Passing random seed/device to a function

Hello! I am trying to pass a seed to several functions, each generating one random number from different distribution, e.g. This does not work, as each iteration (function call) produces the same number. But why? Thank you for your help!

Code:

---

#include <iostream>
#include <random>

using std::cout;
using std::endl;

typedef std::mt19937 rng_type;
rng_type rng;

auto D(rng_type r, int n)
{
        std::uniform_int_distribution<rng_type::result_type> dist(1, n);
        rng_type::result_type random_number = dist(r);
        return random_number;
}

auto Scatter(rng_type r)
{
        std::uniform_real_distribution<rng_type::result_type> dist(0, 1);
        rng_type::result_type random_number = dist(r);
        return random_number;
}

int main()
{
  rng_type::result_type const seedval = std::random_device{}();
  rng.seed(seedval);
 
  for(int i = 0; i <= 10; ++i) cout << D(rng, 6) << ", " << Scatter(rng) << endl;
 
  return 0;
}

---

> auto D(rng_type r, int n)
> auto Scatter(rng_type r)
Because you create brand new random number generators through their copy constructors each time you call the function.

Make each of your r parameters a reference.

It's like you were expecting this to produce different results.

Code:

---

int foo ( int a ) {
  a += 1;
  return a;
}
int main ( ) {
  int x = 42;
  for ( int i = 0 ; i < 10 ; i++ ) cout << foo(x) << endl;
}

---

Quote:

---

Originally Posted by Salem

> auto D(rng_type r, int n)
> auto Scatter(rng_type r)
Because you create brand new random number generators through their copy constructors each time you call the function.

Make each of your r parameters a reference.

---

OK thx that fixed it! The other problem was that the output of Scatter was recast to a fixed 0, so I got rid of the types in the functions.