Construct fixed-size array in variadic template

Question

I want to construct an array of 8 integers in a variadic function. No problem:

template <typename... Ints>
void foo(Ints... ints) {
    static_assert(sizeof...(Ints) < 8, "some useful error");

    const int my_array[8] = {ints...};
}

That even automatically zero initializes the array, so if I call foo(1, 2, 3) I will get an array like {1, 2, 3, 0, 0, 0, 0, 0}.

Now what if I want to default to something other than zero? Like, say -1. This works:

template <int Def>
struct Int {
    Int() : val(Def) { }
    Int(int i): val(i) { }

    inline operator int() const { return val; }

    int val;
};

template <typename... Ints>
void foo(Ints... ints) {
    const Int<-1> my_array_def[8] = {ints...};
    const int* my_array = reinterpret_cast<const int*>(my_array_def);
}

But is there a simpler way that doesn't rely upon having this extra type?

Deduplicator · Accepted Answer · 2014-10-10 22:56:29Z

2

Just use another template:

template <typename... Ints>
auto foo(Ints... ints) -> typename std::enable_if<sizeof...ints==8>::type {
    const int my_array[] = {ints...};
}
template <typename... Ints>
auto foo(Ints... ints) -> typename std::enable_if<sizeof...ints<8>::type {
    return foo(ints..., -1);
}

answered Oct 10, 2014 at 22:56

Deduplicator

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