当`std :: make_index_sequence`和`std :: index_sequence`用于模板参数默认类型时，g ++和clang ++的行为不同 [英] g++ and clang++ different behaviour when `std::make_index_sequence` and `std::index_sequence` are used for a template parameter default type

问题描述

另一个谁在g ++和clang ++之间？ C ++标准专家的问题。

Another "who's right between g++ and clang++?" question for C++ standard gurus.

给出以下代码

#include <utility>

template <std::size_t N, typename = std::make_index_sequence<N>>
struct foo;

template <std::size_t N, std::size_t ... Is>
struct foo<N, std::index_sequence<Is...>>
 { };

template <std::size_t N>
void bar (foo<N> const &)
 { }

int main()
 {
   bar(foo<42u>{});
 }

我看到g ++在clang ++编译时出现以下错误

I see that g++ compile where clang++ gives the following error

tmp_003-14,gcc,clang.cpp:32:4: error: no matching function for call to 'bar'
   bar(foo<42u>{});
   ^~~
tmp_003-14,gcc,clang.cpp:27:6: note: candidate template ignored: could not match
      '__make_integer_seq' against 'integer_sequence'
void bar (foo<N> const &)
     ^
1 error generated.

和往常一样，问题是：谁是对的？ g ++或clang ++？

As usual, the question is: who's right? g++ or clang++?

-编辑-正如HolyBlackCat所指出的（谢谢！），一些较旧版本的clang ++会在以下位置编译此代码：

-- EDIT -- As pointed by HolyBlackCat (thanks!), some older version of clang++ compile this code where the newer don't.

我尝试了Wandbox，发现clang ++从3.4编译（第一个支持 std :: make_index_sequence / std :: index_sequence ）转换为3.8.1。从3.9.1开始会出现上述错误。

I've tried with Wandbox and I see that clang++ compile from 3.4 (the first version supporting std::make_index_sequence/std::index_sequence) to 3.8.1. Starting from 3.9.1 gives the preceding error.

-编辑2-请注意，clang ++编译错误似乎严格限制了使用范围

-- EDIT 2 -- Observe that the clang++ compilation error seems strictly bounded to the use of the first template argument in definition of the default value for the second.

实际上，更改

template <std::size_t N, typename = std::make_index_sequence<N>>
struct foo;

in

// ........................... now doesn't depends from N -->VVV
template <std::size_t N, typename = std::make_index_sequence<10u>>
struct foo;

两个编译器都可以编译。

both compilers compile.

推荐答案

这显然是某种Clang / libc ++错误：类型 std :: make_index_sequence<…> 不是 __ make_integer_seq ，它是… std :: index_sequence<…> 。类型别名（和别名模板）是透明的，并且推导对 std :: vector 一直有效，尽管其默认为（分配器）模板参数。

This is plainly some sort of Clang/libc++ bug: the type std::make_index_sequence<…> isn’t __make_integer_seq, it’s… std::index_sequence<…>. Type aliases (and alias templates) are transparent, and deduction has always worked for std::vector despite its default (allocator) template argument.

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