superstruct 0.1.0

A superclass for value types.


To use this package, run the following command in your project's root directory:

Manual usage
Put the following dependency into your project's dependences section:

SuperStruct

Suppose you've got a few structs representing shapes, say, Circle, Square and Triangle. You want an overarching type to store any one of these shapes.

alias Shape = Algebraic!(Circle, Square, Triangle)
Shape shape = Circle(x, y, radius);

Ok, an Algebraic (or Variant, in general) isn't a bad choice. Now, could you grab the area of that shape for me?

auto area = shape.visit!((Circle c)   => c.area,
                         (Square s)   => s.area,
                         (Triangle t) => t.area);

Alright, that works well enough. How about the perimeter? They all have one of those, don't they?

auto perimeter = shape.visit!((Circle c)   => c.perimeter,
                              (Square s)   => s.perimeter,
                              (Triangle t) => t.perimeter);

Ok, how about the center? Noticing a pattern I started to wonder if some (ab)use of templates could strip away the boilerplate.

What is it?

It's a struct! It's a class! No, its ... SuperStruct!

This bastard child of wrap and variant works like an Algebraic, but exposes members that are common across the source types.

struct Square {
  float size;
  float area() { return size * size; }
}

struct Circle {
  float r;
  float area() { return r * r * PI; }
}

alias Shape = SuperStruct!(Square, Circle);

// look! polymorphism!
Shape sqr = Square(2);
Shape cir = Circle(4);
Shape[] shapes = [ sqr, cir ];

// call functions that are shared between the source types!
assert(shapes.map!(x => x.area).sum.approxEqual(2 * 2 + 4 * 4 * PI));

Notice that there is no explicit interface definition. The 'interface' forms organically from the common members of the source types.

The interface isn't limited to methods -- common fields can be exposed as well:

// `top` is a field of Square
struct Square {
  int top, left, width, height;
}

// but a property of cirle
struct Circle {
  int radius;
  int x, y;

  auto top() { return y - radius; }
  auto top(int val) { return y = val + radius; }
}

alias Shape = SuperStruct!(Square, Circle);

// if a Shape is a Circle, `top` forwards to Circle's top property
Shape cir = Circle(4, 0, 0);
assert(cir.top = 6);
assert(cir.top == 6);

// if a Shape is a Square, `top` forwards to Squares's top field
Shape sqr = Square(0, 0, 4, 4);
assert(sqr.top = 6);
assert(sqr.top == 6);

// Square.left is hidden, as Circle has no such member
static assert(!is(typeof(sqr.left)));

Is it useful?

I don't know, you tell me. I just work here.

If nothing else, its an interesting exercise in what D's compile-time facilities are capable of.

Why not use Variant/Algebraic?

To avoid repeating yourself.

// Compare this...
auto alg = Algebraic!(Rect, Ellipse, Triangle)(someShape);
auto center1 = alg.visit!((Rect r)     => r.center,
                          (Ellipse e)  => e.center,
                          (Triangle t) => t.center)

// To this! Easier, right?
auto center2 = sup.center;

Why not use wrap?

[std.typecons.wrap](http://dlang.org/phobos/std_typecons.html#.wrap) and SuperStruct have similar, but not entirely overlapping uses.

  1. wrap does not currently support structs (but a PR exists to implement this)
  2. wrap allocates a class. SuperStruct is just a struct.
  3. wrap requires an explicitly defined interface. SuperStruct generates an interface automatically.
  4. SuperStruct requires you to specify all sub types in advance. wrap does not.
  5. SuperStruct can expose common fields directly. wrap requires the user to manually wrap fields in getter/setter properties to satisfy an interface.

What does it expose?

  • If all types have a matching field, it gets exposed:
struct Foo { int a; }
struct Bar { int a; }
auto foobar = SuperStruct!(Foo, Bar)(Foo(1));
foobar.a = 5;
assert(foobar.a == 5);
  • If all types have a matching method, all compatible overloads are exposed:
struct Foo {
  int fun(int i) { return i; }
  int fun(int a, int b) { return a + b; }
}
struct Bar {
  int fun(int i) { return i; }
  int fun(int a, int b) { return a + b; }
  int fun(int a, int b, int c) { return a + b + c; }
}

auto foobar = SuperStruct!(Foo, Bar)(Foo());
assert(foobar.fun(1)    == 1);
assert(foobar.fun(1, 2) == 3);
assert(!__traits(compiles, foobar.fun(1,2,3))); // no such overload on Foo
  • If a name refers to a field on one type and a method on another, it is exposed if the field and the method have compatible signatures:
struct Foo { int a; }
struct Bar {
  private int _a;
  int a() { return _a; }
  int a(int val) { return _a = val; }
}

auto foo = SuperStruct!(Foo, Bar)(Foo());
foo.a = 5;          // sets Foo.a
assert(foo.a == 5); // gets Foo.a

auto bar = SuperStruct!(Foo, Bar)(Bar());
bar.a = 5;          // invokes Bar.a(int val)
assert(bar.a == 5); // invokes Bar.a()
  • Private members are not exposed.

How does it work?

Given the types Foo and Bar with members a and b, SuperStruct!(Foo, Bar) looks something like:

struct FooBar {
  private Algebraic!(Foo,Bar) _value;

  auto a(Args...)(Args args) {
    return visitor!"a"(_value, args);
  }
  auto b(Args...)(Args args) {
    return visitor!"b"(_value, args);
  }
}

Where visitor is a helper that tries to forward the call to a matching member on whatever _value is holding. If whatever args you pass don't form a valid call on the given member for every subtype, it won't compile. If they all do form valid calls but there is no common return type for those calls, it won't compile.

This means that 'commonality' of members is checked on a case-by-case basis. It could try to figure out if a member would never be callable and simply omit it, but currently does not (instead it just generates a variadic template that is impossible to instantiate).

Authors:
  • rcorre
Dependencies:
none
Versions:
0.2.0 2015-Nov-23
0.1.0 2015-Oct-18
~master 2015-Nov-23
~wip 2015-Nov-05
~dev 2015-Oct-30
Show all 5 versions
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Short URL:
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