"explicit" 和 "implicit" operator () 调用的区别

Question

标准中是否有一个条款描述从基 classes 调用 operator () 的方法之间的以下区别？

#include <iostream>
#include <type_traits>

#include <cstdlib>
#include <cassert>

template< typename visitor, typename ...visitors >
struct composite_visitor
    : std::decay_t< visitor >
    , composite_visitor< visitors... >
{

    //using std::decay_t< visitor >::operator ();
    //using composite_visitor< visitors... >::operator ();

    composite_visitor(visitor && _visitor, visitors &&... _visitors)
        : std::decay_t< visitor >(std::forward< visitor >(_visitor))
        , composite_visitor< visitors... >{std::forward< visitors >(_visitors)...}
    { ; }

};

template< typename visitor >
struct composite_visitor< visitor >
    : std::decay_t< visitor >
{

    //using std::decay_t< visitor >::operator ();

    composite_visitor(visitor && _visitor)
        : std::decay_t< visitor >(std::forward< visitor >(_visitor))
    { ; }

};

template< typename visitor, typename ...visitors >
composite_visitor< visitor, visitors... >
compose_visitors(visitor && _visitor, visitors &&... _visitors)
{
    return {std::forward< visitor >(_visitor), std::forward< visitors >(_visitors)...};
}

int
main()
{
    struct A {};
    struct B {};
#if 1
    struct { int operator () (A) { return 1; } } x;
    struct { int operator () (B) { return 2; } } y;
    auto xy = compose_visitors(x, y);
#else
    auto xy = compose_visitors([] (A) { return 1; }, [] (B) { return 2; });
#endif
    // "implicit":
    assert(xy(A{}) == 1);
    assert(xy(B{}) == 2);
    // "explicit":
    assert(xy.operator () (A{}) == 1); // error: member 'operator()' found in multiple base classes of different types
    assert(xy.operator () (B{}) == 2);
    return EXIT_SUCCESS;
}

"Implicit" 调用可以正常编译，但 "explicit" 却不行。为什么会这样？

compose_visitors 通过构造一个从所有参数派生的 class 将参数组合成单个 class。

取消对 using 指令的注释可消除硬错误。清楚了。

lambda 函数和仿函数的行为相同。

编译器是 clang 3.6.

Answer 1

"implicit" 调用也是格式错误的。事实上，GCC rejects it，所以它似乎是 Clang 中的一个错误。

标准（N4140，[over.call]/1）明确表示

a call x(arg1,...) is interpreted as x.operator()(arg1, ...) for a class object x of type T if T::operator()(T1, T2, T3) exists and if the operator is selected as the best match function by the overload resolution mechanism (13.3.3).

所以这两个调用的行为必须相同。

更新：这是 Clang 中的 known issue。

"explicit" 和 "implicit" operator () 调用的区别

Difference between "explicit" and "implicit" invocation of operator ()

c++

inheritance

c++11

c++14