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YUY0x7
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 Posted: Sun Jul 31, 2005 1:28 am Post subject: Template specialization of an operator |
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Hi, I am having a bit of trouble with a specialization of operator<<.
Here goes:
class MyStream
{
};
template
MyStream& operator<<(MyStream& lhs, T const &)
{
cout << "operator<< T" << endl;
return lhs;
}
// specialize to work differently for std::string
template <> MyStream& operator<<(MyStream& lhs, string const &)
{
cout << "operator<< const string&" << endl;
return lhs;
}
// specialize to work differently for char const*
template <> MyStream& operator<<(MyStream& lhs, char const * const &)
{
cout << "operator<< char const*" << endl;
return lhs;
}
int main()
{
MyStream strm;
int i = 19;
string mystr = "hello";
strm << "hi" << i << mystr << "n";
return 0;
}
This outputs:
operator<< T
operator<< T
operator<< const string&
operator<< T
Why doesn't strm << "hi" (and << "n") use the specialized version?
Now I tried changing the operator to these:
template
MyStream& operator<<(MyStream& lhs, T)
{
cout << "operator<< T" << endl;
return lhs;
}
// specialize to work differently for std::string
template <> MyStream& operator<<(MyStream& lhs, string const &)
{
cout << "operator<< const string&" << endl;
return lhs;
}
// specialize to work differently for char const*
template <> MyStream& operator<<(MyStream& lhs, char const *)
{
cout << "operator<< char const*" << endl;
return lhs;
}
And the output is:
operator<< char const*
operator<< char const*
operator<< T
operator<< char const*
Now why doesn't strm << mystr use the specialized one? It looks like
it's a perfect match for mystr (and not even using a specialization for
string& works).
Can anyone tell me why this happens, and if there is a way to make this
work for both char const* and string const& ?
Thanks a lot,
George Faraj
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
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Thomas Maeder
Guest
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| Posted: Sun Jul 31, 2005 11:53 am Post subject: Re: Template specialization of an operator |
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"YUY0x7" <gfaraj (AT) gmail (DOT) com> writes:
| Quote: | class MyStream
{
};
template <typename T
MyStream& operator<<(MyStream& lhs, T const &)
{
cout << "operator<< T" << endl;
return lhs;
}
// specialize to work differently for std::string
template <> MyStream& operator<<(MyStream& lhs, string const &)
{
cout << "operator<< const string&" << endl;
return lhs;
}
// specialize to work differently for char const*
template <> MyStream& operator<<(MyStream& lhs, char const * const &)
{
cout << "operator<< char const*" << endl;
return lhs;
}
int main()
{
MyStream strm;
int i = 19;
string mystr = "hello";
strm << "hi" << i << mystr << "n";
return 0;
}
This outputs:
operator<< T
operator<< T
operator<< const string&
operator<< T
Why doesn't strm << "hi" (and << "n") use the specialized version?
|
Because the specialization for T==const char[3] generated from the
primary template is a perfect match, whereas using the specialization
for char const * const & would require a conversion.
Do you really need specialization? If you can do with overloading:
// overload to work differently for std::std::string
MyStream& operator<<(MyStream& lhs, std::string const &)
{
std::cout << "operator<< const std::string&" << std::endl;
return lhs;
}
// overload to work differently for char const*
MyStream& operator<<(MyStream& lhs, char const *)
{
std::cout << "operator<< char const*" << std::endl;
return lhs;
}
, the behavior will be what you expected in the first place.
| Quote: | Now I tried changing the operator to these:
template
MyStream& operator<<(MyStream& lhs, T)
{
cout << "operator<< T" << endl;
return lhs;
}
// specialize to work differently for std::string
template <> MyStream& operator<<(MyStream& lhs, string const &)
{
cout << "operator<< const string&" << endl;
return lhs;
}
// specialize to work differently for char const*
template <> MyStream& operator<<(MyStream& lhs, char const *)
{
cout << "operator<< char const*" << endl;
return lhs;
}
And the output is:
operator<< char const*
operator<< char const*
operator<< T
operator<< char const*
|
Are you sure? I get
operator<< char const*
operator<< T
operator<< T
operator<< char const*
If strm << i invokes the char const * version, your compiler has a
problem.
| Quote: | Now why doesn't strm << mystr use the specialized one? It looks like
it's a perfect match for mystr (and not even using a specialization
for string& works).
|
It's not a perfect match; the argument's type is std::string, whereas
the parameter type is std::string const &. If you change the parameter
type to std::string, the specialization will be selected; otherwise,
the specialization generated from the primary template is a better
match.
| Quote: | Can anyone tell me why this happens, and if there is a way to make
this work for both char const* and string const& ?
|
As mentioned above, consider using overloading instead of
specialization.
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
[ comp.lang.c++.moderated. First time posters: Do this! ]
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YUY0x7
Guest
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| Posted: Mon Aug 01, 2005 10:27 am Post subject: Re: Template specialization of an operator |
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Got it, thanks.
Yeah, the output sample I gave was incorrect, it was what you said.
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
[ comp.lang.c++.moderated. First time posters: Do this! ]
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