Convert template function to generic lambda

问题

I'd like to pass templated functions around as if they were generic lambdas, however this does not work.

#include <iostream>
#include <vector>
#include <tuple>
#include <string>
#include <utility> 


// for_each with std::tuple
// (from https://stackoverflow.com/a/6894436/1583122)
template<std::size_t I = 0, typename FuncT, typename... Tp>
inline typename std::enable_if<I == sizeof...(Tp), void>::type
for_each(std::tuple<Tp...> &, FuncT)
{}

template<std::size_t I = 0, typename FuncT, typename... Tp>
inline typename std::enable_if<I < sizeof...(Tp), void>::type
for_each(std::tuple<Tp...>& t, FuncT f) {
    f(std::get<I>(t));
    for_each<I + 1, FuncT, Tp...>(t, f);
}

// my code
template<class T> auto
print(const std::vector<T>& v) -> void {
    for (const auto& e : v) {
        std::cout << e << "\t";
    }
}

struct print_wrapper {
    template<class T>
    auto operator()(const std::vector<T>& v) {
        print(v);
    }
};

auto print_gen_lambda = [](const auto& v){ print(v); };

auto print_gen_lambda_2 = []<class T>(const std::vector<T>& v){ print(v); }; // proposal P0428R1, gcc extension in c++14/c++17

int main() {
     std::tuple<std::vector<int>,std::vector<double>,std::vector<std::string>> t = { {42,43},{3.14,2.7},{"Hello","World"}};
    for_each(t, print); // case 1: error: template argument deduction/substitution failed: couldn't deduce template parameter 'FuncT'
    for_each(t, print_wrapper()); // case 2: ok
    for_each(t, print_gen_lambda); // case 3: ok
    for_each(t, print_gen_lambda_2); // case 4: ok
}

Note that case 2 and 4 are strictly equivalent. Case 3 is more general but unconstrained (this is a problem for me). I think that case 1 should be treated equivalently to cases 2 and 4 by the language, however this is not the case.

Is there a proposal to implicitly convert a template function to a generic constrained lambda (case 2/4)? If no, is there a fundamental language reason that prevents from doing so?
As of now, I have to use case 2, which is quite cumbersome.
- case 4: not c++14-compliant, even if should be standard in c++20, and still not perfect (verbose since you create a lambda that fundamentally does not add any information).
- case 3: is unconstrained, but I rely (not shown here) on substitution failure for calls to "print" with non-"vector" arguments (P0428R1 mentions this problem). So I guess the subsidiary question is "Can I constrain a generic lambda with some enable_if tricks?"

Is there, in C++14/17/20, a very terse manner to enable the conversion from case 1 to case 2? I am even open to macro hacks.

回答1:

Is there, in C++14/17/20, a very terse manner to enable the conversion from case 1 to case 2? I am even open to macro hacks.

Yes.

// C++ requires you to type out the same function body three times to obtain
// SFINAE-friendliness and noexcept-correctness. That's unacceptable.
#define RETURNS(...) noexcept(noexcept(__VA_ARGS__)) \
     -> decltype(__VA_ARGS__){ return __VA_ARGS__; }

// The name of overload sets can be legally used as part of a function call -
// we can use a macro to create a lambda for us that "lifts" the overload set
// into a function object.
#define LIFT(f) [](auto&&... xs) RETURNS(f(::std::forward<decltype(xs)>(xs)...))

You can then say:

for_each(t, LIFT(print));

Is there a proposal to implicitly convert a template function to a generic constrained lambda?

Yes, look at P0119 or N3617. Not sure about their status.

回答2:

Can I constrain a generic lambda with some enable_if tricks?

If all what you want is to constrain the types of the parameters of your generic lambda, you can do that with a couple of function declarations (no definition required) and a static_assert (so that you get a graceful message error at compile-time in case of failure). No macro at all around (they are so C-ish).

It follows a minimal, working example:

#include<vector>
#include<type_traits>
#include<utility>
#include<list>

template<template<typename...> class C, typename... A>
constexpr std::true_type spec(int, C<A...>);

template<template<typename...> class C, template<typename...> class T, typename... A>
constexpr std::false_type spec(char, T<A...>);

int main() {
    auto fn = [](auto&& v) {
        static_assert(decltype(spec<std::vector>(0, std::declval<std::decay_t<decltype(v)>>()))::value, "!");
        // ...
    };

    fn(std::vector<int>{});
    // fn(std::list<int>{});
    //fn(int{});
}

If you toggle the comments to the last lines, the static_assert will throw an error and the compilation will fail as expected.
See it up and running on wandbox.

Side note.

Of course you can reduce the boilerplate here:

static_assert(decltype(spec<std::vector>(0, std::declval<std::decay_t<decltype(v)>>()))::value, "!");

Add a variable template like the following one:

template<template<typename...> class C, typename T>
constexpr bool match = decltype(spec<C>(0, std::declval<std::decay_t<T>>()))::value;

Then use it in your static_asserts:

static_assert(match<std::vector, decltype(v)>, "!");

Pretty clear, isn't it?

Note.

In C++17 you'll be able to reduce even more the code required to do that by defining your lambda as:

auto fn = [](auto&& v) {
    if constexpr(match<std::vector, decltype(v)>) {
        print(v);
    }
};

See your example code running on wandbox.

来源：https://stackoverflow.com/questions/45187335/convert-template-function-to-generic-lambda

标签

c++

lambda

c++14

enable-if