支持std :: map成员编译器初始化 [英] Braced initialisation of std::map member compiler error

问题描述

以下代码不使用Visual Studio 2013编译。它使用Xcode 6.1（Clang 3.5）编译。

  std :: string s1（one）; 
 std :: string s2（two）; 
 std :: string s3（three）; 
 std :: string s4（four）; 
 
 class X 
 {
 typedef std :: map< std :: string，std :: string> MyMapType; 
 MyMapType map1 = {{s1，s2}，{s3，s4}}; 
 MyMapType map2 = {{std :: make_pair（s1，s2）}，{std :: make_pair（s3，s4）}}; 
}; 
  

报告的两个声明的错误是：

 错误C2664：'std :: map< std :: string，std :: string，std :: less< _Kty>，std :: allocator< std :: pair< const _Kty ，_Ty>>> :: map（std :: initializer_list< std :: pair< const _Kty，_Ty>>，const std :: less< _Ty>&，const std :: allocator< std :: pair< ; const _Kty，_Ty>>&）'：不能将参数2从'initializer-list'转换为'const std :: allocator< std :: pair< const _Kty，_Ty& &'
  

但是，以下操作会编译：

$ b b

  int main（）
 {
 typedef std :: map< std :: string，std :: string> MyMapType; 
 MyMapType map3 = {{s1，s2}，{s3，s4}}; 
 
 return 0; 
} 
  

请有人解释这一点。

他们只是在Visual
Studio 2013 Update 3中的所有情况下的编译器错误（运送实际修订显然被认为对于更新而言风险太大）。

您的代码可以通过 Microsoft的在线编译器进行编译，运行Visual C ++ 2015的预览版本，所以看起来这个问题已修复。

一个解决方法（在上面链接的MSDN页面中注明）也明确指定了RHS上的类型，不实际列表初始化非静态数据成员。

  MyMapType map1 = MyMapType {{s1，s2}，{s3， s4}}; 
 MyMapType map2 = MyMapType {{std :: make_pair（s1，s2）}，{std :: make_pair（s3，s4）}}; 
  

The following code does not compile using Visual Studio 2013. It does compile using Xcode 6.1 (Clang 3.5).

std::string s1("one");
std::string s2("two");
std::string s3("three");
std::string s4("four");

class X
{
    typedef std::map<std::string, std::string> MyMapType;
    MyMapType map1 = { { s1, s2 }, { s3, s4 } };
    MyMapType map2 = { { std::make_pair(s1, s2) }, { std::make_pair(s3, s4) } };
};

The error reported for both declarations is:

error C2664: 'std::map<std::string,std::string,std::less<_Kty>,std::allocator<std::pair<const _Kty,_Ty>>>::map(std::initializer_list<std::pair<const _Kty,_Ty>>,const std::less<_Ty> &,const std::allocator<std::pair<const _Kty,_Ty>> &)' : cannot convert argument 2 from 'initializer-list' to 'const std::allocator<std::pair<const _Kty,_Ty>> &'

However, the following does compile:

int main()
{
    typedef std::map<std::string, std::string> MyMapType;
    MyMapType map3 = { { s1, s2 }, { s3, s4 } };

    return 0;
}

Please can someone explain this.

解决方案

Visual C++ 2013 is known to be buggy with its handling of list initialization in non-static data member initializers and constructor member initializer lists. It was so badly broken - in some cases causing silent bad codegen - that they simply made it a compiler error in all cases in Visual Studio 2013 Update 3 (shipping an actual fix was apparently deemed too risky for an update).

Your code compiles fine with Microsoft's online compiler, which runs a preview version of Visual C++ 2015, so it looks like this has been fixed.

A workaround (noted in the MSDN page linked above) is explicitly specifying the type on the RHS as well, so that you are not actually list-initializing the non-static data member.

MyMapType map1 = MyMapType{ { s1, s2 }, { s3, s4 } };
MyMapType map2 = MyMapType{ { std::make_pair(s1, s2) }, { std::make_pair(s3, s4) } };

这篇关于支持std :: map成员编译器初始化的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！