constexpr 标准函子的正确用法是什么?

What is the proper usage of a constexpr standard functor?

本文关键字:用法 是什么 标准 constexpr      更新时间:2023-10-16

采用以下代码:

#include <iostream>
#include <functional>
template <template<typename> class Op>
auto applyOp(const uint8_t lhs, const uint8_t rhs) {
constexpr Op<uint8_t> op;
return op(lhs, rhs);
}
int main() {
std::cout << +applyOp<std::bit_and>(19, 180) << std::endl;
}

当使用g++时,它可以很好地编译和运行。但是clang++会产生错误:

test.cpp:5:27: error: default initialization of an object of const type 'const bit_and<uint8_t>' (aka 'const bit_and<unsigned char>') without a user-provided default constructor
constexpr Op<uint8_t> op;
^
{}
test.cpp:11:19: note: in instantiation of function template specialization 'applyOp<std::bit_and>' requested here
std::cout << +applyOp<std::bit_and>(19, 180) << std::endl;
^
1 error generated.

所以我看了一下bit_and的源代码:

// Copyright (C) 2001-2016 Free Software Foundation, Inc.
//
// This file is part of the GNU ISO C++ Library.  This library is free
// software; you can redistribute it and/or modify it under the
// terms of the GNU General Public License as published by the
// Free Software Foundation; either version 3, or (at your option)
// any later version.
// This library is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
// GNU General Public License for more details.
// Under Section 7 of GPL version 3, you are granted additional
// permissions described in the GCC Runtime Library Exception, version
// 3.1, as published by the Free Software Foundation.
// You should have received a copy of the GNU General Public License and
// a copy of the GCC Runtime Library Exception along with this program;
// see the files COPYING3 and COPYING.RUNTIME respectively.  If not, see
// <http://www.gnu.org/licenses/>.
/*
*
* Copyright (c) 1994
* Hewlett-Packard Company
*
* Permission to use, copy, modify, distribute and sell this software
* and its documentation for any purpose is hereby granted without fee,
* provided that the above copyright notice appear in all copies and
* that both that copyright notice and this permission notice appear
* in supporting documentation.  Hewlett-Packard Company makes no
* representations about the suitability of this software for any
* purpose.  It is provided "as is" without express or implied warranty.
*
*
* Copyright (c) 1996-1998
* Silicon Graphics Computer Systems, Inc.
*
* Permission to use, copy, modify, distribute and sell this software
* and its documentation for any purpose is hereby granted without fee,
* provided that the above copyright notice appear in all copies and
* that both that copyright notice and this permission notice appear
* in supporting documentation.  Silicon Graphics makes no
* representations about the suitability of this software for any
* purpose.  It is provided "as is" without express or implied warranty.
*/
[ ... content omitted ... ]
template<typename _Arg1, typename _Arg2, typename _Result>
struct binary_function
{
/// @c first_argument_type is the type of the first argument
typedef _Arg1     first_argument_type;
/// @c second_argument_type is the type of the second argument
typedef _Arg2     second_argument_type;
/// @c result_type is the return type
typedef _Result   result_type;
};
[ ... content omitted ... ]
#if __cplusplus > 201103L
template<typename _Tp = void>
struct bit_and;
[ ... content omitted ... ]
#endif
[ ... content omitted ... ]
// _GLIBCXX_RESOLVE_LIB_DEFECTS
// DR 660. Missing Bitwise Operations.
template<typename _Tp>
struct bit_and : public binary_function<_Tp, _Tp, _Tp>
{
_GLIBCXX14_CONSTEXPR
_Tp
operator()(const _Tp& __x, const _Tp& __y) const
{ return __x & __y; }
};
[ ... content omitted ... ]

据我所知,应该在这里生成一个默认构造函数。 有趣的是,错误消息专门要求"用户提供的默认构造函数",而不仅仅是"默认构造函数"。

将违规行更改为使用统一初始化会导致代码与两个编译器一起使用:

- constexpr Op<uint8_t> op;
+ constexpr Op<uint8_t> op { };

我的问题是,要求这些额外的牙套clang++正确,还是g++不正确?

额外信息

使用g++命令编译:

g++ test.cpp -Werror -Wall -pedantic -std=c++14

使用clang++命令编译:

clang++ test.cpp -Werror -Wall -pedantic -std=c++14

运行应用程序命令:

./a.out

bit_and定义的位置:

/usr/include/c++/6.2.0/bits/stl_function.h

关于可能的重复标志,我不认为这是重复的,因为这个问题是在问为什么这个规则存在,而我的问题更多的是关于首先找出规则以及哪个编译器在其执行中是正确的。此外,对所谓的重复的回答并没有回答我的问题。

[dcl.init] 中的原始 C++14 措辞要求:

如果程序调用const限定类型的对象的默认初始化TT应是具有用户提供的默认构造函数的类类型。

因此,仅仅拥有默认构造函数是不够的。它还需要由用户提供。Clang实现了这个规则,这需要constexpr Op<uint8_t> op{};,但gcc没有 - 在这种情况下允许默认初始化。

然而,这个措辞最近在p0490中进行了修订。追溯适用于C++14的新措辞如下:

如果T的默认初始化将调用用户提供的T构造函数(不是从基类继承的)或 if
T的每个直接非变量非静态数据成员M都有一个默认成员初始值设定项,或者,如果M是类类型X(或其数组),TX是 const-default-constructable,
— 如果T是至少有一个非静态数据成员的联合,则只有一个变体成员具有默认成员初始值设定项,
— 如果T不是联合,对于每个具有至少一个非静态数据成员(如果有)的匿名联合成员,只有一个非静态数据成员具有默认成员初始值设定项, 并且每个可能构造的T基类都是 const-default-constructable 的。

如果程序调用 const 限定类型 T 的对象的默认初始化,则 T 应是 const-default-可构造的类类型或其数组。

Op<uint8_t>const-default-constructable(由于它没有非静态数据成员的微不足道的原因),因此您可以默认初始化它。现在,C++14 将允许:

constexpr Op<uint8_t> op;

实际上,我们从一个 gcc 错误(允许默认初始化)变成了一个 clang 错误(不允许它)。