可以包装或合并单独的命名空间？ [英] Possible to wrap or merge separate namespaces?

问题描述

现在，寻找说明我找不到他们 - 甚至搜索使用搜索项组合，分组，合并和包装我不想出任何东西。也许我误解了我之前看到的。

但是，从两个名称空间开始，我们没有一个特定的应用程序，这只是一个好奇心， ...

 命名空间a {int func（）{return 1;}} 
 namespace b {int func ）{return 2;}} 
  

我在寻找语法要么简单地将它们换成另一个名称 - - 事后 -
（是的，我知道我可以重写它在一个嵌套的方式）或将它们合并到一个新的空间。但是，我确实发现，如果我添加到一个命名空间很有效。

  namespace c {namespace a {int func2（）{return 3;}}} 
 
 
 int main（int argc，char ** argv）
 {
 int a = a :: func ）; // normal case 
 int c = c :: a :: func2（）; // wrap并添加到
 
 // int c = c :: func2（）; //不工作
 // int d = a :: func2（）; //不工作
} 
  

问题是：

1）是否有将两个空格合并成一个新空格的语法？

2）

解决方案

您可以这样做：

 命名空间c 
 {
使用命名空间a; 
 using namespace b; 
} 
  

但是如果a和b有相同名称的元素，能够从命名空间c。

中使用它们

I seem to recall seeing notes somewhere on a way to combine multiple namespaces into one.

Now, looking for said notes I am not finding them -- even searching using search terms combing, grouping, merging and wrapping I'm not coming up with anything. Maybe I misunderstood what I saw before. I don't have a specific application for this, it's just a curiosity and it's a bit contrived.

But, starting with two name spaces...

namespace a {int func() {return 1;}}
namespace b {int func() {return 2;}}

I was looking for syntax to either simply wrap them in another name -- after the fact -- (yes, I know I can rewrite it in a nested way) or merge them into one new space. But, I did find that I if I add to one of the namespaces that much works.

namespace c {namespace a{ int func2() {return 3;}} }


int main(int argc, char **argv)
{
    int a = a::func();          // normal case
    int c = c::a::func2();      // wrapped and added to

    //int c = c::func2();       // doesn't work
    //int d = a::func2();       // doesn't work
}

The question are:

1) is there syntax that just combines the two spaces into one new one?

2) is there a syntax to wrap the spaces without adding more to the subspaces?

解决方案

You can do this:

namespace c
{
    using namespace a;
    using namespace b;
}

But if a and b have elements with the same names, you won't be able to use them from namespace c.

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