error: declaration of `operator==' as non-function

[Alex Vinokur]

Hi,

GNU g++ 3.4 detects an error in code below.
What is wrong here?

--------- [C++] foo.cpp : BEGIN ---------

template <typename T>
struct Foo
{
friend bool operator== <T> (const Foo<T>&, const Foo<T>&); // Line#4
};

template <typename T>
bool operator== (const Foo<T>&, const Foo<T>&)
{
return false;
}

int main()
{
Foo<int> foo;
return 0;
}

--------- [C++] foo.cpp : END -----------

------ Compilation with GNU gcc 3.3 (Cygwin) : BEGIN ------

$ g++ --version
g++ (GCC) 3.3.3 (cygwin special)
[---omitted---]

$ g++ foo.cpp
// No errors/warnings

------ Compilation with GNU gcc 3.3 (Cygwin) : END --------

------ Compilation with GNU gcc 3.4 (DJGPP) : BEGIN ------

$ gpp --version
gpp.exe (GCC) 3.4.1
[---omitted---]
$ gpp foo.cpp

foo.cpp:4: error: declaration of `operator==' as non-function
foo.cpp:4: error: expected `;' before '<' token
------ Compilation with GNU gcc 3.4 (DJGPP) : END --------

--
Alex Vinokur
http://mathforum.org/library/view/10978.html
http://sourceforge.net/users/alexvn

[Sharad Kala]

"Alex Vinokur" <al****@big-foot.com> wrote in message

Hi,

GNU g++ 3.4 detects an error in code below.
What is wrong here?

--------- [C++] foo.cpp : BEGIN ---------

template <typename T>
struct Foo
{
friend bool operator== <T> (const Foo<T>&, const Foo<T>&); // Line#4

Shouldn't it be just - friend bool operator== (const Foo<T>&, const
Foo<T>&); // Line#4

Sharad

[Alex Vinokur]

"Sharad Kala" <no************ *****@yahoo.com > wrote in message news:2s******** *****@uni-berlin.de...

"Alex Vinokur" <al****@big-foot.com> wrote in message

Hi,

GNU g++ 3.4 detects an error in code below.
What is wrong here?

--------- [C++] foo.cpp : BEGIN ---------

template <typename T>
struct Foo
{
friend bool operator== <T> (const Foo<T>&, const Foo<T>&); // Line#4

Shouldn't it be just - friend bool operator== (const Foo<T>&, const
Foo<T>&); // Line#4

Sharad

If I put line
friend bool operator== (const Foo<T>&, const Foo<T>&); // // Line#4
instead
friend bool operator== <T> (const Foo<T>&, const Foo<T>&); // Line#4
GNU gpp 3.4.1 produces the following warnings:

foo.cpp:4: warning: friend declaration `bool operator==(cons t Foo<T>&, const
oo<T>&)' declares a non-template function
foo.cpp:4: warning: (if this is not what you intended, make sure the function
template has already been declared and add <> after the function name here) -W
-non-template-friend disables this warning

Because I need template operator== the program should be rewritten as following:

--------- [C++] foo.cpp : BEGIN ---------
template <typename T>
struct Foo;

template <typename T>
bool operator== (const Foo<T>&, const Foo<T>&);

template <typename T>
struct Foo
{
friend bool operator== <> (const Foo<T>&, const Foo<T>&);
};

template <typename T>
bool operator== (const Foo<T>&, const Foo<T>&)
{
return false;
}

int main()
{
Foo<int> foo;
return 0;
}
--------- [C++] foo.cpp : END -----------

Now GNU gpp 3.4.1 accepts it.

Thanks.

--
Alex Vinokur
http://mathforum.org/library/view/10978.html
http://sourceforge.net/users/alexvn