不同 C++ 编译器之间的自动类型推导不匹配

Mismatched deduction of auto types between different c++ compilers

因此,我尝试以现代 C++ 的某种风格实现点积 (https://en.wikipedia.org/wiki/Dot_product),并提出了以下代码:

#include <iostream>

template<class... Args>
auto dot(Args... args)
{
    auto a = [args...](Args...)
    { 
        return [=](auto... brgs)
        {
            static_assert(sizeof...(args) == sizeof...(brgs));

            auto v1 = {args...}, i1 = v1.begin();
            auto v2 = {brgs...}, i2 = v2.begin();
            typename std::common_type<Args...>::type s = 0;

            while( i1 != v1.end() && i2!= v2.end())
            {
                s += *i1++ * *i2++;
            } 
            return s;
        };
    };
  return a(std::forward<Args>(args)...);
}

int main()
{
    auto a = dot(1,3,-5)(4,-2,-1);
    std::cout << a << std::endl;
}

在线:https://gcc.godbolt.org/z/kDSney and also: cppinsights

上面的代码使用 g++ 可以很好地编译和执行,但是 clang(以及 iccmsvc)会卡住它:

clang++ ./funcpp.cpp --std=c++17                                                                                                                                                                                                                                                        
./funcpp.cpp:12:4: error: 'auto' deduced as 'std::initializer_list<int>' in declaration of 
        'v1' and deduced as 'const int *' in declaration of 'i1'
                        auto v1 = {args...}, i1 = v1.begin();
                        ^         ~~~~~~~~~       ~~~~~~~~~~
./funcpp.cpp:28:11: note: in instantiation of function template specialization 
        'dot<int, int, int>' requested here
        auto a = dot(1,3,-5)(4,-2,-1);
                 ^
1 error generated.

现在,如果我将 v1v2i1i2 的定义分解为:

auto v1 = {args...} ;
auto i1 = v1.begin();
auto v2 = {brgs...};
auto i2 = v2.begin();

clangmsvc都没有问题,icc还是卡住了:

<source>(10): error: static assertion failed

                static_assert(sizeof...(args) == sizeof...(brgs));

                ^

          detected during instantiation of "auto dot(Args...) [with Args=<int, int, int>]" at line 30

compilation aborted for <source> (code 2)

Execution build compiler returned: 2

但是,如果我删除有问题的 static_assert,那么 icc 编译代码也没有问题。

除了(典型的)问题:哪个是正确的以及为什么 :) 具体问题是:

根据[dcl.spec.auto]

if the type that replaces the placeholder type is not the same in each deduction, the program is ill-formed

clang 正确识别出在相关行中定义了两种不同的类型:'auto' deduced as 'std::initializer_list<int>' in declaration of 'v1' and deduced as 'const int *' in declaration of 'i1' 所以我想听听您的意见是否:

感谢您阅读这么长的问题。 (作为奖励,如果有人能回答为什么 iccstatic_assert 上失败会很棒。)

根据我的评论展开:

g++ 并不总是这样做,考虑示例 auto i = 0l, f = 0.0;,它给出了错误:

test.cpp: In function ‘int main()’:
test.cpp:4:5: error: inconsistent deduction for ‘auto’: ‘long int’ and then ‘double’
    4 |     auto i = 0l, f = 0.0;

如果我们编译您的程序并打印变量的类型 (with this method),我们将得到以下输出:

v1: std::initializer_list<int>, i1: int const*
v2: std::initializer_list<int>, i2: int const*

使用 gcc 版本 9.2.0,带有标志 -std=c++17 -pedantic -Wall -Wextra,没有任何警告或错误。

根据您对标准的评论,该程序格式错误,标准 specifies 应该发出诊断消息(警告或错误),除非另有说明(在这种情况下不是).因此我会说这是 gcc 中的一个错误。

a known bug.

ICC 上的 static_assert 失败绝对是一个错误。我通过将 static_assert 移动到一个单独的函数中找到了一个简单的解决方法。不是很优雅的解决方案,但它有效。

稍作修改后,这是使用 GCC、Clang 和 ICC 编译的代码:

template<std::size_t size, class... Args>
void args_no_guard(Args... args)
{
    static_assert(sizeof...(args) == size);
}

template<class... Args>
auto dot(Args... args)
{
    return [=](auto... brgs)
    {
        constexpr auto n = sizeof...(args);
        args_no_guard<n>(brgs...);

        using T = std::common_type_t<decltype(args)..., decltype(brgs)...>;
        const T v1[]{static_cast<T>(args)...};
        const T v2[]{static_cast<T>(brgs)...};

        T dot = 0;
        for (std::size_t i = 0; i < n; ++i)
            dot += v1[i] * v2[i];
        return dot;
    };
}