How to use Native Code in Unity

You can use native code in Unity by importing Native Plugins.
Native Plugins are basically libraries of native code written in C/C++. You can import them in C# and make function calls to native code. This way you can gain performance, or use them for whatever reasons you have. I want to note that this isn’t Unity specific thing, I am just using Unity as an example.

How To Create a Native Library

I am working with Visual Studio, so I will show you how to do it in Visual Studio.

You need to create a C++ Dynamic-Link Library (DLL) project. After you write your native code there, you need to build the project and a .dll will be generated. Later you will need to import this .dll in Unity.

Let’s write a simple function so that I can show you how to import it in Unity.

#define DllExport __declspec(dllexport)

extern "C" {
	DllExport float GetFloat() { return 1.0f; }

How To Import It In Unity

First you have to place the .dll somewhere in the Assets folder. You can create a sub-folder specifically fot the native libraries if you want, it doesn’t matter. Let’s say the name of the library is MyLib.dll. This is how you import it.

public class TestBehaviour : MonoBehaviour
    private static extern float GetFloat();

    private void Awake()

How To Import a Class

Lets say you don’t want to import static functions, but an entire class with instance methods instead.

The concept of “instance” method is actually just some sugar that the compiler/framework provides. Actually all methods are static. Instance methods just have an additional hidden parameter “this” that is a reference to the object instance.
If you want to call a native method for a certain native object, you would need to pass the native object reference along to the method call. Usually, you would create a wrapper class in .NET/C# that holds that native pointer (IntPtr) and provides the required methods for the C# environment. Those calls are then forwarded to the native interface. You just have to pass the object along.

Here is an example

// Header
class Person
	Person(const char* name);

	const char* GetName() const;

	char* _name;

// Source
Person::Person(const char* name)
	size_t nameLength = strlen(name);
	_name = new char[nameLength + 1];

	for (int i = 0; i < nameLength; i++)
		_name[i] = name[i];

	_name[nameLength] = '\0';

	delete _name;
	_name = nullptr;

const char* Person::GetName() const
	return _name;

// Export Interface
extern "C" {
	DllExport Person* CreatePerson(const char* name)
		return new Person(name);

	DllExport void DestroyPerson(Person* person)
		delete person;

	DllExport const char* GetPersonName(Person* person)
		return person->GetName();

// C# Wrapper Class
public class Person
    private IntPtr _personPtr = IntPtr.Zero;

    public Person(string name)
        IntPtr namePtr = Marshal.StringToHGlobalAnsi(name);
        _personPtr = CreatePerson(namePtr);

        if (_personPtr != IntPtr.Zero)
            _personPtr = IntPtr.Zero;

    public string GetName()
        IntPtr namePtr = GetPersonName(_personPtr);
        string name = Marshal.PtrToStringAnsi(namePtr);

        return name;

    private static extern IntPtr CreatePerson(IntPtr name);

    private static extern void DestroyPerson(IntPtr person);

    private static extern IntPtr GetPersonName(IntPtr person);

AI Perception in Unreal Engine 4 – How to Setup

For those of you who are wondering what AI Perception is – this is a really nice system in UE4 that provides an easy way for the enemy AI in your game to track hostile/friendly/neutral characters. For example if the enemy AI sees the main character, he will attack it an so on. The AI can become aware of characters through vision, or if they hear them. There are many predefined senses.

The reason for this post is that setting up the AI Perception is very easy, but there is not much documentation yet and it was very hard for me to find a proper way to set it right.

The Setup

There are two components that we need: AIPerceptionComponent and AIPerceptionStimuliSourceComponent. The AIPerceptionComponent is the component that listens for perception stimulants (sight, hearing, etc.). The AIPerceptionStimuliSourceComponent is a stimuli source for the AIPerceptionComponent. The stimuli source stimulates the perception of the enemy AI, so that it can detect the source (our character for example).

I’ve seen many people add the AIPerceptionComponent to the enemy Character, but that is incorrect. It must be added to it’s AI Controller.

AIPerceptionComponent Setup

We create a default AIController, and the only component we add to it is an AIPerception Component. We are going to configure it only for sight sense like so.



I left everything to the default values. As you can see it’s setup only to detect enemies.

AIPerceptionStimuliSourceComponent Setup

The AIPerceptionStimuliSourceComponent must be added to any actors that we want to be stimuli sources for the AIPerceptionComponent. In our case we add it to our character.



Is it working?

Lets add an event to the EnemyAIController Blueprint to see if it’s working.



Well, the “Found” message is not printed so it’s not working. Well I can actually say that it works properly. The message is not being printed on screen because the enemy is not considering us as an enemy. By default all characters are neutral to each other. So how do we tell the enemy AI that the main character is actually an enemy?

Different Teams Setup

We need to place the enemy AI and our character in different teams. By default they are in a team NoTeam which is an enum value that equals to 255. Unfortunately it can’t be done entirely in blueprints. We need to write some code.

In the constructor of the AEnemyAIController we set the team of the AI like so.

AEnemyAIController::AEnemyAIController(const FObjectInitializer& ObjectInitializer)
	: Super(ObjectInitializer)
	// Assign to Team 1

We can’t do the same with the player controller however. Here is the tricky part. A player is controlling a character that is of a certain team, the controller itself doesn’t have a team.

In order to assign our character to a team he needs to implement the IGenericTeamAgentInterface interface. The AIController already implements it. Then we need to override the GetGenericTeamId function. Lets do it.

// KnightCharacter.h
class THEKNIGHT_API AKnightCharacter : public ACharacter, public IGenericTeamAgentInterface

	// ...

