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From: "nickhuang99 at hotmail dot com" <gcc-bugzilla@gcc.gnu.org>
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Subject: [Bug c++/104358] Assignable template lambda as function parameter is
 incorrectly reduced to type of "int"
Date: Sat, 05 Feb 2022 22:46:42 +0000
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https://gcc.gnu.org/bugzilla/show_bug.cgi?id=3D104358

--- Comment #2 from qingzhe huang <nickhuang99 at hotmail dot com> ---
Here are more tests (https://www.godbolt.org/z/5qc8jTGa8) to show that only
msvc is giving the correct result.=20

It just illustrates the use cases of this definition and why it should be
allowed.

1. By using operator "+", here the type "Lambda" is just a free function
pointer.

template<typename T>
using Lambda=3Ddecltype(+[](T){});
static_assert(is_same_v<Lambda<int>, void(*)(int)>);=20

2. The "foo" takes "Lambda" as parameter
template<typename T>
T foo(T&&, Lambda<T>);=20

3. As "Lambda<T>" is just a function pointer, "foo" should allow both free
function pointer or lambda(by conversion operator) to be used as argument. =
And
because GCC mistakenly consider "foo<int>" signature as "int(*)(int&&,
/*Lambda<int>*/int)", this is not possible.

static_assert(is_same_v<decltype(&foo<int>), int(*)(int&&, int)>);=20

These are use cases for "foo":
a) using lambda directly
    foo<int>(5, [](int n){});// lambda conversion operator!
b) using converted function pointer from lambda by operator "+"
    auto funptr=3D+[](int n){};
    foo<int>(5, funptr);
c) directly using a function pointer
    void(*ptr)(int)=3Dfunptr;
    foo<int>(5, ptr);=