如何将右值引用参数传递给 C++ 中的模板 operator() 函数?
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【中文标题】如何将右值引用参数传递给 C++ 中的模板 operator() 函数?【英文标题】:How to pass a rvalue reference parameter to a template operator() function in C++? 【发布时间】:2021-05-23 19:03:31 【问题描述】:我尝试编写一些代码来实现 C++17 中的柯里化函数。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;
;
这是一个测试函数:
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);
到目前为止一切顺利。我可以在我的终端上看到一些日志打印,例如:
$ result : str1 = Hello , str2 = Worldstr2 , str3 = !
这里有两个问题:
第一个是如何通过传递右值引用来调用柯里化函数?我已经尝试了所有可以搜索的内容,但结果要么编译错误,要么什么也没有。 p>
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 测试我的柯里化函数时,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将字符串参数str3
放在字符串引用前一个str4
? 为什么参数定义的顺序很重要让我完全不知所措。
为方便起见,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 9.3.0 版
【问题讨论】:
你试过看看Functional Programming in C++吗? Here's the repo 带有示例,使函数柯里化是示例之一。 顺便说一句,当涉及到作为用户实际需要此功能时(因为您想使用它,而不是编码它),请注意boost::hana::curry
,至少如果您知道要柯里化的最大参数数。
是的,我正在读这本书。而事实上,柯里化类的实现来自于本书中的一个例子。这是一本好书
【参考方案1】:
问题 1
一个问题是在std::apply(m_function, all_args);
行,您将all_args
作为左值传递给std::apply
,这会将它作为左值传递给func_1
的第三个参数,这将失败,因为@987654329 @ 的第三个参数是一个右值引用,它不能绑定到一个左值参数。
确实,将该行更改为std::apply(m_function, std::move(all_args));
会使前两行// Compile Error
实际编译并生成正确的输出。同样,我也会在 all_args
的其他用法上调用 std::move
。
问题 2
看起来std::make_tuple(std::forward<NewArgs>(args)...);
没有按照您的想法进行操作。将其更改为std::tuple<NewArgs&&...>(std::forward<NewArgs>(args)...);
即可解决问题;相当于std::forward_as_tuple(std::forward<NewArgs>(args)...);
。
为什么这个改变起作用的细节在于std::make_tuple
与std::forward_as_tuple
的返回类型:后者返回一个引用元组,而前者返回一个从参数复制/移动的值元组.
现在,按照我的推理:
首先,查看curried(Function function, std::tuple<CapturedArgs...> args)
:它需要args
类型的参数std::tuple<CaptureArgs...>
。我们确定args
有这种类型吗?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从不利用类型推导的优势,因为唯一的调用是在return curried<Function, CapturedArgs..., NewArgs...>(m_function, all_args);
中显式提供模板参数。
所以问题仍然存在:all_args
是构造函数所期望的类型吗?嗯,递归调用中的模板实参CapturedArgs..., NewArgs...
对应了类的class... CapturedArgs
模板形参,用来构成构造函数的实参类型std::tuple<CaptureArgs...>
。
所以这个问题的答案由static_assert
ing 给出,在递归return
之前,all_args
是std::tuple<CapturedArgs..., NewArgs...>
类型:
static_assert(std::is_same_v<decltype(all_args), std::tuple<CapturedArgs..., NewArgs...>>);
return curried<Function, CapturedArgs..., NewArgs...>(m_function, all_args);
不幸的是,您不能将此断言放入代码中,只要您传递包裹在std::ref
/std::cref
中的值,因为那些未通过static_assert
ion 但仍然是有效输入,这正是因为@ 987654356@s 工作。你可以写一个更复杂的断言,或者你可以暂时将std::ref(bla)
更改为bla
等等,并检查我给你的static_assert
在使用std::forward_as_tuple
时是否通过,在使用std::make_tuple
时失败。
感谢您提出这个问题。这是一个很好的机会让我再次深入这个复杂的话题并最终理解它!
还有一点
上面我建议你使用std::forward_as_tuple(std::forward<NewArgs>(args)...);
。
嗯,可能这个建议是错误的。
在第 238 页,作者明确声明他希望元组存储副本,以防止柯里化函数在其参数中幸存的情况。因此,最好改用这个(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)...);
【讨论】:
首先,我试图搜索一些关于右值引用和 std::apply 的东西,这可以给我一些提示,谷歌给了我这个帖子:***.com/questions/34877699/…。你可以看到第二个答案,然后你就会明白我为什么这样做了。感谢您的回答,我的第一个问题已经解决,而第二个问题没有。我已经更正了我的代码。 我也不知道。如果我将std::ref(str3)
和std::ref(str4)
传递给question_2_two_params
和question_2_three_params
,那么一切顺利。然而,这不是我的本意。代码可以按照我的意图工作,即当且仅当 only one 值参数被放置到参数列表中的最后一个位置。
@PhoenixChao,我已经给出了完整的答案。
@PhoenixChao,事后看来,我会更改您的问题的标题,因为事实证明该问题与operator()
的界面无关。
就是这样!!而已!!!太感谢了!我还在思考这个问题,但我还没有想出任何东西。我觉得你的回答是对的。以上是关于如何将右值引用参数传递给 C++ 中的模板 operator() 函数?的主要内容,如果未能解决你的问题,请参考以下文章