班级成员初始化程序在VS 2013中失败 [英] In-class member initializer fails with VS 2013

问题描述

我希望可以编译以下代码，但是Visual Studio 2013 Update 2给我一个错误，而g ++ 4.7可以很好地编译。

I expected the following code to compile, but Visual Studio 2013 Update 2 gives me an error, while g++ 4.7 compiles it fine.

using std::vector;
using std::string;

struct Settings
{
    vector<string> allowable = { "-t", "--type", "-v", "--verbosity" };
};

VS 2013编译失败：

VS 2013 compile fails with:

'std :: vector< std :: string，std :: allocator< _Ty>> :: vector'：没有重载函数需要4参数

'std::vector<std::string,std::allocator<_Ty>>::vector' : no overloaded function takes 4 arguments

如果我按以下方式更改成员，则可以正常编译：

If I change the member as follows then it compiles fine:

vector<string> allowable = vector<string> { "-t", "--type", "-v", "--verbosity" };

我看过建议从Bjarne的常见问题解答，我查看了MSDN 页面关于VS 2013中已完成的C ++ 11功能，但我仍然感到困惑。它应该按原样编译，还是我错了，必须两次指定类型？

I've looked at the proposal linked from Bjarne's FAQ and I've looked at the MSDN page on completed C++11 features in VS 2013 but I'm still confused. Should it compile as is, or am I wrong and must specify the type twice?

推荐答案

• 您显示的示例是完全有效的C ++，但没有适用于VC ++ 2013。

  • The example that you showed is perfectly valid C++, however it doesn't work for VC++2013.

    这是已知的 VC ++ 2013错误，自2013年10月31日以来已报告，其状态仍然有效。

    This is a known VC++2013 bug reported since 31/10/2013 and its status is still active.

    但是，您可以通过解决方法来克服它。正如 @ildjarn 所建议的那样，只需简单地添加一对大括号即可，您可以强制 initializer_list<>引发 std :: vector 的构造函数，而不是其fill构造函数，如下例所示：

    However, you can surmount it by doing a work-around. As @ildjarn suggested, by simply putting an extra pair of curly braces you force initializer_list<> constructor of the std::vector to be evoked instead of its fill constructor, like the example below:

       #include <string>
       #include <vector>
       #include <iostream>
    
       struct Settings {
         std::vector<std::string> allowable = {{"-t", "--type", "-v", "--verbosity"}};
       };
    
       int main() {
         Settings s;
         for (auto i : s.allowable) std::cout << i << " ";
         std::cout << std::endl;
       }
    

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