boost::range::combine with repeated argument

I would like to use boost::range::combine as a cartesian power instead as just a product.

So instead of such expression boost::range::combine(myRange, myRange, myRange); write something like myCombine(myRange, 3);.

How it can be implemented?

Implementing this in C++17 or C++14 would be a lot easier and cleaner, but since you tagged this with c++11 here's a compliant implementation. Here's a generic way of calling a function object f with the same argument repeated N times.

First, we need a way of binding the first argument of a generic function object f and then accepting any number of arguments:

template <typename TF, typename T>
struct bound
{
    TF _f;
    T _x;

    template <typename TFFwd, typename TFwd>
    bound(TFFwd&& f, TFwd&& x)
        : _f{std::forward<TFFwd>(f)}, _x{std::forward<TFwd>(x)}
    {
    }

    template <typename... Ts>
    auto operator()(Ts&&... xs)
        -> decltype(_f(_x, std::forward<Ts>(xs)...))
    {
        return _f(_x, std::forward<Ts>(xs)...);
    }
};

template <typename TF, typename T>
auto bind_first(TF&& f, T&& x)
    -> decltype(bound<TF&&, T&&>(std::forward<TF>(f), std::forward<T>(x)))
{
    return bound<TF&&, T&&>(std::forward<TF>(f), std::forward<T>(x));
}

Then, we need a recursive helper that will bind an argument x multiple TN times:

template <std::size_t TN>
struct helper
{
    template <typename TF, typename T>
    auto operator()(TF&& f, T&& x)
        -> decltype(helper<TN - 1>{}(bind_first(std::forward<TF>(f), x), x))
    {
        return helper<TN - 1>{}(bind_first(std::forward<TF>(f), x), x);
    }
};

template <>
struct helper<0>
{
    template <typename TF, typename T>
    auto operator()(TF&& f, T&& x)
        -> decltype(f(x))
    {
        return f(x);
    }
};

Finally, we can provide a nice interface:

template <std::size_t TN, typename TF, typename T>
auto call_with_same_arg(TF&& f, T&& x)
    -> decltype(helper<TN - 1>{}(std::forward<TF>(f), std::forward<T>(x)))
{
    return helper<TN - 1>{}(std::forward<TF>(f), std::forward<T>(x));
}

Usage:

int add(int a, int b, int c)
{
    return a + b + c;   
}

int main()
{
    assert(call_with_same_arg<3>(add, 5) == 15);
}

live wandbox example

---

Here's a complete C++17 implementation of the same thing:

template <std::size_t TN, typename TF, typename T>
decltype(auto) call_with_same_arg(TF&& f, T&& x)
{
    if constexpr(TN == 1)
    {
        return f(x);
    }
    else
    {
        return call_with_same_arg<TN - 1>(
            [&](auto&&... xs){ return f(x, std::forward<decltype(xs)>(xs)...); }, x);
    }
}

live wandbox example

---

For completeness, C++14 implementation:

template <std::size_t TN>
struct helper
{
    template <typename TF, typename T>
    decltype(auto) operator()(TF&& f, T&& x)
    {
        return helper<TN - 1>{}(
            [&](auto&&... xs){ return f(x, std::forward<decltype(xs)>(xs)...); }, x);
    }
};

template <>
struct helper<0>
{
    template <typename TF, typename T>
    decltype(auto) operator()(TF&& f, T&& x)
    {
        return f(x);
    }
};

template <std::size_t TN, typename TF, typename T>
decltype(auto) call_with_same_arg(TF&& f, T&& x)
{
    return helper<TN - 1>{}(std::forward<TF>(f), std::forward<T>(x));
}

live wandbox example