从另一个名称空间访问标准名称空间数据成员

Question

If the code is written as below:

#include<iostream>

namespace n2    {
    int y = 10;
}

namespace n1    {
    int x =  20;
    int m = ::n2::y;

    std::string str;
}

int main()
{
    std::cout << n1::x << std::endl;
    std::cout << n2::y << std::endl;

    return 0;
}

Then my question is, if i have not make use of the statement,

using namespace std

Then I have to manually mention the data-member along with namespace as std::string (for example), but if I make use of the same data member (std::string) inside the another namespace, eg namespace n1, then will it not be implicitly changed to the namespace n1::std::string?

1. There is no n1::std::string data type, then how it is working. How n1::std::string is getting converted to ::std::string?

2. But why it remains aloof from relative usage inside another namespace. If I wanted to make use of that concept in my code, how do I change the code. If the same concept to be used for my own library, how do i make the changes, to make use of the same concept.

Answer 1

In i1::i2 , i1 is an unqualified identifier and is looked up as any other unqualified identifier. First in the current scope, then in the enclosing one until the global namespace is searched. Thus with std::string as there is no std in n1 , the std under the global namespace is found. In other words, in n1 , you can reference ::n2::y as n2::y as long as you don't have an entity named n2 in n1 .

If you want to protect against a n1::std::string entity, you'll have to use ::std::string instead of std::string .

Answer 2

From within a namespace you can see everything in that namespace, and everything in the enclosing namespace, all the way out to the global namespace. In this case namespace context has a wider meaning than the namespace keyword. It includes any context, such as classes, and functions.

From within n1, you can see std because you can see everything within the global namespace, so it is implicitly ::std

If, within n1 you were to define a namespace std {} then you would entirely obscure that global namespace std from your current context. You could still access it either by doing a 'using' rename or by explicit ::std