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Language Subtlety or Compiler bug?

P: n/a
In VC7.1 the following code outputs:
00000000
Success!

It would seem that the Y::operator<< is hiding the global operator<<
when it is outputting an object from another namespace but not when it
is outputting an object in the global namespace.

I can't help but think one of these 2 behaviors is a bug in VC7.1, but
which one?

Thanks,

Ned.


=========================================
#include <iostream>

namespace X
{
class Abc1
{
};

}
std::ostream& operator<< (std::ostream& strm, const X::Abc1 *node)
{
return strm << "Success!";
}

X::Abc1 * GetDoc() { return NULL; }

class Abc2
{
};

Abc2 * GetDoc2() { return NULL; }

std::ostream& operator<< (std::ostream& strm, const Abc2 *node)
{
return strm << "Success!";
}

namespace Y
{
std::ostream & operator<<(std::ostream& a, const char& b)
{
return a << "This should not run";
}

void DoTest()
{
std::cout << GetDoc() << "\n";
std::cout << GetDoc2() << "\n";
}
}

int main(int argc, char* argv[])
{
Y::DoTest();
return 0;
}
Jul 22 '05 #1
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7 Replies


P: n/a

"Ned Harding" <nh******@extendthereach.com> wrote in message
news:34**************************@posting.google.c om...
In VC7.1 the following code outputs:
00000000
Success!

It would seem that the Y::operator<< is hiding the global operator<<
when it is outputting an object from another namespace but not when it
is outputting an object in the global namespace.

I don't see the operator in the Y namespace coming into play at all here.
What makes you think it is? You're not getting "This should not run" as
your output. Plus, the operator in Y uses a reference to a char, not a
pointer - of *any* type!
=========================================
#include <iostream>

namespace X
{
class Abc1
{
};

}
std::ostream& operator<< (std::ostream& strm, const X::Abc1 *node)
{
return strm << "Success!";
}

X::Abc1 * GetDoc() { return NULL; }

class Abc2
{
};

Abc2 * GetDoc2() { return NULL; }

std::ostream& operator<< (std::ostream& strm, const Abc2 *node)
{
return strm << "Success!";
}

namespace Y
{
std::ostream & operator<<(std::ostream& a, const char& b)
{
return a << "This should not run";
}

void DoTest()
{
std::cout << GetDoc() << "\n";
std::cout << GetDoc2() << "\n";
}
}

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


It looks to me like your operator in the global namespace is correctly
getting called for the case where you've passing it an Abc2*. That's normal
overloading (correct?). In the case where you pass an Abc1*, however, your
code apparently isn't seeing the overloaded operator that uses an Abc1*, but
instead is calling the << normally used for a void* pointer.

I'm no expert on why that is, but if I had to guess (which I do :-)), I'd
say it's because there is nothing specifying that you want to use the
operator << that's in the X namespace. I'm not even sure how you would do
that. Perhaps using X:<< ?

-Howard
Jul 22 '05 #2

P: n/a

"Howard" <al*****@hotmail.com> wrote in message
news:sC3Zc.537788$Gx4.147573@bgtnsc04-> I'm no expert on why that is, but if
I had to guess (which I do :-)), I'd
say it's because there is nothing specifying that you want to use the
operator << that's in the X namespace. I'm not even sure how you would do
that. Perhaps using X:<< ?

I meant to say "X::<<" there. (But as I said, it was just a guess. :-))
-Howard

Jul 22 '05 #3

P: n/a
Howard wrote:
"Howard" <al*****@hotmail.com> wrote in message
news:sC3Zc.537788$Gx4.147573@bgtnsc04-> I'm no expert on why that is, but if
I had to guess (which I do :-)), I'd
say it's because there is nothing specifying that you want to use the
operator << that's in the X namespace. I'm not even sure how you would do
that. Perhaps using X:<< ?

I meant to say "X::<<" there. (But as I said, it was just a guess. :-))


I think that :: is allowed only with the name. << is not a name, so it
probably should be X::operator<<. But I am just adding a guess on top
of your guess...

