http://d.puremagic.com/issues/show_bug.cgi?id=9949

           Summary: template initialization when alias cannot be read at
                    compile time
           Product: D
           Version: D2
          Platform: All
        OS/Version: All
            Status: NEW
          Severity: normal
          Priority: P2
         Component: DMD
        AssignedTo: nobody@puremagic.com
        ReportedBy: luka8088@owave.net


--- Comment #0 from luka8088 <luka8088@owave.net> 2013-04-17 06:17:01 PDT ---
The following code compiles aldo s is not readable at compile time:

struct S (alias T) {
  typeof(T) value;
}

void main () {
  auto s = "some";
  s ~= "string";
  S!s value;
}


The side effect of this is the following error:

// Error: function literal __lambda3 (S!(s) a) is not
// callable using argument types (S!(s))

module program;

import std.stdio;

struct S (alias T) {
  typeof(T) value;
}

void f (alias l = x => 1) (string s) {
  l(S!(s).init);
}

void main () {
  auto s = "some";
  s ~= "string";
  f!((S!s a) { return 1; })(s);
}


I am not sure what the correct behavior should be, I think that the first code should not compile. But if it should, and this is the correct behavior than the error message should definitely be improved because.

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Andrej Mitrovic <andrej.mitrovich@gmail.com> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
                 CC|                            |andrej.mitrovich@gmail.com


--- Comment #1 from Andrej Mitrovic <andrej.mitrovich@gmail.com> 2013-04-18 05:23:37 PDT ---
(In reply to comment #0)
> The following code compiles aldo s is not readable at compile time:
> 
> struct S (alias T) {
>   typeof(T) value;
> }
> 
> void main () {
>   auto s = "some";
>   s ~= "string";
>   S!s value;
> }

That code isn't reading 's', it is only using it to get its type. And all types are known at compile-time. The code is ok.

> 
> 
> The side effect of this is the following error:
> 
> // Error: function literal __lambda3 (S!(s) a) is not
> // callable using argument types (S!(s))
> 
> module program;
> 
> import std.stdio;
> 
> struct S (alias T) {
>   typeof(T) value;
> }
> 
> void f (alias l = x => 1) (string s) {
>   l(S!(s).init);
> }
> 
> void main () {
>   auto s = "some";
>   s ~= "string";
>   f!((S!s a) { return 1; })(s);
> }

It doesn't compile because 's' within 'main' and 's' within 'f' are two different variables. You'd have to pass 's' from within main as an alias parameter to 'f'. E.g.:

----
import std.stdio;

struct S(alias T)
{
    typeof(T) value;
}

void f(alias lambda = x => 1, alias str)()
{
    lambda(S!str.init);
}

void main()
{
    string str = "some";
    f!((S!str a) { return 1; }, str)();
}
----

But I'd recommend changing the struct S definition to take a type and use typeof() at the call site to avoid having to use aliases everywhere.

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--- Comment #2 from luka8088 <luka8088@owave.net> 2013-04-18 07:03:26 PDT ---
(In reply to comment #1)
> (In reply to comment #0)
> > The following code compiles aldo s is not readable at compile time:
> > 
> > struct S (alias T) {
> >   typeof(T) value;
> > }
> > 
> > void main () {
> >   auto s = "some";
> >   s ~= "string";
> >   S!s value;
> > }
> 
> That code isn't reading 's', it is only using it to get its type. And all types are known at compile-time. The code is ok.
> 

Oh, I see, value can be used in runtime but compile time is only using it's type:

import std.stdio;

struct S (alias T) {
  typeof(T) value;
  void print () { writeln(T); }
}

void main () {
  auto s = "some";
  s ~= "string";
  S!s value;
  value.print();
}

> > 
> > 
> > The side effect of this is the following error:
> > 
> > // Error: function literal __lambda3 (S!(s) a) is not
> > // callable using argument types (S!(s))
> > 
> > module program;
> > 
> > import std.stdio;
> > 
> > struct S (alias T) {
> >   typeof(T) value;
> > }
> > 
> > void f (alias l = x => 1) (string s) {
> >   l(S!(s).init);
> > }
> > 
> > void main () {
> >   auto s = "some";
> >   s ~= "string";
> >   f!((S!s a) { return 1; })(s);
> > }
> 
> It doesn't compile because 's' within 'main' and 's' within 'f' are two different variables. You'd have to pass 's' from within main as an alias parameter to 'f'. E.g.:
> 

I agree, but shouldn't in that case 's' in error message be fully qualified? Or distinguishable in some way?

> ----
> import std.stdio;
> 
> struct S(alias T)
> {
>     typeof(T) value;
> }
> 
> void f(alias lambda = x => 1, alias str)()
> {
>     lambda(S!str.init);
> }
> 
> void main()
> {
>     string str = "some";
>     f!((S!str a) { return 1; }, str)();
> }
> ----
> 
> But I'd recommend changing the struct S definition to take a type and use typeof() at the call site to avoid having to use aliases everywhere.

I wanted to only take a type without using typeof() so I wrote:

template S(alias T) if (!is(T)) {
  alias S = S!(typeof(T));
}

struct S (T) {
  // ...
}

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--- Comment #3 from Andrej Mitrovic <andrej.mitrovich@gmail.com> 2013-04-18 10:35:19 PDT ---
(In reply to comment #2)
> I agree, but shouldn't in that case 's' in error message be fully qualified? Or distinguishable in some way?

Yep. This is covered by Issue 9631, but you can add your test-case there so it isn't missed.

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luka8088 <luka8088@owave.net> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|NEW                         |RESOLVED
         Resolution|                            |WONTFIX


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