Getting value of member depending on underlying object type

Question

I have two structs from a third party library so I can't change their layout. But I have two structs I wrote on my own which match their signature so I can add functions to them (again, but not their layout, therefore also no virtual functions). The structs are called Foo1 and Foo2. Both have the same member called _b, but at different offsets.

struct FooBase { };

struct Foo1 : FooBase
{
    int _a = 2;
    int _b = 1;
};

struct Foo2 : FooBase
{
    int _b = 2;
};

struct Wrap
{
    Wrap(const FooBase& x) : _x(x) { }
    FooBase& _x;
    int GetValue() { return /* MISSING */; }
};

I have a wrapper class called Wrap. I am looking for a way to return the value of _b without using virtual functions in the foo classes, since I can't change their size anymore.

Foo1 f1 = Foo1();
Wrap x = f1;
int a = x.GetValue(); // should return 1

Foo2 f2 = Foo2();
Wrap y = f2;
int b = y.GetValue(); // should return 2

Any ideas? I posted an approach in the comments but curious for a better solution.

`Foo1::_b` and `Foo2::_b` are totally unrelated. It seems like there should be a way to do this statically, since they have the same name, but that's not how it works. If all `Foo` have a `int _b` data member, then it should be a member of `FooBase` instead. — François Andrieux, Feb 27 '18 at 19:15
I could add a `std::function` to `Wrap` which would contain the function of `Foo1` or `Foo2` which returns the member. I don't feel like that's the best solution though — HelloWorld, Feb 27 '18 at 19:15
Beware [slicing](https://stackoverflow.com/q/274626/10077)! Your `_x` needs to be a reference, not a value. A `FooBase` instance has neither `_a` nor `_b`. — Fred Larson, Feb 27 '18 at 19:15
Yes, they would have the same name but otherwise are unrelated. That's what makes it so difficult to wrap my head around this — HelloWorld, Feb 27 '18 at 19:16
@HelloWorld Just pretend they have different names. Or even give them different names if it helps. — François Andrieux, Feb 27 '18 at 19:18
@FrancoisAndrieux: I can't change their structure anymore, these are not my types — HelloWorld, Feb 27 '18 at 19:20
When you construct a `Wrap` instance, do you have a concrete type to give the constructor, or just a `FooBase&`? — Fred Larson, Feb 27 '18 at 19:21
@HelloWorld Then there is nothing you can do with a `FooBase`. You need to capture the concrete type in some way. — François Andrieux, Feb 27 '18 at 19:22
@FrancoisAndrieux: I updated the first part of my question accordingly to your comment. Thanks a lot! @FredLarson: Wrap could have information, if it is `Foo1` or `Foo2` and the wrap class can also look how I want it too look like — HelloWorld, Feb 27 '18 at 19:24

score 6 · Accepted Answer · edited Feb 27 '18 at 19:30

6

struct Wrap
{
    Wrap(const Foo1& x) : _b(x._b) { }
    Wrap(const Foo2& x) : _b(x._b) { }
    int& _b;
    int GetValue() { return _b; }
};

Unless a class is needed, you can use a function template to the same effect.

template <typename Foo> int getValue(Foo const& foo)
{
   return foo._b;
}

edited Feb 27 '18 at 19:30

R Sahu

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answered Feb 27 '18 at 19:23

Ishamael

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This assumes the user has knowledge of the real type of `x`. – François Andrieux Feb 27 '18 at 19:24
2

Seems like the return type of GetValue in the original snippet was already influenced by it? – Ishamael Feb 27 '18 at 19:24
The return type of `GetValue` is clearly `int` in all cases. – François Andrieux Feb 27 '18 at 19:26
2

@RSahu: Nice one! – Fred Larson Feb 27 '18 at 19:32
I feel so silly now. I focused too much by getting Foo into the Wrap class instead of the actual member as a reference. I accept this as an answer. The int-ref member approach solves my issue. Thanks a lot! – HelloWorld Feb 27 '18 at 19:33

Getting value of member depending on underlying object type

1 Answers1