Contents
LLVM avoids using C++’s built in RTTI. Instead, it pervasively uses its own hand-rolled form of RTTI which is much more efficient and flexible, although it requires a bit more work from you as a class author.
A description of how to use LLVM-style RTTI from a client’s perspective is given in the Programmer’s Manual. This document, in contrast, discusses the steps you need to take as a class hierarchy author to make LLVM-style RTTI available to your clients.
Before diving in, make sure that you are familiar with the Object Oriented Programming concept of “is-a”.
This section describes how to set up the most basic form of LLVM-style RTTI (which is sufficient for 99.9% of the cases). We will set up LLVM-style RTTI for this class hierarchy:
class Shape {
public:
Shape() {}
virtual double computeArea() = 0;
};
class Square : public Shape {
double SideLength;
public:
Square(double S) : SideLength(S) {}
double computeArea() override;
};
class Circle : public Shape {
double Radius;
public:
Circle(double R) : Radius(R) {}
double computeArea() override;
};
The most basic working setup for LLVM-style RTTI requires the following steps:
In the header where you declare Shape, you will want to #include "llvm/Support/Casting.h", which declares LLVM’s RTTI templates. That way your clients don’t even have to think about it.
#include "llvm/Support/Casting.h"
In the base class, introduce an enum which discriminates all of the different concrete classes in the hierarchy, and stash the enum value somewhere in the base class.
Here is the code after introducing this change:
class Shape {
public:
+ /// Discriminator for LLVM-style RTTI (dyn_cast<> et al.)
+ enum ShapeKind {
+ SK_Square,
+ SK_Circle
+ };
+private:
+ const ShapeKind Kind;
+public:
+ ShapeKind getKind() const { return Kind; }
+
Shape() {}
virtual double computeArea() = 0;
};
You will usually want to keep the Kind member encapsulated and private, but let the enum ShapeKind be public along with providing a getKind() method. This is convenient for clients so that they can do a switch over the enum.
A common naming convention is that these enums are “kind”s, to avoid ambiguity with the words “type” or “class” which have overloaded meanings in many contexts within LLVM. Sometimes there will be a natural name for it, like “opcode”. Don’t bikeshed over this; when in doubt use Kind.
You might wonder why the Kind enum doesn’t have an entry for Shape. The reason for this is that since Shape is abstract (computeArea() = 0;), you will never actually have non-derived instances of exactly that class (only subclasses). See Concrete Bases and Deeper Hierarchies for information on how to deal with non-abstract bases. It’s worth mentioning here that unlike dynamic_cast<>, LLVM-style RTTI can be used (and is often used) for classes that don’t have v-tables.
Next, you need to make sure that the Kind gets initialized to the value corresponding to the dynamic type of the class. Typically, you will want to have it be an argument to the constructor of the base class, and then pass in the respective XXXKind from subclass constructors.
Here is the code after that change:
class Shape {
public:
/// Discriminator for LLVM-style RTTI (dyn_cast<> et al.)
enum ShapeKind {
SK_Square,
SK_Circle
};
private:
const ShapeKind Kind;
public:
ShapeKind getKind() const { return Kind; }
- Shape() {}
+ Shape(ShapeKind K) : Kind(K) {}
virtual double computeArea() = 0;
};
class Square : public Shape {
double SideLength;
public:
- Square(double S) : SideLength(S) {}
+ Square(double S) : Shape(SK_Square), SideLength(S) {}
double computeArea() override;
};
class Circle : public Shape {
double Radius;
public:
- Circle(double R) : Radius(R) {}
+ Circle(double R) : Shape(SK_Circle), Radius(R) {}
double computeArea() override;
};
Finally, you need to inform LLVM’s RTTI templates how to dynamically determine the type of a class (i.e. whether the isa<>/dyn_cast<> should succeed). The default “99.9% of use cases” way to accomplish this is through a small static member function classof. In order to have proper context for an explanation, we will display this code first, and then below describe each part:
class Shape {
public:
/// Discriminator for LLVM-style RTTI (dyn_cast<> et al.)
enum ShapeKind {
SK_Square,
SK_Circle
};
private:
const ShapeKind Kind;
public:
ShapeKind getKind() const { return Kind; }
Shape(ShapeKind K) : Kind(K) {}
virtual double computeArea() = 0;
};
class Square : public Shape {
double SideLength;
public:
Square(double S) : Shape(SK_Square), SideLength(S) {}
double computeArea() override;
+
+ static bool classof(const Shape *S) {
+ return S->getKind() == SK_Square;
+ }
};
class Circle : public Shape {
double Radius;
public:
Circle(double R) : Shape(SK_Circle), Radius(R) {}
double computeArea() override;
+
+ static bool classof(const Shape *S) {
+ return S->getKind() == SK_Circle;
+ }
};
The job of classof is to dynamically determine whether an object of a base class is in fact of a particular derived class. In order to downcast a type Base to a type Derived, there needs to be a classof in Derived which will accept an object of type Base.
