c++ - std::bind，无法使具有单个参数的方法起作用

Question

以下代码有效:

struct Foo {
    void print(int n1, int n2)
    {
        std::cout << n1 << " " << n2 << std::endl;
    }
};

Foo foo;
auto f = std::bind(&Foo::print, &foo, 95, _1);
f(5);

但是如果我想绑定(bind)到一个只有一个参数的方法:

struct Foo {
    void print(int n1)
    {
        std::cout << n1 << std::endl;
    }
};

Foo foo;
auto f3 = std::bind(&Foo::print, &foo, 95, _1);
f3();

我得到一长串错误:

g++ -Wall -O3 -fno-rtti -pedantic -Wextra -pthread -std=c++17 -g   -c -o main.o main.cpp
In file included from main.cpp:3:
/usr/include/c++/8/functional: In instantiation of 'struct std::_Bind_check_arity<void (Foo::*)(int), Foo*, int, const std::_Placeholder<1>&>':
/usr/include/c++/8/functional:787:12:   required from 'struct std::_Bind_helper<false, void (Foo::*)(int), Foo*, int, const std::_Placeholder<1>&>'
/usr/include/c++/8/functional:808:5:   required by substitution of 'template<class _Func, class ... _BoundArgs> typename std::_Bind_helper<std::__is_socketlike<_Func>::value, _Func, _BoundArgs ...>::type std::bind(_Func&&, _BoundArgs&& ...) [with _Func = void (Foo::*)(int); _BoundArgs = {Foo*, int, const std::_Placeholder<1>&}]'
main.cpp:152:50:   required from here
/usr/include/c++/8/functional:775:7: error: static assertion failed: Wrong number of arguments for pointer-to-member
       static_assert(_Varargs::value
                     ~~~~~~~~~~~~~~~
       ? sizeof...(_BoundArgs) >= _Arity::value + 1
       ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
       : sizeof...(_BoundArgs) == _Arity::value + 1,
       ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
main.cpp: In function 'int main()':
main.cpp:153:8: error: no match for call to '(std::_Bind<void (Foo::*(Foo*, int, std::_Placeholder<1>))(int)>) ()'
     f3();
        ^
In file included from main.cpp:3:
/usr/include/c++/8/functional:480:2: note: candidate: 'template<class ... _Args, class _Result> _Result std::_Bind<_Functor(_Bound_args ...)>::operator()(_Args&& ...) [with _Args = {_Args ...}; _Result = _Result; _Functor = void (Foo::*)(int); _Bound_args = {Foo*, int, std::_Placeholder<1>}]'
  operator()(_Args&&... __args)
  ^~~~~~~~
/usr/include/c++/8/functional:480:2: note:   template argument deduction/substitution failed:
/usr/include/c++/8/functional:491:2: note: candidate: 'template<class ... _Args, class _Result> _Result std::_Bind<_Functor(_Bound_args ...)>::operator()(_Args&& ...) const [with _Args = {_Args ...}; _Result = _Result; _Functor = void (Foo::*)(int); _Bound_args = {Foo*, int, std::_Placeholder<1>}]'
  operator()(_Args&&... __args) const
  ^~~~~~~~
/usr/include/c++/8/functional:491:2: note:   template argument deduction/substitution failed:
/usr/include/c++/8/functional:509:2: note: candidate: 'template<class ... _Args, class _Result> _Result std::_Bind<_Functor(_Bound_args ...)>::operator()(_Args&& ...) volatile [with _Args = {_Args ...}; _Result = _Result; _Functor = void (Foo::*)(int); _Bound_args = {Foo*, int, std::_Placeholder<1>}]'
  operator()(_Args&&... __args) volatile
  ^~~~~~~~
/usr/include/c++/8/functional:509:2: note:   template argument deduction/substitution failed:
/usr/include/c++/8/functional:521:2: note: candidate: 'template<class ... _Args, class _Result> _Result std::_Bind<_Functor(_Bound_args ...)>::operator()(_Args&& ...) const volatile [with _Args = {_Args ...}; _Result = _Result; _Functor = void (Foo::*)(int); _Bound_args = {Foo*, int, std::_Placeholder<1>}]'
  operator()(_Args&&... __args) const volatile

Answer 1

_1、_2 等是生成的可调用对象的参数表示。例如

struct Foo {
    void print(int n1)
    {
        std::cout << n1 << " " << n2 << std::endl;
    }
};

Foo foo;
auto f = std::bind(&Foo::print, &foo, _2, _1);
f(5,4); //would call foo.print(4,5)

因此，如果您需要将 void (Foo::*)(int) 绑定(bind)到带有签名 void (*)(int) 的可调用对象，您必须编写

auto f = std::bind(&Foo::print, &foo, _1);

请注意，在 C++11(和一些 C++03 编译器，例如 Microsoft VS2010+)和更高版本中，您可以使用 lambda 表达式来执行相同的操作。