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
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Jump to PostThe operators like . , -> and * can not be overloaded.
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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
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