struct B; struct A { operator B(); }; struct B { B operator + (B) {return B();} }; A::operator B() { return B(); } int main() { A x; B y; y+x; // Как запретить неявное преобразование для данного оператора/ группы операторов? };

    1 answer 1

    Implicit conversion can be disabled as follows:

     struct A { explicit operator B(); };

    Now it’s only possible to convert A to B

    If you want to leave the implicit conversion as a whole, but in particular cases it is undesirable, then you can use this trick:

     #include <type_traits> //... struct B { template<typename TB, typename=std::enable_if_t<std::is_same<TB, B>::value>> B operator + (TB) { return B(); } };
    • @hero, you can explicitly C-cast type ( static_cast or C-cast ) - ixSci
    • Why in the first case write expilicit operator B() if you can simply not write it? - hero
    • @hero, if you don’t write it, you cannot convert A to B , and if you make explicit , you can convert it, but only explicitly (as I indicated in the previous comment) - ixSci
    • Now it's clear, thanks - hero
    • Writes [Error] only declarations of constructors can be 'explicit' - hero