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digitalmars.D.bugs - [Issue 9555] New: Type deduction for new lambda syntax literals breaks with templates

reply d-bugmail puremagic.com writes:
http://d.puremagic.com/issues/show_bug.cgi?id=9555

           Summary: Type deduction for new lambda syntax literals breaks
                    with templates
           Product: D
           Version: D2
          Platform: All
        OS/Version: All
            Status: NEW
          Severity: normal
          Priority: P2
         Component: DMD
        AssignedTo: nobody puremagic.com
        ReportedBy: m.strashun gmail.com



Simple motivating example:

--- test.d ---
import std.functional;

void main()
{
    auto deleg = toDelegate(a => a > 2);
}
------

--- shell ---
$ rdmd test.d 
test.d(5): Error: template std.functional.toDelegate does not match any
function template declaration. Candidates are:
/usr/include/dmd/phobos/std/functional.d(722):       
std.functional.toDelegate(F)(auto ref F fp) if (isCallable!(F))
test.d(5): Error: template std.functional.toDelegate(F)(auto ref F fp) if
(isCallable!(F)) cannot deduce template function from argument types !()(void)
------

Type of lambda was deduced as "void" here.

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next sibling parent d-bugmail puremagic.com writes:
http://d.puremagic.com/issues/show_bug.cgi?id=9555


Maxim Fomin <maxim maxim-fomin.ru> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
             Status|NEW                         |RESOLVED
                 CC|                            |maxim maxim-fomin.ru
         Resolution|                            |INVALID



---
Actually type of lambda was not deduced to void, void here is pseudo type of
non-instantiated template, because a => a > 2 is a lambda template. If you
append type of a parameter, this would work, for ex:

import std.functional;

void main()
{
    auto deleg = toDelegate( (int a) => a > 2);
}

Since there is no guesses what type of "a" can be in the original code,
template cannot be instantiated. 

By the way, idea mentioned in forum discussion that there is problem with new
(lambda) syntax is also wrong, because the code can be rewritten with delegate
template:

import std.functional;

void main()
{
    auto deleg = toDelegate( delegate (a) { return a > 2; } );
}

with the same problem and same error message.

Close this as invalid.

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prev sibling next sibling parent d-bugmail puremagic.com writes:
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bioinfornatics <bioinfornatics gmail.com> changed:

           What    |Removed                     |Added
----------------------------------------------------------------------------
                 CC|                            |bioinfornatics gmail.com



04:39:49 PST ---
but if your delegate look like : in bool delegate(in size_t) lambda

you can't use (in int a) => a > 2)
or (const int a) => a > 2)

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prev sibling next sibling parent d-bugmail puremagic.com writes:
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04:41:32 PST ---
skip my comment that is allowed to do ((in int a) => a > 2)

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prev sibling next sibling parent d-bugmail puremagic.com writes:
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Waa, it is a template? Unexpected. What about turning this into enhancement
request to improve error message though? Something like "not enough context to
deduce lambda type".

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prev sibling next sibling parent d-bugmail puremagic.com writes:
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---


 Waa, it is a template? Unexpected. What about turning this into enhancement
 request to improve error message though? Something like "not enough context to
 deduce lambda type".


Enough context can be provided later. 

template get(alias a)
{
    alias a!int get;
}

template foo(fun...)
{
    alias get!fun foo;
}

void main()
{
    alias foo!(a=>a) a;
    assert(a(1) == 1);
    assert(a(1.0) is 1.0); //NG
}

Note, this is reduced from how map and unaryfun works. If you turn it into
context-independent error, this would mean code break for some usages which are
currently accepted.

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prev sibling parent d-bugmail puremagic.com writes:
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I was speaking exactly about the case when it is used and context is lacking.
This message "cannot deduce template function from argument types !()(void)" is
very misleading for someone who is not aware of lambda implementation inner
details.

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