全局作用域与全局命名空间

In C++, every name has its scope outside which it doesn't exist. A scope can be defined by many ways : it can be defined by namespace, functions, classes and just { }.

所以一个命名空间，无论是全局的还是其他的，定义了一个范围.全局命名空间是指使用::，在这个命名空间中定义的符号被称为具有全局作用域.默认情况下，符号存在于全局命名空间中，除非它定义在以关键字 namespace 开头的块中，或者它是类的成员，或者函数的局部变量:

So a namespace, global or otherwise, defines a scope. The global namespace refers to using ::, and the symbols defined in this namespace are said to have global scope. A symbol, by default, exists in a global namespace, unless it is defined inside a block starts with keyword namespace, or it is a member of a class, or a local variable of a function:

int a; //this a is defined in global namespace
       //which means, its scope is global. It exists everywhere.

namespace N
{
     int a;  //it is defined in a non-global namespace called `N`
             //outside N it doesn't exist.
}
void f()
{
   int a;  //its scope is the function itself.
           //outside the function, a doesn't exist.
   {
        int a; //the curly braces defines this a's scope!
   }
}
class A
{
   int a;  //its scope is the class itself.
           //outside A, it doesn't exist.
};

另请注意，name 可以被命名空间、函数或类定义的内部作用域隐藏.因此，名称空间 N 中的名称 a 在全局命名空间中隐藏了名称 a.同样，函数和类中的名称隐藏了全局命名空间中的名称.如果你遇到这种情况，那么你可以使用::a来引用全局命名空间中定义的名字:

Also note that a name can be hidden by inner scope defined by either namespace, function, or class. So the name a inside namespace N hides the name a in the global namspace. In the same way, the name in the function and class hides the name in the global namespace. If you face such situation, then you can use ::a to refer to the name defined in the global namespace:

int a = 10;

namespace N
{
    int a = 100;

    void f()
    {
         int a = 1000;
         std::cout << a << std::endl;      //prints 1000
         std::cout << N::a << std::endl;   //prints 100 
         std::cout << ::a << std::endl;    //prints 10
    }
}

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