filter!(not!(predicate))(someInputRange) does not compile

[Andrei Alexandrescu]

On 1/19/11 7:19 PM, Jens Mueller wrote:
> Andrei Alexandrescu wrote:
>> On 1/19/11 5:53 PM, Jens Mueller wrote:
>>> Hi,
>>>
>>> I cannot make the following compile.
>>>
>>> import std.functional;
>>> import std.array;
>>> import std.algorithm;
>>> import std.stdio;
>>>
>>> void main() {
>>>      auto numbers = [0, 1, 2, 3, 4, 5];
>>>
>>>      bool alwaysTrue(uint a) { return true; }
>>>      alias not!(alwaysTrue) alwaysFalse;
>>>
>>>      numbers = array(filter!(alwaysTrue)(numbers));
>>>      writeln(numbers);
>>>      numbers = array(filter!(alwaysFalse)(numbers)); // does not compile
>>>      writeln(numbers);
>>> }
>>>
>>> The line with alwaysFalse fails with:
>>> /path/to/../src/phobos/std/algorithm.d(854): Error: constructor
>>> std.algorithm.Filter!(not,int[]).Filter.this cannot get frame pointer to
>>> not
>>>
>>> Any ideas what I'm doing wrong or workarounds?
>>>
>>> Jens
>>
>> Place the call to not!alwaysTrue in a local function inside main:
>>
>>      bool alwaysFalse(uint a) { return not!alwaysTrue(a); }
>
> Thanks. Can you elaborate a bit please? I wonder why the alias won't
> work.

I thought of it for a bit. It's a limitation of the compiler that's 
worth a bug report. The explanation is a bit involved. Let me start by 
remarking that if you prepend "static" to alwaysTrue, the alias works as 
expected:

static bool alwaysTrue(uint a) { return true; }
alias not!(alwaysTrue) alwaysFalse;

Without "static", alwaysTrue has access to a hidden pointer to the stack 
frame of the function is in.

In theory a template should be unable to manipulate alwaysTrue because 
of that frame pointer. But Walter has had this great idea of 
instantiating templates in the context of the function they're in, so 
they gain access to the frame pointer too. However, that instantiation 
mechanism still has a few limitations. I think the code above runs into 
one of them.


Andrei