Victor
Jul 22 '05 #4

P: n/a
nh******@extendthereach.com (Ned Harding) wrote:
In VC7.1 the following code outputs:
00000000
Success!

It would seem that the Y::operator<< is hiding the global operator<<
when it is outputting an object from another namespace but not when it
is outputting an object in the global namespace.
Actually the behaviour is correct.
Also the Y::operator<< is irrelevant, the code will behave the same
without it.

[Abbreviated, equivalent code]: #include <iostream>

namespace X { class Abc1 {} };

std::ostream& operator<< (std::ostream& strm, const X::Abc1 *node)
{ return strm << "Success!"; }

class Abc2 { };

std::ostream& operator<< (std::ostream& strm, const Abc2 *node)
{ return strm << "Success!"; }

namespace Y {
void DoTest() {
std::cout << (X::Abc1 *)0 << "\n";
std::cout << (Abc2 *)0 << "\n";
}
}

int main() {
Y::DoTest();
}
The first example looks for:
operator<< (std::ostream &, X::Abc1 *)
in the namespaces 'std' and 'X'. It finds neither.
It doesn't look in the global namespace because none of its
parameter types are in the global namespace.

However there is an implicit conversion (X::Abc1 *) --> (void *)
so it finds std::operator<<(std::ostream &, void *).

The second example looks for:
operator<< (std::ostream &, ::Abc2 *)
in the namespace 'std' and the global namespace. It finds
your std::ostream& operator<< (std::ostream& strm, const Abc2 *node)
from the global namespace.
namespace Y {
std::ostream & operator<<(std::ostream& a, const char& b)
};


Why did you include that function? It has nothing to do with
anything else in this program because you never go
std::cout << (something with char type or convertible to char).
Jul 22 '05 #5

P: n/a
ol*****@inspire.net.nz (Old Wolf) wrote in message
Why did you include that function? It has nothing to do with
anything else in this program because you never go
std::cout << (something with char type or convertible to char).


That's the weird part. If you take out Y::operator << (which I agree
never gets called) then I get

Success!
Success!

I'm not sure why y::operator << is hiding anything, but if it hides
one of the ::operator<<, why doesn't it hide both?

ned.
Jul 22 '05 #6

P: n/a

"Ned Harding" <nh******@extendthereach.com> wrote in message
news:34**************************@posting.google.c om...
ol*****@inspire.net.nz (Old Wolf) wrote in message
Why did you include that function? It has nothing to do with
anything else in this program because you never go
std::cout << (something with char type or convertible to char).


That's the weird part. If you take out Y::operator << (which I agree
never gets called) then I get

Success!
Success!

I'm not sure why y::operator << is hiding anything, but if it hides
one of the ::operator<<, why doesn't it hide both?

ned.


Try moving the operator that takes the X::Abc1* parameter INSIDE the X
namespace. Then you won't have any problem.

BTW, I have no idea how it's possible to "hide" the operator like you've
described, which I've verified with all kinds of variations of that Y::
operator. But even though it's never called, it does indeed prevent VC7
from "seeing" the global namespace operator that takes a parameter type from
another namespace. But I tend not to sweat stuff like that, once I've found
a solution! :-)

-Howard


Jul 22 '05 #7

P: n/a
nh******@extendthereach.com (Ned Harding) wrote:
ol*****@inspire.net.nz (Old Wolf) wrote in message
Why did you include that function? It has nothing to do with
anything else in this program because you never go
std::cout << (something with char type or convertible to char).
That's the weird part. If you take out Y::operator << (which I agree
never gets called) then I get

Success!
Success!


I don't (gcc 3.4.1). Perhaps it is a compiler bug in MSVC.
(On BCC 5.5.1, both versions give Success! Success!, I suppose it
looks in the global namespace when it should not.. BCC is not
famous for its lack of bugs)
I'm not sure why y::operator << is hiding anything, but if it hides
one of the ::operator<<, why doesn't it hide both?

Jul 22 '05 #8

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