What would be a good implementation of iota_n (missing algorithm from the STL)

Question

With C++11, the STL has now a std::iota function (see a reference). In contrast to std::fill_n, std::generate_n, there is no std

Answer 1

You're so focused on code generation that you forgot to get the interface right.

You correctly require OutputIterator, but what happens if you want to call it a second time?

list list(2 * N);
iota_n(list.begin(), N, 0);
// umm...
iota_n(list.begin() + N, N, 0); // doesn't compile!
iota_n(list.rbegin(), N, 0); // works, but create 0..N,N-1..0, not 0..N,0..N
auto it = list.begin();
std::advance(it, N);
iota_n(it, N, 0); // works, but ... yuck and ... slow (O(N))

inside iota_n, you still know where you are, but you've thrown that information away, so the caller cannot get at it in constant time.

General principle: don't throw away useful information.

template 
auto iota_n(OutputIterator dest, SizeType N, ValueType value) {
    while (N) {
        *dest = value;
        ++dest;
        ++value;
        --N;
    }
    // now, what do we know that the caller might not know?
    // N? No, it's zero.
    // value? Maybe, but it's just his value + his N
    // dest? Definitely. Caller cannot easily compute his dest + his N (O(N))
    //       So, return it:
    return dest;
}

With this definition, the above example becomes simply:

list list(2 * N);
auto it = iota_n(list.begin(), N, 0);
auto end = iota_n(it, N, 0);
assert(end == list.end());