在 C++11 中获取函数指针的语法

Question

我正在尝试理解函数指针，我有以下测试代码：

#include <iostream>

using namespace std;

class Test {
public:
    void test_func() {
        cout << "Test func called.";
    }
};

void outer_test_func(Test &t, void (Test::*func)()) {
    (t.*func)();
}

int main(int argc, char *argv[]) {
    auto t = Test();
    outer_test_func(t, &Test::test_func);
}

这行得通。但据我了解 Test::test_func 和 &Test::test_func 都会产生指针。那么为什么我不能使用前者而不是后者呢？如果我尝试它，g++ 会抱怨。

Answer 1

But from what I understand Test::test_func and &Test::test_func both result in pointers.

Test::test_func 不是创建成员函数指针或数据成员指针的有效语法。它从来都不是有效的 C++ 语法。

来自 cppreference（强调我的），

A pointer to non-static member object m which is a member of class C can be initialized with the expression &C::m exactly. Expressions such as &(C::m) or &m inside C's member function do not form pointers to members.

在 C++11 中获取函数指针的语法

Syntax to get function pointers in C++11

c++

c++11

member-function-pointers