Question on friend functions and accessing private members

[Hans De Winter]

Why does my compiler complain with the message "`int*My::Test::_shape'
is private" when I try to compile the following code? It seems it has
something to do with the namespace since when I leave that out then
everything works fine.

Many thanks for your help.
Hans

// START CODE
#include <iostream>

namespace My
{ // Begin of namespace 'My'

class Test
{
private:
int* _shape;
public:

friend std:stream& operator<<(std:stream&, Test&);
};

} // End of namespace 'My'



std:stream& operator<<(std:stream& s, My::Test& t)
{
t._shape = 0;
return s;
}



int main(int argc, const char* argv[])
{
return 0;
}

// END CODE

 

[Jonathan Turkanis]

Why does my compiler complain with the message "`int*My::Test::_shape'
is private" when I try to compile the following code? It seems it has
something to do with the namespace since when I leave that out then
everything works fine.

Many thanks for your help.
Hans

// START CODE
#include <iostream>

namespace My
{ // Begin of namespace 'My'

class Test
{
private:
int* _shape;
public:

friend std:stream& operator<<(std:stream&, Test&);
};

} // End of namespace 'My'

std:stream& operator<<(std:stream& s, My::Test& t)
{
t._shape = 0;
return s;
}

The problem is the operator<< you've decalred to be a friend is in the
namespace My.

Try:

namespace My { class Test; }
friend std:stream& operator<<(std:stream&, My::Test&);

class Test {
friend std:stream& :perator<<(std:stream&, Test&);
// ...
};

HTH

Jonathan

 

[Ian]

The problem is the operator<< you've decalred to be a friend is in the
namespace My.

Try:

namespace My { class Test; }
friend std:stream& operator<<(std:stream&, My::Test&);

class Test {
friend std:stream& :perator<<(std:stream&, Test&);
// ...
};

Or just


std:stream& My:perator<<(std:stream& s, My::Test& t)
{
t._shape = 0;
return s;
}

Ian

 

[Sumit Rajan]

The problem is the operator<< you've decalred to be a friend is in the
namespace My.

Try:

namespace My { class Test; }
friend std:stream& operator<<(std:stream&, My::Test&);

class Test {
friend std:stream& :perator<<(std:stream&, Test&);
// ...
};

HTH

Jonathan

Or you could define << in the namespace itself.

Something like:

#include <iostream>

namespace My {

class Test {
private:
int shape;
public:
Test():shape(0){}
Test(int i):shape(i){}

friend std:stream& operator<<(std:stream&, Test&);
};


std:stream& operator<<(std:stream& s, Test& t)
{
s << t.shape;
return s;
}
}



int main(int argc, const char* argv[])
{
My::Test t(100);
std::cout << t << '\n';
}




Regards,
Sumit.

 

[Sumit Rajan]

Or you could define << in the namespace itself.

Or you could separate the implementation from the interface by declaring <<
withing the within the namespace and defining it at a different location.

#include <iostream>

namespace My {

class Test {
private:
int shape;
public:
Test():shape(0){}
Test(int i):shape(i){}

friend std:stream& operator<<(std:stream&, Test&);
};


std:stream& operator<<(std:stream& s, Test& t);

}


std:stream& My:perator<<(std:stream& s, My::Test& t)
{
s << t.shape;
return s;
}

int main(int argc, const char* argv[])
{
My::Test t(100);
std::cout << t << '\n';
}

Regards,
Sumit.