Mixed list initialization in C++11

Question

With reference to c++11 list initialization, may I initialize a list with an element and another list?

Let's say I have the following code:

#include <vector>

class Foo
{
    public:
        Foo(int value){m_v=value;}
    private:
        int m_v = 0;
};

int main()
{
   std::vector<Foo> v1, v2, v3;
   v1 = {Foo(1)}; //ok
   v2 = {Foo(2), Foo(3)}; //ok
   v3 = {Foo(3), v2}; //error: no match for ‘operator=’ (operand types are ‘std::vector’ and ‘’)
}

Is there a way to create in one line of code, using list initialization, a vector made of the element of another vector plus a new element (a prepend, in the example above).

Answer 1

We can create some template infrastructure to allow creation of vectors through optional concatenation of objects and other vectors.

This is very much a first cut:

#include <utility>
#include <vector>

namespace extended
{
    template<class T>
    struct appender
    {
        template<class V, class A, class Arg>
        void operator()(std::vector<V, A>& vec, Arg&& arg) const
        {
            vec.push_back(std::forward<Arg>(arg));
        }
    };

    template<class V2, class A2>
    struct appender<std::vector<V2, A2>>
    {
        template<class V, class A, class X>
        void operator()(std::vector<V, A>& vec, X&& arg) const
        {
            vec.insert(end(vec), begin(std::forward<X>(arg)), end(std::forward<X>(arg)));
        }
    };

    template<class V, class A, class T>
    auto append(std::vector<V, A>& target, T&& x) -> decltype(auto)
    {
        auto op = appender<std::decay_t<T>>();
        op(target, std::forward<T>(x));
        return target;
    }
}

template<class T, class...Args>
auto make_vector(Args&&...args)
{
    using extended::append;
    std::vector<T> result;
    using expand = int[];
    expand {0,
        (append(result, std::forward<Args>(args)), 0)...
    };
    return result;
}

class Foo
{
    public:
        Foo(int value){m_v=value;}
    private:
        int m_v = 0;
};

int main()
{
   auto v1 = make_vector<Foo>(Foo(1)); //ok
   auto v2 = make_vector<Foo>(Foo(2), Foo(3)); //ok
   auto v3 = make_vector<Foo>(Foo(3), v2); //ok
}

Of course, by looking for common interfaces we can start to push the boundaries a little:

#include <utility>
#include <iterator>
#include <vector>
#include <list>
#include <set>

namespace extended
{
    // The general case of an appender.
    // simply calls emplace_back
    template<class T, class Diff = void>
    struct appender
    {
        template<class V, class A, class Arg>
        void operator()(std::vector<V, A>& vec, Arg&& arg) const
        {
            vec.emplace_back(std::forward<Arg>(arg));
        }
    };

    // specific specialisation for an appender where the
    // source object supports begin() and end() (i.e. a container)
    //
    template<class T>
    struct appender
    <
        T, 
        decltype(
            std::begin(std::declval<T>()), 
            std::end(std::declval<T>()),
            void()
        )
    >
    {
        template<class V, class A, class X>
        void operator()(std::vector<V, A>& vec, X&& arg) const
        {
            vec.insert(std::end(vec), std::begin(std::forward<X>(arg)), std::end(std::forward<X>(arg)));
        }
    };

    template<class V, class A, class T>
    auto append(std::vector<V, A>& target, T&& x) -> decltype(auto)
    {
        auto op = appender<std::decay_t<T>>();
        op(target, std::forward<T>(x));
        return target;
    }
}

template<class T, class...Args>
auto make_vector(Args&&...args)
{
    using extended::append;
    std::vector<T> result;
    using expand = int[];
    expand {0,
        (append(result, std::forward<Args>(args)), 0)...
    };
    return result;
}

class Foo
{
    public:
        Foo(int value){m_v=value;}

        bool operator<(const Foo& r) const { return m_v < r.m_v; }
    private:
        int m_v = 0;
};



int main()
{
   auto v1 = make_vector<Foo>(Foo(1)); //ok
   auto v2 = make_vector<Foo>(Foo(2), Foo(3)); //ok
   auto v3 = make_vector<Foo>(Foo(3), v2); //ok
   auto v4 = make_vector<Foo>(Foo(1), 
    std::list<Foo> { Foo(2), Foo(3) }, 
    make_vector<Foo>(4, make_vector<Foo>(8, 9, 10)),
    std::set<Foo> {Foo(6), Foo(7) }); // bizzare but ok
}

Answer 2

std::vector<Foo> means a std::vector of Foo instances. This means it cannot arbitrarily store other std::vector instances, which is what you're asking the compiler when writing

v3 = {Foo(3), v2};

std::initializer_list<T> is a homogeneous collection of T instances. std::vector<Foo>'s list constructor takes std::initializer_list<Foo>. There's no way to achieve what you want without manually unpacking v2's elements inside the curly braces.

---

Is there a way to create in one line of code, using list initialization, a vector made of the element of another vector plus a new element (a prepend, in the example above).

Using list initialization, no. You can write your own function to achieve the same thing, however.