我正在研究关于lambda继承的问题，我尝试在这个stackoverflow链接下更改下面的示例。我修改了代码使用模板推演指南和折叠表达式。
这段代码
#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，我得到以下错误
<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, trimRight, capitalize};
| ^
<source>:5:7: note: candidate: 'constexpr FunctionSequence<main()::<lambda(std::string&)>, main()::<lambda(std::string&)>, main()::<lambda(std::string&)> >::FunctionSequence(const FunctionSequence<main()::<lambda(std::string&)>, 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&)>, main()::<lambda(std::string&)> >::FunctionSequence(FunctionSequence<main()::<lambda(std::string&)>, main()::<lambda(std::string&)>, main()::<lambda(std::string&)> >&&)'
<source>:5:7: note: candidate expects 1 argument, 3 provided
ASM generation compiler returned: 1
哪里出了问题？
代码示例--&>；演示
using Ts::operator()...;
这将所有
这显然不是你想要的。看起来您希望按顺序执行组成基类操作。
下面是如何使用逗号运算符以一种巧妙的方式写出它的:
template<class... Ts>
struct FunctionSequence : Ts... {
auto operator()(std::string& str) const {
return (Ts::operator()(str), ...);
}
};
住在科里鲁
#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;
}
打印
Before TrimAndCapitalize: str = " what a Hullabaloo "
After TrimAndCapitalize: str = "WHAT A HULLABALOO"
我建议私下继承，并使函数要么对参数进行void-操作，要么使它们纯净（获取const-references并返回副本）。另外，请参见C++/C++11中的函数组合。
