c++ namespace 疑问

[sx666777888]

代码如下
namespace.h

namespace my
{
        doule fetch = 1.2;
}

main1.cpp

using namespace std;

using namespace my;
double fetch = 1.3;
int main ()
{
    cout<<"fetch "<<fetch<<endl;
}

main2.cpp

using namespace std;
int main ()
{
    using namespace my;
    double fetch = 1.3;
    cout<<"fetch "<<fetch<<endl;
}

其中main1编译会报错fetch重定义了，main2确可以编译通过，其实main2我大概明白，局部变量隐藏了名称空间的变量，但是main1为什么不是一样的机制呢？
请大家指教，多谢

[xskxzr]

http://en.cppreference.com/w/cpp/language/namespace#Using-directives

引用

From the point of view of unqualified name lookup of any name after a using-directive and until the end of the scope in which it appears, every name from namespace-name is visible as if it were declared in the nearest enclosing namespace which contains both the using-directive and namespace-name.

因此就这两种情况而言，my::fetch 都如同在全局空间声明。所以main1冲突，main2是隐藏。

[sx666777888]

引用 1 楼 xsklld 的回复:

http://en.cppreference.com/w/cpp/language/namespace#Using-directives

引用

From the point of view of unqualified name lookup of any name after a using-directive and until the end of the scope in which it appears, every name from namespace-name is visible as if it were declared in the nearest enclosing namespace which contains both the using-directive and namespace-name.

因此就这两种情况而言，my::fetch 都如同在全局空间声明。所以main1冲突，main2是隐藏。

多谢大师指点，明白了，就是说在main的作用域内，main2中的using namespace my的作用和在外部是一样的效果，但是非main作用域的效果还是不一样的。