How does sizeof work? How can I write my own?

后端 未结 10 2334
粉色の甜心
粉色の甜心 2021-02-08 07:29

I know C++ and know the function sizeof itself but I need to write my own sizeof function so please explain how it works exactly? What does it do with the parameter

10条回答
  •  小蘑菇
    小蘑菇 (楼主)
    2021-02-08 08:21

    You haven't provided any meaningful details about what it is you want to do, so it is hard to figure out what you need.

    You can "wrap" sizeof by you own template function like

    template  size_t my_sizeof() {
      return sizeof(T);
    }
    

    and then use it as

    size_t s = my_sizeof();
    

    From time to time one can come across a request to implement sizeof-like functionality without using sizeof. Requests like that make no practical sense whatsoever, yet sometimes are used as homework assignments. One can probably do it as follows

    template  size_t my_sizeof() {
      T t;
      return (char *) (&t + 1) - (char *) &t;
    }
    

    which would require a default-constructible T. A less restricting but formally illegal solution (a hack) would be something like

    template  size_t my_sizeof() {
      return (char *) ((T *) NULL + 1) - (char *) (T *) NULL;
    }
    

    The above implementations implement type-based sizeof.

    An attempt to emulate the functionality of value-based sizeof might look as follows

    template  size_t my_sizeof(const T& obj) { 
      return my_sizeof();
    }
    

    but this will not be even remotely equivalent to the built-in sizeof, at least because the built-in sizeof does not evaluate its argument.

    Finally, neither of these implementations will produce integral constant expressions (ICE), as the built-in sizeof does. Producing an ICE that way is impossible to achieve in the current version of the language.

    In any case this all, of course, is totally devoid of any practical value. Just use sizeof when you want to know the size.

提交回复
热议问题