C++ operator overloading and implicit conversion

cimplicit-conversionoperator-overloading

I have a class that encapsulates some arithmetic, let's say fixed point calculations. I like the idea of overloading arithmetic operators, so I write the following:

class CFixed
{
   CFixed( int   );
   CFixed( float );
};

CFixed operator* ( const CFixed& a, const CFixed& b )
{ ... }

It all works. I can write 3 * CFixed(0) and CFixed(3) * 10.0f. But now I realize, I can implement operator* with an integer operand much more effective. So I overload it:

CFixed operator* ( const CFixed& a, int b )
{ ... }
CFixed operator* ( int a, const CFixed& b )
{ ... }

It still works, but now CFixed(0) * 10.0f calls overloaded version, converting float to int ( and I expected it to convert float to CFixed ). Of course, I can overload a float versions as well, but it seems a combinatorial explosion of code for me. Is there any workaround (or am I designing my class wrong)? How can I tell the compiler to call overloaded version of operator* ONLY with ints?

Best Answer

You should overload with float type as well. Conversion from int to user-specified type (CFixed) is of lower priority than built-in floating-integral conversion to float. So the compiler will always choose function with int, unless you add function with float as well.

For more details, read 13.3 section of C++03 standard. Feel the pain.

It seems that I've lost track of it too. :-( UncleBens reports that adding float only doesn't solve the problem, as version with double should be added as well. But in any case adding several operators related to built-in types is tedious, but doesn't result in a combinatorial boost.

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