unnamed namespace results in value being improper
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I am on windows 7 machine with visual studio 2015 community edition .
Consider the following code:
#include "stdafx.h"
#include "iostream"
using namespace std;
namespace
{
int y=4;
int x=6;
}
int u = ::y;
int y = ::y;
int x = 567;
int main()
{
cout << u << "n";
cout << ::x << "n";
//cout << y << "n";
cout<< ::y << "n";
int y2;
cin >> y2;
return 0;
}
The result of the program is :
4
567
0
If we see u and x are right but y is wrong . why ?
c++
edited Nov 25 '18 at 5:40
rsjaffe
3,98671632
asked Nov 25 '18 at 2:57
MAGMAG
1,00321630
add a comment |
I am on windows 7 machine with visual studio 2015 community edition .
Consider the following code:
#include "stdafx.h"
#include "iostream"
using namespace std;
namespace
{
int y=4;
int x=6;
}
int u = ::y;
int y = ::y;
int x = 567;
int main()
{
cout << u << "n";
cout << ::x << "n";
//cout << y << "n";
cout<< ::y << "n";
int y2;
cin >> y2;
return 0;
}
The result of the program is :
4
567
0
If we see u and x are right but y is wrong . why ?
c++
edited Nov 25 '18 at 5:40
rsjaffe
3,98671632
asked Nov 25 '18 at 2:57
MAGMAG
1,00321630
add a comment |
I am on windows 7 machine with visual studio 2015 community edition .
Consider the following code:
#include "stdafx.h"
#include "iostream"
using namespace std;
namespace
{
int y=4;
int x=6;
}
int u = ::y;
int y = ::y;
int x = 567;
int main()
{
cout << u << "n";
cout << ::x << "n";
//cout << y << "n";
cout<< ::y << "n";
int y2;
cin >> y2;
return 0;
}
The result of the program is :
4
567
0
If we see u and x are right but y is wrong . why ?
c++
edited Nov 25 '18 at 5:40
rsjaffe
3,98671632
asked Nov 25 '18 at 2:57
MAGMAG
1,00321630
I am on windows 7 machine with visual studio 2015 community edition .
Consider the following code:
#include "stdafx.h"
#include "iostream"
using namespace std;
namespace
{
int y=4;
int x=6;
}
int u = ::y;
int y = ::y;
int x = 567;
int main()
{
cout << u << "n";
cout << ::x << "n";
//cout << y << "n";
cout<< ::y << "n";
int y2;
cin >> y2;
return 0;
}
The result of the program is :
4
567
0
If we see u and x are right but y is wrong . why ?
c++
c++
edited Nov 25 '18 at 5:40
rsjaffe
3,98671632
asked Nov 25 '18 at 2:57
MAGMAG
1,00321630
edited Nov 25 '18 at 5:40
rsjaffe
3,98671632
asked Nov 25 '18 at 2:57
MAGMAG
1,00321630
edited Nov 25 '18 at 5:40
rsjaffe
3,98671632
edited Nov 25 '18 at 5:40
rsjaffe
3,98671632
edited Nov 25 '18 at 5:40
rsjaffe
3,98671632
3,98671632
asked Nov 25 '18 at 2:57
MAGMAG
1,00321630
asked Nov 25 '18 at 2:57
MAGMAG
1,00321630
asked Nov 25 '18 at 2:57
MAGMAG
1,00321630
1,00321630
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
int y = ::y
Here the second y is the same as the first - the variable is initialized with its own value. Since it's a global variable, its own value is zero.
Once a declaration introduces a name y into the global namespace, there's no syntax to access y from the unnamed namespace. A variable declaration is visitble from its own initializer.
answered Nov 25 '18 at 2:59
Igor TandetnikIgor Tandetnik
33.5k33659
But the u is properly referencing properly to 4 . Shouldn't compiler either complain or give right value 4.
