How can I create a C++ basic type that self-initializes?

Question

Related question: std::map default value for build-in type -- the subtle difference is I want more than to know whether the value is initialized to 0 or garbage, I want to specify a "constructor". I don't even care if it involves overhead with a class definition, I just want a clean "special" basic type. Even a syntactical hack would do. A non basic type is very easy to do this for, it is the entire job of the constructor.

I'd like to have a hashmap unordered_map<void *, int> but to have all its values default-initialized to -1 instead of 0 or garbage. This is because zero is a valid index, and I would prefer to default-initialize with a certainly invalid value.

I think I see a few sloppy ways this might be done with:

struct minus1 {
    int i; 
    minus1() i(-1) {}
};
unordered_map<void*, minus1>

But I don't like this because I have to use .i to access the int, and it really just needs to be an int.

Okay so maybe I can have my map handle this:

struct PointerToIDHash {
    std::unordered_map<void *, int> h;
    PointerToIDHash() {
        // ctor is powerless to affect the initialized values of future insertions into h
    }
};

Well, crap now I have a .h too. Uhhhh. Can I inherit from a template? (sounds scary, but this might be a clean way if it can be pulled off)

How can I make a type that transparently acts like an int but is always initialized to -1?

I would prefer to know both how to do this with and without C++11.

Have a look at a.lasram's answer. That's basically what I meant. — Zeta, Aug 25 '13 at 21:49
Standard library containers typically don't default-initialize their elements, but rather *value*-initialize them. So you never get garbage as long as your element types are sane. — Kerrek SB, Aug 25 '13 at 21:59
Don't solve the initialization problem. Instead, solve the problem that you want an `int` to be initialized with an invalid index value. This capability already exists, and it's called `boost::optional`. Then you don't have to use special int values to signal that it's uninitialized, you just check if the optional has been initialized. — Mark B, Aug 25 '13 at 23:51
*"Can I inherit from a template?"* - Yes, of course you can. *"But can I also inherit from a standard library container (well, why am I even asking this, but just to be sure)?"* - Yes, of course you can (contrary to what others may tell you). So yes, your second solution with a dedicated `PointerToIDHash` would work perfectly, too. — Christian Rau, Aug 26 '13 at 08:07
@MarkB I'm not sure I prefer putting things into std::pair et. al. unless the data really can't fit in its usual type. I'd rather give up one bit of my integer than expand it to a 64-bit object. That said I'm sure taking such an approach is more maintainable. — Steven Lu, Aug 26 '13 at 14:24
@Steven Lu Unless you're on embedded hardware, code clarity and future maintainability will be much more important than a few extra bits of space. — Mark B, Aug 27 '13 at 12:42

Andrew Tomazos · Accepted Answer · 2013-08-26T21:16:35.437

11

#include <unordered_map>
#include <iostream>

using namespace std;

template<typename T, T default_value>
class SelfInitializer
{
public:
    SelfInitializer(T x = default_value) : x(x) {}
    operator T&() { return x; }
    operator const T&() const { return x; }
private:
    T x;
};

// demo
int main()
{
    using minus1 = SelfInitializer<int, -1>;

    unordered_map<int, minus1> m;

    m[7] = 3; // assignment works

    minus1 x = 3;

    int y = x; // conversion to int works

    int z = int(x); // explicit conversion works

    cout << m[7] << endl;
}

edited Aug 26 '13 at 21:16

answered Aug 25 '13 at 23:01

Andrew Tomazos

66,139
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Would `operator=` be useful? `my_map[7] = 3` would then work. – Yakk - Adam Nevraumont Aug 26 '13 at 20:02
@user1131467 Can you show me how to use it in a `ostream& operator << (ostream& os, const minus1& i)` function? I'm using `os << int(i);` but it seems unclean. – Steven Lu Aug 27 '13 at 02:01
@StevenLu: You don't need to define such a function, the `operator<<(ostream&, int)` will already pick it up from `operator T&`. See last line of demo for demo. – Andrew Tomazos Aug 27 '13 at 11:04

score 8 · Answer 2 · answered Aug 25 '13 at 21:49

8

add a conversion operator to int& so that your struct minus1 behaves like an int

struct minus1 {
    int i; 
    minus1() : i(-1) {}
    operator int&() { return i; }
};

answered Aug 25 '13 at 21:49

a.lasram

4,371
1
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2

You'll probably want a converting constructor too. `minus1(int n) : i(n) {}` [Or combined with the default constructor, `minus1(int n=-1) : i(n) {}`.] – aschepler Aug 25 '13 at 21:55
@aschepler Yep, compiler told me as much :) – Steven Lu Aug 26 '13 at 06:28
1

`int& operator=(int o)` so you can assign to it, copy-construct, a `.get()` for when you really need to convert explicitly, `operator int const&() const`. – Yakk - Adam Nevraumont Aug 26 '13 at 20:04
@Yakk yes this would complete user1131467 answer; mine isn't elaborated and limited to the "not using" `.i` – a.lasram Aug 26 '13 at 20:46
@Yakk: Assignment already works in both solutions. To explicitly convert you can use `int(x)` in both solutions. – Andrew Tomazos Aug 26 '13 at 21:17

How can I create a C++ basic type that self-initializes?

2 Answers2