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Message: [QUOTE="Nikolai Borissov, post: 1462468"] Nicolay Korslund wrote: There is a rule that in a conversion sequence only one use-defined conversion operator is allowed. My guess is that the rule covers overloaded operators too. t1[0] works because there is only one use-defined conversion is involved : operator mytype () {return i;} It returns object of double* type for which built-in [] operator is defined. t2[0], however, casts through the above user-defined conversion to Test2 type, which, in turn, has only use-defined [] operator. Of course, when you remove one of user-defined operators from implicit sequence by explicit typecasting, everything works: ((Test2)t2)[0] I would suggest adding [] operator to the *Test* class, like this int operator[](int ind) {return i[ind];} Then you don't have to think about typecasting. Nikolai Borissov [/QUOTE]