如何复制initializer_list？

Question

The following doesn't output anything for me. From what I understand, the lifetime of an initializer_list is very limited and that copying an initializer_list doesn't copy the underlying elements. I thought that passing by value would allow a move, and even an explicit move doesn't work. I can't bind it to a reference and using a separate reference variable doesn't work here. Even using a vector doesn't work??

template <typename T, typename Pred>
void for_each(T t, const Pred& pred)
{
    auto local = std::vector<typename T::value_type>{t.begin(), t.end()};

    for (typename decltype(local)::size_type i = 0; i < local.size(); ++i)
        pred(*(i + std::begin(local)));
}

void pred(char c)
{
    std::cout << c << " ";
}

int main()
{
    auto arr { 1, 2, 3 };
    for_each(arr, pred);

    return 0;
}

Answer 1

Your predicate is wrong : you pass it int values, not char , so your predicate is passed the characters 0x01, 0x02 and 0x03, which are not printable.

You could use :

void pred(int c)
{
    std::cout << c << " ";
}

And your code now works fine .

You could also just write a function template:

template<typename T>
void pred(T c)
{
    std::cout << c << " ";
}

---

Note:

Consider writing your loop with a for-range :

for (auto&& v : t)
    pred(v);