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digitalmars.D.bugs - [Issue 9949] New: template initialization when alias cannot be read at compile time

reply d-bugmail puremagic.com writes:
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



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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Apr 17 2013
next sibling parent d-bugmail puremagic.com writes:
http://d.puremagic.com/issues/show_bug.cgi?id=9949


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

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



05:23:37 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;
 }


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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prev sibling next sibling parent d-bugmail puremagic.com writes:
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 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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prev sibling next sibling parent d-bugmail puremagic.com writes:
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10:35:19 PDT ---


 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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prev sibling parent d-bugmail puremagic.com writes:
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luka8088 <luka8088 owave.net> changed:

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


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