[英]How to pass a rvalue reference parameter to a template operator() function in C++?
我试图编写一些代码来在 C++17 中实现一个咖喱 function。 我的当前实现在下面(我将在这个问题的底部给你一个最小的工作示例)。
template <class Function, class... CapturedArgs>
class curried{
private:
using CapturedArgsTuple = std::tuple<std::decay_t<CapturedArgs>...>;
template <class... Args>
static auto capture_by_value(Args&&... args){
return std::tuple<std::decay_t<Args>...>(std::forward<Args>(args)...);
}
public:
curried(Function function, CapturedArgs&&... args)
: m_function(function), m_capture(capture_by_value(std::move(args)...)){}
curried(Function function, std::tuple<CapturedArgs...> args)
: m_function(function), m_capture(std::move(args)){}
template <class... NewArgs>
auto operator()(NewArgs&&... args){
auto new_args = capture_by_value(std::forward<NewArgs>(args)...);
auto all_args = std::tuple_cat(m_capture, new_args);
if constexpr(std::is_invocable_v<Function, CapturedArgs..., NewArgs...>){
return std::apply(m_function, all_args);
}else{
return curried<Function, CapturedArgs..., NewArgs...>(m_function, all_args);
}
}
private:
Function m_function;
std::tuple<CapturedArgs...> m_capture;
};
这是一个测试 function:
void func(const string& str1, string& str2, string str3){
str2 += "str2 ";
cout << "str1 = " << str1
<< ", str2 = " << str2
<< ", str3 = " << str3 << endl;
}
int main(){
string str1 = "Hello ", str2 = "World", str3 = "!";
auto test = curried(func);
auto test_two = test(std::cref(str1))(std::ref(str2));
cout << "result : ";
test_two(str3);
}
到目前为止,一切都很好。 我可以在终端上看到一些日志打印,例如:
$ result : str1 = Hello , str2 = Worldstr2 , str3 = !
我在这里有两个问题:
第一个是如何通过传递右值引用来调用 curried function? 我已经尝试了所有可以搜索的内容,但结果要么是编译错误,要么什么都没有。
void func_1(const string& str1, string& str2, string&& str3){
str2 += "str2 ";
cout << "str1 = " << str1
<< ", str2 = " << str2
<< ", str3 = " << str3 << endl;
}
int main(){
string str1 = "Hello ", str2 = "World", str3 = "!";
auto test = curried(func_1);
auto test_two = test(std::cref(str1))(std::ref(str2));
cout << "result : ";
// test_two(std::move(str3)); Compile Error
// test_two(string("!")); Compile Error
test_two(std::bind(std::move<string&>, str3)); // Compile successfully, but there's nothing output
}
在解决第一个问题的过程中,我发现了一些奇怪的东西。 这是一个例子:
void func_2(const string& str1, string& str2, string str3, string& str4){
str2 += "str2 ";
cout << "str1 = " << str1
<< ", str2 = " << str2
<< ", str3 = " << str3
<< ", str4 = " << str4 << endl;
}
int main(){
string str1 = "Hello ", str2 = "World", str3 = "!", str4 = "abc";
auto test = curried(func_2);
auto test_two = test(std::cref(str1))(std::ref(str2))(str3);
cout << "result : ";
test_two(std::ref(str4));
}
当我使用 func_2 测试我的咖喱 function 时,我收到了一些错误消息:
$ g++ curried.cc -std=c++17
curried.cc: In instantiation of ‘auto curried<Function, CapturedArgs>::operator()(NewArgs&& ...) [with NewArgs = {std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&}; Function = void (*)(const std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>, std::__cxx11::basic_string<char>&); CapturedArgs = {std::reference_wrapper<const std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::reference_wrapper<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >}]’:
curried.cc:60:15: required from here
curried.cc:28:11: error: no matching function for call to ‘curried<void (*)(const std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>, std::__cxx11::basic_string<char>&), std::reference_wrapper<const std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::reference_wrapper<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&>::curried(void (*&)(const std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>, std::__cxx11::basic_string<char>&), std::tuple<std::reference_wrapper<const std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::reference_wrapper<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >&)’
28 | return curried<Function, CapturedArgs..., NewArgs...>(m_function, all_args);
| ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
curried.cc:18:2: note: candidate: ‘curried<Function, CapturedArgs>::curried(Function, std::tuple<_Elements ...>) [with Function = void (*)(const std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>, std::__cxx11::basic_string<char>&); CapturedArgs = {std::reference_wrapper<const std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::reference_wrapper<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&}]’
18 | curried(Function function, std::tuple<CapturedArgs...> args)
| ^~~~~~~
curried.cc:18:57: note: no known conversion for argument 2 from ‘tuple<std::reference_wrapper<const std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::reference_wrapper<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >>’ to ‘tuple<std::reference_wrapper<const std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::reference_wrapper<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&>’
