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Initializing std::vector with iterative function calls

In many languages, there are generators that help to initialize collections. In C++, if one wants to initialize a vector uniformly, one can write:

std::vector<int> vec(10, 42); // get 10 elements, each equals 42

What if one wants to generate different values on the fly? For example, initialize it with 10 random values, or consecutive numbers from 0 to 9? This syntax would be convenient, but it does not work in C++11:

int cnt = 0;
std::vector<int> vec(10, [&cnt]()->int { return cnt++;});

Is there a nice way to initialize a collection by iterative function calls? I currently use this ugly pattern (not much more readable/short than a loop):

std::vector<int> vec;
int cnt = 0;
std::generate_n(std::back_inserter(vec), 10, [&cnt]()->int { return cnt++;});

There is a thing that would help, and it would explain the lack of the first constructor. I can imagine an iterator that takes a function and number of calls, so that the constructor

vector ( InputIterator first, InputIterator last);

would be applicable. But I did not find anything like this in the standard library. Did I miss it? Is there another reason why the first constructor did not make it to the standard?

like image 818
Roman Shapovalov Avatar asked Sep 20 '12 11:09

Roman Shapovalov


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3 Answers

Sadly, there is no standard facility to do this.

For your specific example, you could use Boost.Iterator's counting_iterator like this:

std::vector<int> v(boost::counting_iterator<int>(0),
    boost::counting_iterator<int>(10));

Or even with Boost.Range like this:

auto v(boost::copy_range<std::vector<int>>(boost::irange(0,10)));

(copy_range will basically just return std::vector<int>(begin(range), end(range)) and is a great way to adopt full range construction to exisiting containers that only support range construction with two iterators.)


Now, for the general purpose case with a generator function (like std::rand), there is the function_input_iterator. When incremented, it calls the generator and saves the result, which is then returned when dereferencing it.

#include <vector>
#include <iostream>
#include <cmath>
#include <boost/iterator/function_input_iterator.hpp>

int main(){
  std::vector<int> v(boost::make_function_input_iterator(std::rand, 0),
      boost::make_function_input_iterator(std::rand,10));
  for(auto e : v)
    std::cout << e << " ";
}

Live example.

Sadly, since function_input_iterator doesn't use Boost.ResultOf, you need a function pointer or a function object type that has a nested result_type. Lambdas, for whatever reason, don't have that. You could pass the lambda to a std::function (or boost::function) object, which defines that. Here's an example with std::function. One can only hope that Boost.Iterator will make use of Boost.ResultOf someday, which will use decltype if BOOST_RESULT_OF_USE_DECLTYPE is defined.

like image 83
Xeo Avatar answered Oct 13 '22 19:10

Xeo


The world is too large for C++ to ship a solution for everything. However, C++ doesn't want to be a huge supermarket full of ready meals for every conceivable palate. Rather, it is a small, well-equipped kitchen in which you, the C++ Master Chef, can cook up any solution you desire.

Here's a silly and very crude example of a sequence generator:

#include <iterator>

struct sequence_iterator : std::iterator<std::input_iterator_tag, int>
{
    sequence_iterator() : singular(true) { }
    sequence_iterator(int a, int b) : singular(false) start(a), end(b) { }
    bool singular;
    int start;
    int end;

    int operator*() { return start; }
    void operator++() { ++start; }

    bool operator==(sequence_iterator const & rhs) const
    {
        return (start == end) == rhs.singular;
    }
    bool operator!=(sequence_iterator const & rhs) const
    {
        return !operator==(rhs);
    }
};

Now you can say:

std::vector<int> v(sequence_iterator(1,10), sequence_iterator());

In the same vein, you can write a more general gadget that "calls a given functor a given number of times", etc. (e.g. take a function object by templated copy, and use the counters as repetition counters; and dereferencing calls the functor).

like image 29
Kerrek SB Avatar answered Oct 13 '22 20:10

Kerrek SB


If you're using a compiler that supports lambdas as you use in your question, then chances are pretty good it also includes std::iota, which at least makes the counting case a little cleaner:

std::vector <int> vec(10);
std::iota(begin(vec), end(vec), 0);

For this scenario (and quite a few others, I think) we'd really prefer an iota_n though:

namespace stdx {
template <class FwdIt, class T>
void iota_n(FwdIt b, size_t count, T val = T()) {
    for ( ; count; --count, ++b, ++val)
        *b = val;
}
}

Which, for your case you'd use like:

std::vector<int> vec;

stdx::iota_n(std::back_inserter(vec), 10);

As to why this wasn't included in the standard library, I really can't even guess. I suppose this could be seen as an argument in favor of ranges, so the algorithm would take a range, and we'd have an easy way to create a range from either a begin/end pair or a begin/count pair. I'm not sure I completely agree with that though -- ranges do seem to work well in some cases, but in others they make little or no sense. I'm not sure that without more work, we have an answer that's really a lot better than a pair of iterators.

like image 2
Jerry Coffin Avatar answered Oct 13 '22 18:10

Jerry Coffin