Re: [Boost-users] function template with a universal reference to a specific type

I think what you're asking is "how can I constrain a universal reference to be one primary type?" One way is to implement the general interface in terms of a specialised function object (this is my favourite). example: http://coliru.stacked-crooked.com/a/317cadb674e175b7 This is my favourite because it allows selection of specialisations based on pattern matching rather than explicit overloads. #include #include <iostream> template<class T> struct foo_op { template void operator()(Prefix&& p, Arg&& /* a */) const { p(); } }; template <typename T> auto foo(T &&t) { auto op = foo_op(); return op([funcname = __PRETTY_FUNCTION__] { std::cout << "Primary: " << funcname << std::endl; }, std::forward<T>(t)); } // then specialise struct Bar {}; template<> struct foo_op<Bar> { template void operator()(Prefix&& /*p*/, Arg&& /*bar*/) const { std::cout << "It's a Bar\n"; } }; int main() { foo(6); // default case foo(Bar()); // bar case } On Sun, 30 Dec 2018 at 16:05, Ireneusz Szcześniak via Boost-users < boost-users@lists.boost.org> wrote:

Hi,

I'm writing to ask a C++ question, not specifically Boost-related, but this list is the best place I know.

How can I write a function template, which has a parameter of the universal reference to a specific type?

I can write a function template like this:

template <typename T> void foo(T &&t) { // The primary template implementation. cout << "Primary: " << __PRETTY_FUNCTION__ << endl; }

And so I can call this function with an expression of any value category, and any type, cv-qualified or not, like this:

int main() { foo(1); int i = 1; foo(i); const int ci = 1; foo(ci); }

In my code I need to provide different implementations of foo for different types (and these types are templated). How can I accomplish that?

Best, Irek _______________________________________________ Boost-users mailing list Boost-users@lists.boost.org https://lists.boost.org/mailman/listinfo.cgi/boost-users

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