`boost::function<int(const char*)> f` 是指针还是对象？

Question

根据文章(https://theboostcpplibraries.com/boost.function)，其中说：

boost::function makes it possible to define a pointer to a function with a specific signature. boost::function<int(const char*)> f defines a pointer f that can point to functions that expect a parameter of type const char* and return a value of type int.

如何理解boost::function<int(const char*)> f定义了一个可以指向函数的指针f

难道f不是boost::function<int(const char*)>类型的对象吗，确实不是指针

Answer 1

@sehe Sorry, I forgot. The link is added to the post. theboostcpplibraries.com/boost.function – John 37 mins ago

感谢您添加 link。

虽然我已经出版了他的书并且重视他对 Boost 介绍材料的贡献，但我坚持我的判断，这里的措辞不必要地令人困惑，甚至误导：它似乎完全忽略了非函数可调用对象这一事实可以分配。在我看来，这就是进行类型擦除的全部意义所在。

如果你只是想方便地定义函数指针，你可以直接使用函数指针：

static int my_function(const char*) { return 42; }

// regular function pointer stuff
{
    using MyFunction = int(const char*);
    MyFunction* pf = &my_function;

    if (pf) {
        std::cout << "pf(): " << pf(argument) << "\n";
    }
}

但是使用boost::function（或std::function）你也可以这样做：

// gereralized assignable callables:
boost::function<int(const char*)> f = &my_function;
if (f) {
    std::cout << "same with f(): " << f(argument) << "\n";
}
struct MyCallable {
    int operator()(const char*) const { return _retval; }
    int _retval = 42;
};

f = MyCallable{};
std::cout << f(argument) << "\n"; // prints 42

f = MyCallable{99};
std::cout << f(argument) << "\n"; // prints 99

MyCallable instance { 10 };
f = std::ref(instance);
while (instance._retval--)
    std::cout << f(argument) << "\n"; // prints 9 .. 0

现场演示

Live On Coliru

#include <boost/function.hpp>
#include <iostream>

static int my_function(const char*) { return 42; }

int main() {
    auto argument = "dummy";
    // regular function pointer stuff
    {
        using MyFunction = int(const char*);
        MyFunction* pf = &my_function;

        if (pf) {
            std::cout << "pf(): " << pf(argument) << "\n";
        }
    }

    // gereralized assignable callables:
    boost::function<int(const char*)> f = &my_function;
    if (f) {
        std::cout << "same with f(): " << f(argument) << "\n";
    }
    struct MyCallable {
        int operator()(const char*) const { return _retval; }
        int _retval = 42;
    };

    f = MyCallable{};
    std::cout << f(argument) << "\n"; // prints 42

    f = MyCallable{99};
    std::cout << f(argument) << "\n"; // prints 99

    MyCallable instance { 10 };
    f = std::ref(instance);
    while (instance._retval--)
        std::cout << f(argument) << "\n"; // prints 9 .. 0
}

版画

pf(): 42
same with f(): 42
42
99
9
8
7
6
5
4
3
2
1
0

`boost::function<int(const char*)> f` 是指针还是对象？

Is `boost::function<int(const char*)> f` a pointer or an object?

c++

boost

现场演示