thanks very much. I should have noticed this before. T.T

mogu <mogucpp 163.com> writes:

```
struct S(T) {}

static assert(is (typeof(S) == void));
```

Why S's type isn't something like `S: (T) -> S`?

Adam D. Ruppe <destructionator gmail.com> writes:

On Wednesday, 20 July 2016 at 01:14:05 UTC, mogu wrote:
 Why S's type isn't something like `S: (T) -> S`?

Because S isn't a type... think of a template as being like a 
function that returns a type.

int foo(int) { return 0; }

There, you wouldn't expect typeof(foo) to be int, no, typeof(foo) 
is a function that returns an int.

The template is the same thing - it isn't a type, it is a 
template that returns a type.

mogu <mogucpp 163.com> writes:

On Wednesday, 20 July 2016 at 01:50:37 UTC, Adam D. Ruppe wrote:
 On Wednesday, 20 July 2016 at 01:14:05 UTC, mogu wrote:
 Why S's type isn't something like `S: (T) -> S`?

 Because S isn't a type... think of a template as being like a 
 function that returns a type.

 int foo(int) { return 0; }

 There, you wouldn't expect typeof(foo) to be int, no, 
 typeof(foo) is a function that returns an int.

 The template is the same thing - it isn't a type, it is a 
 template that returns a type.

So it's a higher kinded type aka type class in Haskell. But D's 
reflection cannot get enough information about template's kind. 
Only a `void` given. It may be inconvenient in distinction 
between an alias of template and void. The only solution AFAIK is 
by string of the type property .stringof.

Lodovico Giaretta <lodovico giaretart.net> writes:

On Wednesday, 20 July 2016 at 05:54:41 UTC, mogu wrote:
 On Wednesday, 20 July 2016 at 01:50:37 UTC, Adam D. Ruppe wrote:
 On Wednesday, 20 July 2016 at 01:14:05 UTC, mogu wrote:
 Why S's type isn't something like `S: (T) -> S`?

 Because S isn't a type... think of a template as being like a 
 function that returns a type.

 int foo(int) { return 0; }

 There, you wouldn't expect typeof(foo) to be int, no, 
 typeof(foo) is a function that returns an int.

 The template is the same thing - it isn't a type, it is a 
 template that returns a type.

 So it's a higher kinded type aka type class in Haskell. But D's 
 reflection cannot get enough information about template's kind. 
 Only a `void` given. It may be inconvenient in distinction 
 between an alias of template and void. The only solution AFAIK 
 is by string of the type property .stringof.

Note that void is a type, while S is not. So you can do:

assert(is(void)) // is(type) returns true
assert(!is(S))   // is(template) returns false;

mogu <mogucpp 163.com> writes:

On Wednesday, 20 July 2016 at 08:01:01 UTC, Lodovico Giaretta 
wrote:
 Note that void is a type, while S is not. So you can do:

 assert(is(void)) // is(type) returns true
 assert(!is(S))   // is(template) returns false;

Thanks very much. I should have noticed this before. T.T