– MAG
Nov 25 '18 at 3:04
2
At the point whereuis declared, only the declaration ofyfrom unnamed namespace is visible, so that's the one being used.
– Igor Tandetnik
Nov 25 '18 at 3:14
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
int y = ::y
Here the second y is the same as the first - the variable is initialized with its own value. Since it's a global variable, its own value is zero.
Once a declaration introduces a name y into the global namespace, there's no syntax to access y from the unnamed namespace. A variable declaration is visitble from its own initializer.
answered Nov 25 '18 at 2:59
Igor TandetnikIgor Tandetnik
33.5k33659
But the u is properly referencing properly to 4 . Shouldn't compiler either complain or give right value 4.
– MAG
Nov 25 '18 at 3:04
2
At the point whereuis declared, only the declaration ofyfrom unnamed namespace is visible, so that's the one being used.
– Igor Tandetnik
Nov 25 '18 at 3:14
add a comment |
int y = ::y
Here the second y is the same as the first - the variable is initialized with its own value. Since it's a global variable, its own value is zero.
Once a declaration introduces a name y into the global namespace, there's no syntax to access y from the unnamed namespace. A variable declaration is visitble from its own initializer.
answered Nov 25 '18 at 2:59
Igor TandetnikIgor Tandetnik
33.5k33659
But the u is properly referencing properly to 4 . Shouldn't compiler either complain or give right value 4.
– MAG
Nov 25 '18 at 3:04
2
At the point whereuis declared, only the declaration ofyfrom unnamed namespace is visible, so that's the one being used.
– Igor Tandetnik
Nov 25 '18 at 3:14
add a comment |
int y = ::y
Here the second y is the same as the first - the variable is initialized with its own value. Since it's a global variable, its own value is zero.
Once a declaration introduces a name y into the global namespace, there's no syntax to access y from the unnamed namespace. A variable declaration is visitble from its own initializer.
answered Nov 25 '18 at 2:59
Igor TandetnikIgor Tandetnik
33.5k33659
int y = ::y
Here the second y is the same as the first - the variable is initialized with its own value. Since it's a global variable, its own value is zero.
Once a declaration introduces a name y into the global namespace, there's no syntax to access y from the unnamed namespace. A variable declaration is visitble from its own initializer.
answered Nov 25 '18 at 2:59
Igor TandetnikIgor Tandetnik
33.5k33659
answered Nov 25 '18 at 2:59
Igor TandetnikIgor Tandetnik
33.5k33659
answered Nov 25 '18 at 2:59
Igor TandetnikIgor Tandetnik
33.5k33659
answered Nov 25 '18 at 2:59
Igor TandetnikIgor Tandetnik
33.5k33659
33.5k33659
But the u is properly referencing properly to 4 . Shouldn't compiler either complain or give right value 4.
– MAG
Nov 25 '18 at 3:04
2
At the point whereuis declared, only the declaration ofyfrom unnamed namespace is visible, so that's the one being used.
– Igor Tandetnik
Nov 25 '18 at 3:14
add a comment |
But the u is properly referencing properly to 4 . Shouldn't compiler either complain or give right value 4.
– MAG
Nov 25 '18 at 3:04
2
At the point whereuis declared, only the declaration ofyfrom unnamed namespace is visible, so that's the one being used.
– Igor Tandetnik
Nov 25 '18 at 3:14
But the u is properly referencing properly to 4 . Shouldn't compiler either complain or give right value 4.
– MAG
Nov 25 '18 at 3:04
But the u is properly referencing properly to 4 . Shouldn't compiler either complain or give right value 4.
– MAG
Nov 25 '18 at 3:04
2
2
At the point where
u is declared, only the declaration of y from unnamed namespace is visible, so that's the one being used.– Igor Tandetnik
Nov 25 '18 at 3:14
At the point where
u is declared, only the declaration of y from unnamed namespace is visible, so that's the one being used.– Igor Tandetnik
Nov 25 '18 at 3:14
add a comment |
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