使用包扩展的 Lambda 继承

问题描述

我正在研究 lambda 继承并尝试更改以下示例 在这个stackoverflow link下。我使用模板推导指南和折叠表达式修改了代码。

代码片段

#include <iostream>
using namespace std;

template<class... Ts>
class FunctionSequence :Ts...
{
    using Ts::operator()...;
};
template <class...Ts> FunctionSequence(Ts...) -> FunctionSequence<Ts...>; // (1)

int main(){

    //note: these lambda functions are bug ridden. Its just for simplicity here.
    //For correct version,see the one on coliru,read on.
    auto trimLeft = [](std::string& str) -> std::string& { str.erase(0,str.find_first_not_of(' ')); return str; };
    auto trimRight = [](std::string& str) -> std::string& { str.erase(str.find_last_not_of(' ')+1); return str; };
    auto capitalize = [](std::string& str) -> std::string& { for(auto& x : str) x = std::toupper(x); return str; };

    auto trimAndCapitalize = FunctionSequence{trimLeft,trimRight,capitalize};
    std::string str = " what a Hullabaloo     ";

    std::cout << "Before TrimAndCapitalize: str = \"" << str << "\"\n";
    trimAndCapitalize(str);
    std::cout << "After TrimAndCapitalize:  str = \"" << str << "\"\n";

    return 0;
}

但是我收到以下错误

<source>: In function 'int main()':
<source>:19:78: error: no matching function for call to 'FunctionSequence<main()::<lambda(std::string&)>,main()::<lambda(std::string&)>,main()::<lambda(std::string&)> >::FunctionSequence(<brace-enclosed initializer list>)'
   19 |     auto trimAndCapitalize = FunctionSequence{trimLeft,capitalize};
      |                                                                              ^
<source>:5:7: note: candidate: 'constexpr FunctionSequence<main()::<lambda(std::string&)>,main()::<lambda(std::string&)> >::FunctionSequence(const FunctionSequence<main()::<lambda(std::string&)>,main()::<lambda(std::string&)> >&)'
    5 | class FunctionSequence :Ts...
      |       ^~~~~~~~~~~~~~~~
<source>:5:7: note:   candidate expects 1 argument,3 provided
<source>:5:7: note: candidate: 'constexpr FunctionSequence<main()::<lambda(std::string&)>,main()::<lambda(std::string&)> >::FunctionSequence(FunctionSequence<main()::<lambda(std::string&)>,main()::<lambda(std::string&)> >&&)'
<source>:5:7: note:   candidate expects 1 argument,3 provided
ASM generation compiler returned: 1

出了什么问题？

代码示例 --> demo

解决方法

using Ts::operator()...;

这将所有 operator() 从基类“提升”到类命名空间中。

这显然不是您想要的。看起来您想按顺序执行组成基类的操作。

以下是如何使用逗号运算符以漂亮的方式编写：

template<class... Ts>
struct FunctionSequence : Ts...  {
    auto operator()(std::string& str) const {
        return (Ts::operator()(str),...);
    }
};

Live On Coliru

#include <iostream>
using namespace std;

template<class... Ts>
struct FunctionSequence : Ts...  {
    auto operator()(std::string& str) const {
        return (Ts::operator()(str),...);
    }
};
template <class...Ts> FunctionSequence(Ts...) -> FunctionSequence<Ts...>; // (1)

int main(){
    //note: these lambda functions are bug ridden. Its just for simplicity
    //here.  For correct version,see the one on coliru,read on.
    auto trimLeft = [](std::string& str) -> std::string& { str.erase(0,str.find_first_not_of(' ')); return str; };
    auto trimRight = [](std::string& str) -> std::string& { str.erase(str.find_last_not_of(' ')+1); return str; };
    auto capitalize = [](std::string& str) -> std::string& { for(auto& x : str) x = std::toupper(x); return str; };

    auto trimAndCapitalize = FunctionSequence{trimLeft,trimRight,capitalize};
    std::string str = " what a Hullabaloo     ";

    std::cout << "Before TrimAndCapitalize: str = \"" << str << "\"\n";
    trimAndCapitalize(str);
    std::cout << "After TrimAndCapitalize:  str = \"" << str << "\"\n";

    return 0;
}

印刷品

Before TrimAndCapitalize: str = " what a Hullabaloo     "
After TrimAndCapitalize:  str = "WHAT A HULLABALOO"

注意事项

我建议私有继承，并使函数要么对参数进行无效操作，要么或使它们纯（采用常量引用并返回副本）。另外，请参阅 function composition in C++ / C++11 例如