使用std :: function的所有参数传递成员函数

Question

如何从成员函数创建一个std :: function而无需键入std :: placeholders :: _ 1,std :: placeholders :: _ 2等 - 我想"放置"所有参数,只保存对象本身.

struct Foo{
  int bar(int,float,bool) {return 0;}
};
int baz(int,float,bool) {return 0;}
int main() {
  Foo object;
  std::function<int(int,float,bool)> fun1 = baz; // OK
  std::function<int(int,float,bool)> fun2 = std::bind(&Foo::bar, object); // WRONG, needs placeholders
}

我不想在这个阶段提供参数,我只想在某处存储函数+对象.例如,我想让std :: vector同时具有全局函数和成员函数.使用FastDelegate(fastdelegate::MakeDelegate(object, &Class::function))很容易.

我不想使用lambda,因为它也需要我重新输入参数.我只想要旧的FastDelegate行为.

Answer 1

您可以使用函数模板来推导出所有成员函数参数类型,如下所示:

template<typename Obj, typename Result, typename ...Args>
auto make_delegate(const Obj &x, Result (Obj::*fun)(Args...)) -> // ...

并将返回特殊的委托对象,它将包含您的对象(或指向它的指针),并将所有传递的参数转发给底层对象的成员函数:

template<typename Obj, typename Result, typename ...Args>
struct Delegate
{
    Obj x;
    Result (Obj::*f)(Args...);

    template<typename ...Ts>
    Result operator()(Ts&&... args)
    {
        return (x.*f)(forward<Ts>(args)...);
    }
};

您将获得以下用法语法:

function<int(int,float,bool)> fun = make_delegate(object, &Foo::bar);

这是完整的例子:

#include <functional>
#include <iostream>
#include <utility>

using namespace std;

struct Foo
{
    int bar(int x, float y, bool z)
    {
        cout << "bar: " << x << " " << y << " " << z << endl;
        return 0;
    }
};

int baz(int x, float y, bool z)
{
    cout << "baz: " << x << " " << y << " " << z << endl;
    return 0;
}

template<typename Obj, typename Result, typename ...Args>
struct Delegate
{
    Obj x;
    Result (Obj::*f)(Args...);

    template<typename ...Ts>
    Result operator()(Ts&&... args)
    {
        return (x.*f)(forward<Ts>(args)...);
    }
};

template<typename Obj, typename Result, typename ...Args>
auto make_delegate(const Obj &x, Result (Obj::*fun)(Args...))
    -> Delegate<Obj, Result, Args...>
{
    Delegate<Obj, Result, Args...> result{x, fun};
    return result;
}

int main()
{
    Foo object;
    function<int(int,float,bool)> fun[] =
    {
        baz,
        make_delegate(object, &Foo::bar) // <---- usage
    };
    for(auto &x : fun)
        x(1, 1.0, 1);
}

输出是:

baz: 1 1 1
bar: 1 1 1

Coliru现场演示