I have the code:

vector<vector<int128_t>> resh(n, vector<int128_t>(n, 0)); int128_t S = 0; int128_t R = 0; int128_t qs = (1019 - 1) / 2; int64_t x = 0;

When calculating x :

 x = int64_t(-resh[k.convert_to<size_t>()][0] * S - R) % qs;

Gives an error message:

impossible to convert
"boost :: multiprecision :: number, boost :: multiprecision :: et_off>" in "int64_t"

I just can not understand why?

  • because int64_t knows nothing about the boost class, like the boost itself. Therefore, it is so easy to take and casting is impossible - KoVadim
  • and how to be? How to translate from boost to normal? - Alexander
  • 3
    you converted to size_t with, convert_to<size_t> , what do you think, how now to int64_t to int64_t ? - Fat-Zer
  • thanks a lot) - Alexander

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