Hey men is it possible to build your own operator with your own definition and your own symbol for every type?
massivefermion 3 Junior Poster
kbshibukumar 1 Junior Poster in Training
The operators like . , -> and * can not be overloaded.
Narue 5,707 Bad Cop Team Colleague
>it possible to build your own operator with your own definition and your own symbol for every type?
Yes, it's called a member function with an informative name. But if you're thinking about creating something like ** for exponentiation or ^^ for a logical exclusive OR, you're out of luck. You can only overload existing C++ operators that support overloading.
ArkM 1,090 Postaholic
It's possible (to some extent) in Algol 68.
Download Algol 68 interpreter and try to make the most cryptic code in the World: ;)
http://www.xs4all.nl/~jmvdveer/algol.html
Be a part of the DaniWeb community
We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.