To be concrete, consider the following code:
Shape *S = ...;
if (isa<Circle>(S)) {
/* do something ... */
}
The code of the isa<> test in this code will eventually boil down—after template instantiation and some other machinery—to a check roughly like Circle::classof(S). For more information, see The Contract of classof.
The argument to classof should always be an ancestor class because the implementation has logic to allow and optimize away upcasts/up-isa<>‘s automatically. It is as though every class Foo automatically has a classof like:
class Foo {
[...]
template <class T>
static bool classof(const T *,
::std::enable_if<
::std::is_base_of<Foo, T>::value
>::type* = 0) { return true; }
[...]
};
Note that this is the reason that we did not need to introduce a classof into Shape: all relevant classes derive from Shape, and Shape itself is abstract (has no entry in the Kind enum), so this notional inferred classof is all we need. See Concrete Bases and Deeper Hierarchies for more information about how to extend this example to more general hierarchies.
Although for this small example setting up LLVM-style RTTI seems like a lot of “boilerplate”, if your classes are doing anything interesting then this will end up being a tiny fraction of the code.
For concrete bases (i.e. non-abstract interior nodes of the inheritance tree), the Kind check inside classof needs to be a bit more complicated. The situation differs from the example above in that
Say that SpecialSquare and OtherSpecialSquare derive from Square, and so ShapeKind becomes:
enum ShapeKind {
SK_Square,
+ SK_SpecialSquare,
+ SK_OtherSpecialSquare,
SK_Circle
}
Then in Square, we would need to modify the classof like so:
- static bool classof(const Shape *S) {
- return S->getKind() == SK_Square;
- }
+ static bool classof(const Shape *S) {
+ return S->getKind() >= SK_Square &&
+ S->getKind() <= SK_OtherSpecialSquare;
+ }
The reason that we need to test a range like this instead of just equality is that both SpecialSquare and OtherSpecialSquare “is-a” Square, and so classof needs to return true for them.
This approach can be made to scale to arbitrarily deep hierarchies. The trick is that you arrange the enum values so that they correspond to a preorder traversal of the class hierarchy tree. With that arrangement, all subclass tests can be done with two comparisons as shown above. If you just list the class hierarchy like a list of bullet points, you’ll get the ordering right:
| Shape
| Square
| SpecialSquare
| OtherSpecialSquare
| Circle
The example just given opens the door to bugs where the classofs are not updated to match the Kind enum when adding (or removing) classes to (from) the hierarchy.
Continuing the example above, suppose we add a SomewhatSpecialSquare as a subclass of Square, and update the ShapeKind enum like so:
enum ShapeKind {
SK_Square,
SK_SpecialSquare,
SK_OtherSpecialSquare,
+ SK_SomewhatSpecialSquare,
SK_Circle
}
Now, suppose that we forget to update Square::classof(), so it still looks like:
static bool classof(const Shape *S) {
// BUG: Returns false when S->getKind() == SK_SomewhatSpecialSquare,
// even though SomewhatSpecialSquare "is a" Square.
return S->getKind() >= SK_Square &&
S->getKind() <= SK_OtherSpecialSquare;
}
As the comment indicates, this code contains a bug. A straightforward and non-clever way to avoid this is to introduce an explicit SK_LastSquare entry in the enum when adding the first subclass(es). For example, we could rewrite the example at the beginning of Concrete Bases and Deeper Hierarchies as:
enum ShapeKind {
SK_Square,
+ SK_SpecialSquare,
+ SK_OtherSpecialSquare,
+ SK_LastSquare,
SK_Circle
}
...
// Square::classof()
- static bool classof(const Shape *S) {
- return S->getKind() == SK_Square;
- }
+ static bool classof(const Shape *S) {
+ return S->getKind() >= SK_Square &&
+ S->getKind() <= SK_LastSquare;
+ }
Then, adding new subclasses is easy:
enum ShapeKind {
SK_Square,
SK_SpecialSquare,
SK_OtherSpecialSquare,
+ SK_SomewhatSpecialSquare,
SK_LastSquare,
SK_Circle
}
Notice that Square::classof does not need to be changed.
To be more precise, let classof be inside a class C. Then the contract for classof is “return true if the dynamic type of the argument is-a C”. As long as your implementation fulfills this contract, you can tweak and optimize it as much as you want.
For example, LLVM-style RTTI can work fine in the presence of multiple-inheritance by defining an appropriate classof. An example of this in practice is Decl vs. DeclContext inside Clang. The Decl hierarchy is done very similarly to the example setup demonstrated in this tutorial. The key part is how to then incorporate DeclContext: all that is needed is in bool DeclContext::classof(const Decl *), which asks the question “Given a Decl, how can I determine if it is-a DeclContext?”. It answers this with a simple switch over the set of Decl “kinds”, and returning true for ones that are known to be DeclContext‘s.