Is it possible to initialize a vector with increasing values in a single line?

Question

Is it possible to merge the two initialization lines into a single statement with the help of initializer lists or other C++ features? The vector values always increment with one, but the size n is not fixed.

#include <numeric>
#include <vector>
#include <iostream>

int main()
{
    int n = 10;

    // Can the two lines below be combined into a single statement?
    std::vector<int> v(n);
    std::iota(v.begin(), v.end(), 1);

    for (int i : v)
        std::cout << i << std::endl;

    return 0;
}

Answer 1

You can use Boost.counting_iterator for this:

std::vector<int> v(boost::counting_iterator<int>(1),
                    boost::counting_iterator<int>(n + 1));

(Live) Now whether this is worth it and easier to read than what you already have is for you to decide.

Answer 2

Not really, no. If n is a runtime variable, the best you could probably do is to just throw this in a function somewhere:

std::vector<int> ints(int n) {
    std::vector<int> v;
    v.reserve(n);
    for (int i = 0; i < n; ++i) {
        v.push_back(n+1);
    }
    return v;
}

// now it's one line?
std::vector<int> v = ints(n);

If it's compile time, you can use std::index_sequence to provide an initializer list:

template <int... Is>
std::vector<int> ints(std::integer_sequence<int, Is...> ) {
    return std::vector<int>{ (Is+1)... };
}


template <int N>
std::vector<int> ints() {
    return ints(std::make_integer_sequence<int, N>{});
}

But either way, you need a helper function.