18 | curried(Function function, std::tuple<CapturedArgs...> args)
| ~~~~~~~~~~~~~~~~~~~~~~~~~~~~^~~~
curried.cc:15:2: note: candidate: ‘curried<Function, CapturedArgs>::curried(Function, CapturedArgs&& ...) [with Function = void (*)(const std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>, std::__cxx11::basic_string<char>&); CapturedArgs = {std::reference_wrapper<const std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::reference_wrapper<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&}]’
15 | curried(Function function, CapturedArgs&&... args)
| ^~~~~~~
curried.cc:15:2: note: candidate expects 4 arguments, 2 provided
curried.cc:7:7: note: candidate: ‘constexpr curried<void (*)(const std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>, std::__cxx11::basic_string<char>&), std::reference_wrapper<const std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::reference_wrapper<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&>::curried(const curried<void (*)(const std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>, std::__cxx11::basic_string<char>&), std::reference_wrapper<const std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::reference_wrapper<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&>&)’
7 | class curried{
| ^~~~~~~
curried.cc:7:7: note: candidate expects 1 argument, 2 provided
curried.cc:7:7: note: candidate: ‘constexpr curried<void (*)(const std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>, std::__cxx11::basic_string<char>&), std::reference_wrapper<const std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::reference_wrapper<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&>::curried(curried<void (*)(const std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>&, std::__cxx11::basic_string<char>, std::__cxx11::basic_string<char>&), std::reference_wrapper<const std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::reference_wrapper<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&>&&)’
curried.cc:7:7: note: candidate expects 1 argument, 2 provided
curried.cc: In function ‘int main()’:
curried.cc:60:10: error: void value not ignored as it ought to be
60 | test_two(str3)(std::ref(str4));
| ~~~~~~~~^~~~~~
所以第二个问题是为什么我不能把字符串参数str3
放在字符串引用一个str4
之前? 为什么参数定义的顺序很重要让我完全不知所措。
为了您的方便,这里是一个最小的工作示例:
#include <iostream>
#include <functional>
#include <tuple>
using namespace std;
template <class Function, class... CapturedArgs>
class curried{
private:
using CapturedArgsTuple = std::tuple<std::decay_t<CapturedArgs>...>;
template <class... Args>
static auto capture_by_value(Args&&... args){
return std::tuple<std::decay_t<Args>...>(std::forward<Args>(args)...);
}
public:
curried(Function function, CapturedArgs&&... args)
: m_function(function), m_capture(capture_by_value(std::move(args)...)){}
curried(Function function, std::tuple<CapturedArgs...> args)
: m_function(function), m_capture(std::move(args)){}
template <class... NewArgs>
auto operator()(NewArgs&&... args){
auto new_args = std::make_tuple(std::forward<NewArgs>(args)...);
auto all_args = std::tuple_cat(m_capture, std::move(new_args));
if constexpr(std::is_invocable_v<Function, CapturedArgs..., NewArgs...>){
return std::apply(m_function, all_args);
}else{
return curried<Function, CapturedArgs..., NewArgs...>(m_function, all_args);
}
}
private:
Function m_function;
std::tuple<CapturedArgs...> m_capture;
};
void func_1(const string& str1, string& str2, string&& str3){
str2 += "str2 ";
cout << "str1 = " << str1
<< ", str2 = " << str2
<< ", str3 = " << str3 << endl;
}
void func_2(const string& str1, string& str2, string str3, string& str4){
str2 += "str2 ";
cout << "str1 = " << str1
<< ", str2 = " << str2
<< ", str3 = " << str3
<< ", str4 = " << str4 << endl;
}
int main()
{
/* code */
string str1 = "Hello ", str2 = "World", str3_for_func_1 = "!",
str3_for_func_2 = "!", str4 = "abc";
auto question_1 = curried(func_1); // For the first question
auto question_2 = curried(func_2); // For the second question
auto question_1_two_params = question_1(std::cref(str1))(std::ref(str2));
auto question_2_two_params = question_2(std::cref(str1))(std::ref(str2));
cout << "result : ";
//question_1_two_params(std::move(str3_for_func_1)); // Compile Error
//question_1_two_params(string("abc")); // Compile Error
//auto question_2_three_params = question_2_two_params(str3_for_func_2); // Compile Error
//question_2_three_params(std::ref(str4)); // It should output some log like "result : str1 = Hello, balabala..."
