const成员函数的std函数包装器

Question

#include <functional>
#include <iostream>

struct Foo {
    Foo(int num) : num_(num) {}
    void print_add(int i) const { std::cout << num_+i << '\n'; }
    int get_num(int i) { return num_;}
    void set_num(int i) { num_ = i;}
    int num_;
};
int main() {
    std::function<int(const Foo *, int)> f_get_num;
    f_get_num = &Foo::get_num;
    return 0;
}

This will generate a error, error: invalid conversion from 'const Foo*' to 'Foo*' [-fpermissive] at line f_get_num = &Foo::get_num . Foo::get_num type is int (Foo:: *fp)(int) . Can anybody explain it? Thanks.

Answer 1

You cannot call non- const functions on const objects. You can pass const Foo* to f_get_num , but Foo::get_num takes non-const implicit this .

The following two calls are just as illegal:

Foo foo;
Foo const* const_ptr = &foo;

const_ptr->get_num(42);
f_get_num(const_ptr, 42); // results in const_ptr->get_num(42)

You can declare your get_num to be const :

int get_num(int i) const { return num_;}

And then your code will work correctly.

The other way is to make your f_get_num take non- const parameter, but that's not the way to go when your function is a getter and shouldn't modify the object.

std::function<int(Foo*, int)> f_get_num;
f_get_num = &Foo::get_num;