Question

我想包装一些类成员函数，并做一些准备和清理工作。

我尝试复制一些其他线程池代码，但得到一些我无法解决的错误。如何正确地做到这一点？

#include <iostream>
#include <string>
using namespace std;
class A {
public:
    void connect() {};
    void close() {};
    template<typename F, typename ... Args>
    auto wrapper(F&& f, Args&& ... args) -> typename std::result_of<F(Args...)>::type {
        using return_type = typename std::result_of<F(Args...)>::type;
        connect();
        return_type ret = f(args...);
        close();
        return ret;
    }
    bool c(int a, string b) {}
    string c(string b) {return b;}
    bool r(int a, string b) {}
};
int main() {
    A a;
    a.connect();
    a.c(1, "abc");
    a.close(); // equal to a.wrapper(a.c, 1, "abc"); but compling error, how to write it correctly?
    cout << "result of a is: " << a.wrapper(a.c, 1, "abc") ? "successful" : "fail" << endl;
    cout << "result of another a is: " << a.wrapper(a.c, "abc") << endl;
    cout << "result of r is:" << a.wrapper(a.r, 1, "abc") << endl;
}

我得到的错误是这样的：

main.cpp: In function ‘int main()’:
main.cpp:25:58: error: no matching function for call to ‘A::wrapper(<unresolved overloaded function type>, int, const char [4])’
     cout << "result of a is: " << a.wrapper(a.c, 1, "abc") ? "successful" : "fail" << endl;
                                                          ^
main.cpp:25:58: note: candidate is:
main.cpp:9:10: note: template<class F, class ... Args> typename std::result_of<_Functor(_ArgTypes ...)>::type A::wrapper(F&&, Args&& ...)
     auto wrapper(F&& f, Args&& ... args) -> typename std::result_of<F(Args...)>::type {
          ^
main.cpp:9:10: note:   template argument deduction/substitution failed:
main.cpp:25:58: note:   couldn't deduce template parameter ‘F’
     cout << "result of a is: " << a.wrapper(a.c, 1, "abc") ? "successful" : "fail" << endl;
                                                          ^
main.cpp:25:87: error: invalid operands of types ‘const char [5]’ and ‘<unresolved overloaded function type>’ to binary ‘operator<<’
     cout << "result of a is: " << a.wrapper(a.c, 1, "abc") ? "successful" : "fail" << endl;
                                                                                       ^
main.cpp:26:63: error: no matching function for call to ‘A::wrapper(<unresolved overloaded function type>, const char [4])’
     cout << "result of another a is: " << a.wrapper(a.c, "abc") << endl;
                                                               ^
main.cpp:26:63: note: candidate is:
main.cpp:9:10: note: template<class F, class ... Args> typename std::result_of<_Functor(_ArgTypes ...)>::type A::wrapper(F&&, Args&& ...)
     auto wrapper(F&& f, Args&& ... args) -> typename std::result_of<F(Args...)>::type {
          ^
main.cpp:9:10: note:   template argument deduction/substitution failed:
main.cpp:26:63: note:   couldn't deduce template parameter ‘F’
     cout << "result of another a is: " << a.wrapper(a.c, "abc") << endl;



                                                                   ^

Answer 1

表达式a.c实际上不是指向可调用成员函数的指针。 &A::c是，但是它需要调用A的实例。

有两种方法可以解决这个问题：

使用std::bind。与std::bind(&A::c, a)
使用lambda。与[&a](int i, std::string const& s) { return a.c(i, s); }

Answer 2

后面的代码将通过编译。然而，它仍然比预期的复杂得多。

using sig1 = bool(A::*)(int, string);
cout << "result of a is: " << (a.wrapper(bind(static_cast<sig1>(&A::c), a, _1, _2), 1, string("abc")) ? "successful" : "fail") << endl;
//cout << "result of another a is: " << a.wrapper(a.c, "abc") << endl;
//cout << "result of r" << a.wrapper(a.r, 1, "abc") << endl;

顺便说一句，clang的错误报告优于gcc。当您遇到一些模板编程问题时，您需要更清晰的错误消息，FYI

如何编写C ++类成员函数包装器？

2 个答案: