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From: Ben Bacarisse <ben.usenet@bsb.me.uk>
Newsgroups: comp.lang.c++
Subject: Re: What's decltype doing here?
Date: Tue, 07 Nov 2023 15:16:42 +0000
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Bo Persson <bo@bo-persson.se> writes:

> On 2023-11-07 at 11:11, Ben Bacarisse wrote:
>> This question got a bit lost in another thread, so I am re-posting it on
>> its own.  What is decltype doing in the following short program?
>>    #include <iostream>
>>    #include <memory>
>>        int main()
>>    {
>>       int x = 42;
>>       std::unique_ptr<int, decltype([](int *p) {
>>          std::cout << "delete (" << *p << ")\n";
>>       })> upi(&x);
>>    }
>> I expected decltype to supply just a type in the template and for a
>> second argument to be required when constructing upi, but no.  The
>> program compiles (gcc and clang) and prints "delete (42)".
>> The equivalent using a named function works as I expect:
>>    #include <iostream>
>>    #include <memory>
>>        void deleter(int *p) { std::cout << "deleter (" << *p << ")\n"; };
>>        int main()
>>    {
>>       int x = 42;
>>       std::unique_ptr<int, decltype(&deleter)> upi(&x, deleter);
>>    }
>> In this case it's an error to omit the second argument to the
>> constructor.  (I used decltype just to make the comparison clear but of
>> course one could simply write the type (void (*)(int *)) in the template
>> instead.)
>> I am pretty sure that the answer is in some way related to the fact that
>> C++ lambdas have unique types, but it's not clear to me what's really
>> going on.  Explanations welcome.  Pointers to explanatory text in the
>> standard even more so.
>
> You *have* the explanation right there. :-)
>
> The lambda has a unique type and can be default constructed, so that is
> what happens.

Ah.  OK.  It seems a little mysterious, but it's obvious when you see
it.  Here's a more direct example:

  #include <iostream>

  int main()
  {
     decltype([](int i){std::cout << "i=" << i << "\n";}) f;
     f(42);
  }

Thank you.

-- 
Ben.