将模板函数转换为通用Lambda [英] Convert template function to generic lambda
问题描述
我想像模板化函数那样传递模板化函数，但这不起作用。
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
}
请注意，情况2和4严格等效。情况3更一般，但不受限制（这对我来说是个问题）。我认为用语言应将情况1与情况2和4等效，但是情况并非如此。
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.
- 提议将模板函数隐式转换为通用约束的lambda（案例2/4）？如果没有，那么是否有根本的语言原因阻止这样做？
- 到目前为止，我必须使用情况2，这很麻烦。
- 情况4：不符合c ++ 14，即使在 c ++ 20 ，但仍然不够完美（冗长，因为您创建的Lambda基本上可以不添加任何信息。）
- 情况3：不受约束，但我依靠（此处未显示）对使用非 vector参数调用 print的替换失败（P0428R1提到这个问题）。因此，我猜想附属的问题是我可以使用一些enable_if技巧来约束通用lambda吗？
- 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?"
在C ++ 14/17/20中，是否存在一种非常简洁的方式来实现从案例1到案例2的转换？我什至对宏黑客持开放态度。
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.
推荐答案
在C ++ 14/17中存在/ 20，这是一种非常简洁的方式，可实现从案例1到案例2的转换？我什至对宏黑客持开放态度。
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.
是。
// 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)...))然后您可以说：
for_each(t, LIFT(print));
是否有提议将模板函数隐式转换为通用约束的lambda？
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.
这篇关于将模板函数转换为通用Lambda的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
