" :: " before a variable name c++

Question

#include <iostream>

using namespace std;

int d = 10;

int main()
{
    int d = 20;

    {
        int d = 30;
        cout << d << endl << ::d; // what does it mean?
    }

    return 0;
}

output is:

30
10

I don't understand why "::d" gives 10? Can someone explain it to me please?

Because the C++ group wanted to make a language so complex and confusing they would ensure their extreme hourly rates a decade from now because they would be the only ones able to work on the software.. (snark snark). — Andrew T Finnell
– Andrew T Finnell, Commented Apr 16, 2012 at 23:47
Same reason why \file.txt is not the same as \subdirectory\file.txt. — MSalters
– MSalters, Commented Apr 17, 2012 at 7:39

qehgt · Accepted Answer · 2012-04-17 00:10:22Z

22

::d means d from global namespace

EDIT: There are three different variables with similar name d. One is in global namespace d=10, one is inside scope of main function (20), and the last one is inside internal block of the main function (30).

Inside every block you have access (by name) to corresponding variable and always have access to the global namespace (by ::).

edited Apr 17, 2012 at 0:10

answered Apr 16, 2012 at 23:44

qehgt

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4 Comments

Thunfische Over a year ago

But there is 'd=20', why doesnt it change to its value ?

Andrew T Finnell Over a year ago

@İbrahimYazıcı There are three scoping levels happening in that code. The global scope, the scope in main() and the scope in the inner most {}. You have three separate variables declared in that program. Not one that you keep reassigning a value to.

Andrew T Finnell Over a year ago

@qehgt While you are correct, it might be extremely helpful to others that come across this question if you go into some detail as to the scoping rules and what is really happening in that piece of code.

qehgt Over a year ago

@Andrew Finnell thanks for the comment, I added more explanations.

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