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Home/ Questions/Q 3637400
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Editorial Team
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Editorial Team
Asked: May 19, 20262026-05-19T01:09:13+00:00 2026-05-19T01:09:13+00:00

For example: shared_ptr<const shared_ptr<const int> > pp; is fairly intimidating… whereas const int ^

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For example:

shared_ptr<const shared_ptr<const int> > pp;

is fairly intimidating… whereas

const int ^ const ^ pp;

is instantly reminiscent of the raw pointer equivalent

const int * const * pp;

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  1. Editorial Team
    Editorial Team
    2026-05-19T01:09:14+00:00Added an answer on May 19, 2026 at 1:09 am

    Everyone missed the real reason 🙂

    The nice notation T^ suggests action like a real pointer, which suggests conversions would work the same way. This is not the case. Templates can’t support conversions. For example consider

    T^ --> T const^
    

    If you take ^ as *, this will work. However by default for a template:

    p<T> --> p<T const>
    

    will not work: templates are not covariant in their arguments. It is not possible to fix this in C++ AFAIK. You can handle that one conversion, but don’t forget the others:

    derived --> base
    T ** --> T const * const*
    

    just for example. If you want a real type system it has to be designed my mathematicians not compiler writers.

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