return 0;
}
编译命令:
$ g++ curryied.cc -std=c++17 -o curried
我的工作环境是:
操作系统:Ubuntu-20.04 编译器:gcc 版本 9.3.0
一个问题在于std::apply(m_function, all_args);
您将all_args
作为左值传递给std::apply
,这会将其作为左值传递给func_1
的第三个参数,这将失败,因为func_1
的第三个参数是一个右值引用,它不能绑定到左值参数。
实际上,将该行更改为std::apply(m_function, std::move(all_args));
使前两个// Compile Error
行实际编译并生成正确的 output。 同样,我也会在all_args
的其他用法上调用std::move
。
它看起来像std::make_tuple(std::forward<NewArgs>(args)...);
没有做你认为它做的事情。 将其更改为std::tuple<NewArgs&&...>(std::forward<NewArgs>(args)...);
解决问题; 相当于std::forward_as_tuple(std::forward<NewArgs>(args)...);
.
为何此更改起作用的细节在于std::make_tuple
与std::forward_as_tuple
的返回类型:后者返回引用元组,而前者返回已从 arguments 复制/移动的值元组。
现在,按照我的推理:
curried(Function function, std::tuple<CapturedArgs...> args)
:它需要参数args
应该是std::tuple<CaptureArgs...>
类型。 我们确定args
有那种类型吗? 好吧,如果发生模板类型推导,那么答案显然是肯定的。 但是,对该构造函数的调用永远不会利用类型推导,因为唯一的调用是return curried<Function, CapturedArgs..., NewArgs...>(m_function, all_args);
其中明确提供了模板 arguments。all_args
是构造函数期望的类型吗? 嗯,递归调用中的模板arguments CapturedArgs..., NewArgs...
对应的是class... CapturedArgs
模板参数class的模板参数,用来形成参数的构造函数的类型std::tuple<CaptureArgs...>
。static_assert
给出,在递归return
之前, all_args
的类型是std::tuple<CapturedArgs..., NewArgs...>
: static_assert(std::is_same_v<decltype(all_args), std::tuple<CapturedArgs..., NewArgs...>>); return curried<Function, CapturedArgs..., NewArgs...>(m_function, all_args);
std::ref
/ std::cref
中的值,因为这些值无法通过static_assert
离子但仍然是有效输入,这正是因为std::reference_wrapper
的工作。 您可以编写更复杂的断言,或者您可以暂时将std::ref(bla)
更改为bla
等等,并检查我给您的static_assert
在使用std::forward_as_tuple
时是否通过,在使用std::make_tuple
时是否失败。感谢您提出这个问题。 这对我来说是一个很好的机会,可以再次深入这个复杂的话题并最终理解它!
上面我建议你使用std::forward_as_tuple(std::forward<NewArgs>(args)...);
.
好吧,可能这个建议是错误的。
在第 238 页,作者明确指出他希望元组存储副本,以防止柯里化的 function 在其 arguments 中幸存的情况。 因此,最好用这个代替 go (注意,传递给std::tuple
的模板参数中没有&&
):
auto new_args = std::tuple<NewArgs...>(std::forward<NewArgs>(args)...);
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