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decltype(auto) function return type does not deduce && types


decltype(auto) function return type does not deduce && types

By : Mihir Sabnis
Date : October 25 2020, 07:10 AM
seems to work fine This seems to be a bug in GCC. Since decltype(ret) is int&&, foo1 should have return type int&&. However, this immediately renders foo1 ill-formed, since a function that returns int&& cannot have its return value initialized from ret, which is an lvalue (you would need std::move to make it work properly). Note that Clang gets this right (see link in comments to the question).
code :


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Unable to instantiate function templates which uses decltype to deduce return type, if called from inside a lambda?

Unable to instantiate function templates which uses decltype to deduce return type, if called from inside a lambda?


By : user1794256
Date : March 29 2020, 07:55 AM
Any of those help I'm trying to use C++0x, and in particular lambda expression and decltype to simplify some of my code, using the MSVC10 RC compiler. , These are just some test cases for people to observe.
Works
code :
template <typename F>
auto foo(F f) -> decltype(f())
{
  return f();
}

void dummy() {}

int main()
{
    auto x = []()
            {   // non-lambda parameter
                foo(dummy);
            };
}
template <typename F>
auto foo(F f) -> decltype(f())
{
  return f();
}

int main()
{
    auto f = [](){};
    auto x = [&]()
            {    // pre-defined lambda
                foo(f);
            };
}
template <typename F>
auto foo(F f) -> decltype(f())
{
  return f();
}

int main()
{
    auto x = []()
            {   // in-argument lambda
                foo([]{});
            };
}
template <typename F>
auto foo(F f) -> decltype(f())
{
  return f();
}

int main()
{
    auto x = []()
            {   // in-scope lambda
                auto f = []{};
                foo(f);
            };
}
template <typename F>
auto foo(F f) -> decltype(f())
{
  return f();
}

int main()
{
    auto x = []()
            {   // in-scope lambda, explicit return
                // (explicit return type fails too, `-> void`)
                auto f = [](){ return; };
                foo(f);
            };
}
template <typename F>
auto foo(F f) -> decltype(f())
{
  return f();
}

int main()
{
    auto x = []()
            {   // in-argument lambda, explicit return non-void
                // (explicit return type fails too, `-> int`)
                foo([]{ return 5; }); 
            };
}
decltype(auto) vs auto&& to perform generic handling of function's return type

decltype(auto) vs auto&& to perform generic handling of function's return type


By : Md Nafiul Alam
Date : March 29 2020, 07:55 AM
it helps some times auto&& is already optimal for capturing function return values, such that the differences of decltype(auto) can only be disadvantages. In your example, lifetime extension is applied to the otherwise-temporary object returned from the function. This causes it to behave essentially the same as a directly named object, with the effect that the reference qualifier gets "erased."
Using decltype(auto) with a return-by-value function causes its return value object to be moved into the local. Depending what's inside the function, copy elision may apply which removes the distinction between the local and the temporary. But that only applied sometimes, whereas reference-bound lifetime extension is unconditional.
Using auto/decltype to deduce array type

Using auto/decltype to deduce array type


By : Fico
Date : March 29 2020, 07:55 AM
wish of those help As pointed out by @nwp the simpler solution is remember that std::initializer_list is created when doing auto foo = {,}. So a more succinct solution is
code :
auto connectors = {std::make_pair(std::wstring(L"send_order"),std::bind(&RESTListener::send_order, this, std::placeholders::_1)),
                   std::make_pair(std::wstring(L"on_cancel_order"), std::bind(&RESTListener::on_cancel_all, this, std::placeholders::_1)) };
What is difference between decltype(auto) and decltype(returning expr) as return type?

What is difference between decltype(auto) and decltype(returning expr) as return type?


By : JUNJIE CAI
Date : March 29 2020, 07:55 AM
it fixes the issue Yes there is a difference. The first one will detect the return type based on the return expression in the body of the function.
The second one will not also set the return type to the type of the expression inside decltype(), but will also apply expression sfinae on it. That means that if the expression inside decltype is not valid, the compiler will search for another valid overload. Whereas the first version will be a hard error.
code :
template<typename T>
auto fun(T a) -> decltype(a.f()) { return a.f(); }

template<typename T>
auto fun(T a) -> decltype(a.g()) { return a.g(); }

struct SomeType {
    int g() { return 0; }
};

fun(SomeType{});
Using auto and decltype for making a function return the type of its class. How can I make it return a value, instead of

Using auto and decltype for making a function return the type of its class. How can I make it return a value, instead of


By : Arun Misra
Date : March 29 2020, 07:55 AM
To fix this issue The following is off the cuff, and not fully tested. However I find myself wondering if the well known handle/body idiom might be a good match for your requirements here. It works by making Matrix nothing but a pointer to an implementation (what you currently call Matrix). And then derived clients derive from the implementation, not from Matrix itself.
Here's a small sketch:
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