Can a functor provide std :: generate be stateful?

Question

I recently read that some STL algorithms have undefined behavior if the passed functor has a state (has internal side effects). I used a function std::generatewith a similar functor (less trivial):

class Gen
{
public:
    explicit Gen(int start = 0)
        : next(start)
    {
    }

    int operator() ()
    {
        return next++;
    }

private:
    int next;
};

Is it safe to use with std::generate? Is the order of generating values ​​guaranteed?

Edit: Claim made here Functional Functions and STL: undefined behavior

Answer 1

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#include <iostream>
#include <algorithm>

template<class id_type>
struct id_generator {
  id_type operator() () {
    return ++idx;
  }

  id_type next_id () const {
    return idx + 1;
  }

  id_type idx {};
};

int main () {
  id_generator<int> gen;

  std::vector<int>  vec1 (5);
  std::vector<int>  vec2 (5);

  std::generate (vec1.begin (), vec1.end (), gen);
  std::generate (vec2.begin (), vec2.end (), gen);

  std::cout << gen.next_id () << std::endl; // will print '1'
}

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# 1

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std::generate (vec1.begin (), vec1.end (), std::ref (gen));
std::generate (vec2.begin (), vec2.end (), std::ref (gen));

std::cout << gen.next_id () << std::endl; // will print '11'

№ 2

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std::generate<decltype(vec1.begin()), id_generator<int>&>(vec1.begin(), vec1.end(), gen);

Answer 2

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