从函数隐式转换为函数指针？

Question

我有一个接受函数指针作为参数的函数。令人惊讶的是，我可以同时传入函数指针和普通函数：

#include <iostream>
#include <functional>

int triple(int a) {
    return 3*a;
}

int apply(int (*f)(int), int n) {
    return f(n);
}

int main() {
    std::cout << apply(triple, 7) << "\n";
    std::cout << apply(&triple, 7) << "\n";
}

我不明白为什么会这样。是否存在从函数到函数指针的隐式转换？

Answer 1

是的，有 function-to-pointer implicit conversion:

An lvalue of function type T can be implicitly converted to a prvalue pointer to that function. This does not apply to non-static member functions because lvalues that refer to non-static member functions do not exist.

和

A pointer to function can be initialized with an address of a non-member function or a static member function. Because of the function-to-pointer implicit conversion, the address-of operator is optional:
void f(int);
void (*p1)(int) = &f;
void (*p2)(int) = f; // same as &f

这意味着当在需要函数指针的上下文中使用时，函数（非静态成员函数除外）将隐式转换为函数指针，operator&的用法是可选的。

从函数隐式转换为函数指针？

Implicit cast from function to function pointer?

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