Re: template operator== not working
* 3DCoderGuy:
"Alexander Grigoriev" <alegr@earthlink.net> wrote in message
news:udZOD8LsIHA.4848@TK2MSFTNGP05.phx.gbl...
And, by the way, you want the operator== to be const. Not returning const
bool, which doesn't make sense.
bool operator==(const XYZ_POINT<T> &xyzTest) const;
"3DCoderGuy" <nobody@nospam.com> wrote in message
news:uzopyzJsIHA.4476@TK2MSFTNGP06.phx.gbl...
This is my XYZ_POINT.h file
template<typename T>
class XYZ_POINT
{
private:
T x;
T y;
T z;
public:
XYZ_POINT(void) {};
XYZ_POINT(const T allVals) : x(allVals), y(allVals), z(allVals) {};
bool const operator==(const XYZ_POINT<T> &xyzTest);
}
In my cpp file I have this.
template<typename T>
bool const XYZ_POINT<T>::operator ==(const XYZ_POINT<T> &xyzTest)
{
return ((x == xyzTest.GetX()) &&(y == xyzTest.GetY()) &&(z ==
xyzTest.GetZ()));
}
In the main file I have this
int _tmain(int arc, _TCHAR* argv[])
{
XYZ_POINT<double> dXYZPnt1(1,0,0);
XYZ_POINT<double> dXYZPnt2(3,0,0);
if (dXYZPnt1 == dXYZPnt2)
{
bool b = true;
b = false;
}
}
Doesn't compile with the following error
error LNK2019: unresolved external symbol "public: bool const __thiscall
XYZ_POINT<double>::operator==(class XYZ_POINT<double> const &)"
(??8?$XYZ_POINT@N@@QAE?B_NABV0@@Z) referenced in function _main
If I move the implimentation to the header file it compiles fine.
What am I doing wrong?
The actual implimentation is more complex then I'm demonstrating and I
need to include some header files that I don't want in the declaration.
Thanks
Mark
Thanks Alexander for your reply.
The use of const on the return comes from Scott Meyers' Effective C++ Third
Edition Item 3
"const Rational operator*(const Rational& lhs, const Rational& rhs);
Many programmers squint when they first see this.
And rightly.
With the current language definition it prevents optimizations like Andrei's
Mojo move semantics.
C++0x is another matter.
Why should the result of operator* be a const object?
Ideally it shouldn't, but there is no great harm, just a loss of possible
optimization.
Because if it weren't, clents would be able to
commit atrocities like this:
Rational a, b, c;
...
(a * b) = c; // invoke operator= on the result of a*b!
..."
You might not see that example as typed but if( a * b = c) ... does occure
because of typing mistakes. The compiler will catch this.
And why would you want to prevent that?
Now this might not apply to the operator== but it is a practice I've picked
up and just haven't changed.
It would be a good idea to change it (see above).
Cheers, & hth.,
- Alf
--
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