Make conversion to a native type explicit in C++
- by Tal Pressman
I'm trying to write a class that implements 64-bit ints for a compiler that doesn't support long long, to be used in existing code. Basically, I should be able to have a typedef somewhere that selects whether I want to use long long or my class, and everything else should compile and work.
So, I obviously need conversion constructors from int, long, etc., and the respective conversion operators (casts) to those types. This seems to cause errors with arithmetic operators. With native types, the compiler "knows" that when operator*(int, char) is called, it should promote the char to int and call operator*(int, int) (rather than casting the int to char, for example). In my case it gets confused between the various built-in operators and the ones I created.
It seems to me like if I could flag the conversion operators as explicit somehow, that it would solve the issue, but as far as I can tell the explicit keyword is only for constructors (and I can't make constructors for built-in types).
So is there any way of marking the casts as explicit? Or am I barking up the wrong tree here and there's another way of solving this? Or maybe I'm just doing something else wrong...
