Why can't I use vector with mpz_t in GMP?

Question

I don't know why this program goes wrong, because it is simple; however, if I can't use vector in GMP, the task will be so hard.

This is my code:

vector<mpz_t> elements;
  cout << elements.size() << endl;
  mpz_t a;
  mpz_init(a);
  elements.push_back(a);

This works well, but it is not as useful as vector:

// mpz_t b[10];
  // for (int i = 0; i < 10; i++){
  //   mpz_init(b[i]);
  // }

enter image description here

I don't know what `mpz_t` is but it probably doesn't have a copy constructor? — Kaldrr, Jun 05 '20 at 06:46
yeah,i think this is the point too. because '=' can't use with mpz_t.so i solved it with the mpz_t *. — chris yang, Jun 06 '20 at 07:13

score 1 · Accepted Answer · answered Jun 05 '20 at 06:58

mpz_t is a typedef to an array type. Instead of a vector of mpz_t, in C++ I would advise to either encapsulate mpz_t in your own class (that will also allow to manage memory) or use a vector of pointers to mpz_t.

As for the first option, it already exists in the name of mpz_class.

why i can't use vector with mpz_t in gmp?

mpz_t is a typedef to an array type. (From the answer of this question): arrays are not copy constructible and not assignable, so you can't do std::vector<int[1]>.

Why can't I use vector with mpz_t in GMP?

1 Answers1