malat malat - 5 months ago 69
C++ Question

std::iota is very limited

Coming from a

world, I find the function
very limited. Why is the interface restricted to not take any

For instance I can convert

>>> x = range(0, 10)


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

But how would one do:

>>> x = range(0,20,2)

or even

>>> x = range(10,0,-1)

I know this is trivial to write one such function or use Boost, but I figured that C++ committee must have picked this design with care. So clearly I am missing something from C++11.

Answer Source

But how would one do:

x = range(0,20,2)

Alternatively to std::generate() (see other answer), you can provide your own unary function to std::iota(), it just have to be called operator++():

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

template<class T>
struct IotaWrapper
    typedef T type;
    typedef std::function<type(const type&)> IncrFunction;

    type value;
    IncrFunction incrFunction;

    IotaWrapper() = delete;
    IotaWrapper(const type& n, const IncrFunction& incrFunction) : value(n), incrFunction(incrFunction) {};

    operator type() { return value; }
    IotaWrapper& operator++() { value = incrFunction(value); return *this; }

int main()
    IotaWrapper<int> n(0, [](const int& n){ return n+2; });
    std::vector<int> v(10);
    std::iota(v.begin(), v.end(), n);

    for (auto i : v)
        std::cout << i << ' ';
    std::cout << std::endl;

Output: 0 2 4 6 8 10 12 14 16 18


Here is an idea of how one could implement Range():

struct Range
    template<class Value, class Incr>
    std::vector<Value> operator()(const Value& first, const Value& last, const Incr& increment)
        IotaWrapper<Value> iota(first, [=](const int& n){ return n+increment; });
        std::vector<Value> result((last - first) / increment);
        std::iota(result.begin(), result.end(), iota);
        return result;


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