使用 std::function 调用非指针对象的成员函数

Question

代码如下

std::string::empty()应该以this指针为参数，类型为指针、std::string *.

2 和 3 行的调用怎么可能正常？

#include <iostream>
#include <functional>

int main() {
    std::string str{"A small pond"};

    std::function<bool(std::string*)> fp = &std::string::empty;
    std::cout << fp(&str) << std::endl; // 1

    std::function<bool(std::string)> f = &std::string::empty;
    std::cout << f(str) << std::endl; // 2

    std::function<bool(std::string&)> fr = &std::string::empty;
    std::cout << fr(str) << std::endl; // 3
}

/*
output:
0
0
0
*/

clang version 9.0.0-2~ubuntu18.04.2 (tags/RELEASE_900/final)
g++ (Ubuntu 8.4.0-1ubuntu1~18.04) 8.4.0

Answer 1

std::function 可以接受任何匹配其类型签名的 Callable。调用时，可调用对象和参数根据以下规则进行评估（引用 cppreference）：

If f is a pointer to member function of class T:

If std::is_base_of<T, std::decay_t<decltype(t1)>>::value is true, then INVOKE(f, t1, t2, ..., tN) is equivalent to (t1.*f)(t2, ..., tN)

If std::decay_t<decltype(t1)> is a specialization of std::reference_wrapper, then INVOKE(f, t1, t2, ..., tN) is equivalent to (t1.get().*f)(t2, ..., tN)

If t1 does not satisfy the previous items, then INVOKE(f, t1, t2, ..., tN) is equivalent to ((*t1).*f)(t2, ..., tN).

所以第一种情况的评估方式类似于 (*t1).*f()，而其他两种情况的评估方式类似于 t1.*f()。

使用 std::function 调用非指针对象的成员函数

Member functions called on non-pointer objects with std::function

c++

std-function

this-pointer