	FGenericTeamId TeamId;

	virtual FGenericTeamId GetGenericTeamId() const override;

	// ...

// KnightCharacter.cpp
AKnightCharacter::AKnightCharacter(const FObjectInitializer& ObjectInitializer)
	: Super(ObjectInitializer)
	// ...
	TeamId = FGenericTeamId(0);
	// ...

FGenericTeamId AKnightCharacter::GetGenericTeamId() const
	return TeamId;

Now if we test it again, the enemy is considering us as an enemy and is seeing us.


The reason our character is “Found” if we exit the field of view of the enemy is that the perceptions is updated when we exit the sight of the enemy too.

A Description of the C++ “typename” keyword

Hi guys, I found a really cool post about the description of the typename keyword in C++, so I decided to repost. Here is the original post

A Secondary Use

There is a use of typename that is entirely distinct from the main focus of this discussion. I will present it first because it is easy. It seems to me that someone said “hey, since we’re adding typename anyway, why not make it do this” and people said “that’s a good idea.”

Most older C++ books, when discussing templates, use syntax such as the following:

template <class T> ...

I know when I was starting to learn templates, at first I was a little thrown by the fact that T was prefaced by class, and yet it was possible to instantiate that template with primitive types such as int. The confusion was very short-lived, but the use of class in that context never seemed to fit entirely right. Fortunately for my sensibilities, it is also possible to use typename:

template <typename T> ...

This means exactly the same thing as the previous instance. The typename and class keywords can be used interchangeably to state that a template parameter is a type variable (as opposed to a non-type template parameter).

I personally like to use typename in this context because I think it’s ever-so-slightly clearer. And maybe not so much “clearer” as just conceptually nicer. (I think that good names for things are very important.) Some C++ programmers share my view, and use typename for templates. (However, later we will see how it’s possible that this decision can hurt readability.) Some programmers make a distinction between templates that are fully generic (such as the STL containers) and more special purpose ones that can only take certain classes, and use typename for the former category and class for the latter. Others use class exclusively. This is just a style choice.

However, while I use typename in real code, I will stick to class in this document to reduce confusion with the other use of typename.

The real reason for typename

This discussion I think follows fairly closely appendix B from the book C++ Template Metaprogramming: Concepts, Tools, and Techniques from Boost and Beyond by David Abrahams and Aleksey Gurtovoy, though I don’t have it in front of me now. If there are any deficiencies in my discussion of the issues, that book contains the clearest description of them that I’ve seen.

Some definitions

There are two key concepts needed to understand the description of typename, and they are qualified and dependent names.

Qualified and unqualified names

A qualified name is one that specifies a scope. For instance, in the following C++ program, the references to cout and endl are qualified names:

#include <iostream>

int main()  {
   std::cout << "Hello world!" << std::endl;

In both cases, the use of cout and endl began with std::.

Had I decided to bring cout and endl into scope with a using declaration or directive*, and used just cout by itself, they would have been unqualified names, because they would lack the std::.

(* Remember, a using declaration is like using std::cout;, and actually introduces the name cout into the scope that the using appears in. A using directive is of the form using namespace std; and makes names visible but doesn’t introduce anything. [I’m not sure this is true. Just a warning.])

Note, however, that if I had brought them into scope with using but still used std::cout, it remains a qualified name. The qualified-ness of a name has nothing to do with what scope it’s used in, what names are visible at that point of the program etc.; it is solely a statement about the name that was used to reference the entity in question. (Also note that there’s nothing special about std, or indeed about namespaces at all. vector::iterator is a nested name as well.)

Dependent and non-dependent names

A dependent name is a name that depends on a template parameter. Suppose we have the following declaration (not legal C++):

template <class T>
class MyClass {
   int i;
   vector<int> vi;
   vector<int>::iterator vitr;

   T t;
   vector<T> vt;
   vector<T>::iterator viter;

The types of the first three declarations are known at the time of the template declaration. However, the types of the second set of three declarations are not known until the point of instantiation, because they depend on the template parameter T.

The names T, vector, and vector::iterator are called dependent names, and the types they name are dependent types. The names used in the first three declarations are called non-dependent names, at the types are non-dependent types.

The final complication in what’s considered dependent is that typedefs transfer the quality of being dependent. For instance:

typedef T another_name_for_T;

another_name_for_T is still considered a dependent name despite the type variable T from the template declaration not appearing.

Some other issues of wording

Note that while there is a notion of a dependent type, there is not a notion of a qualified type. A type can be unqualified in one instance, and qualified the next; the qualification is a property of a particular naming of a type, not of the type itself. (Indeed, when a type is first defined, it is always unqualified.)

However, it will be useful to refer to a qualified type; what I mean by this is a qualified name that refers to a type. I will switch back to the more precise wording when I talk about the rules of typename.

The problem

So now we can consider the following example:

template <class T>
void foo() {
   T::iterator * iter;

What did the programmer intend this bit of code to do? Probably, what the programmer intended was for there to be a class that defined a nested type called iterator:

class ContainsAType {
   class iterator { ... }:

and for foo to be called with an instantiation of T being that type:


In that case, then line 3 would be a declaration of a variable called iter that would be a pointer to an object of type T::iterator (in the case of ContainsAType, int*, making iter a double-indirection pointer to an int). So far so good.

However, what the programmer didn’t expect is for someone else to come up and declare the following class:

class ContainsAValue {
   static int iterator;

and call foo instantiated with it:


In this case, line 3 becomes a statement that evaluates an expression which is the product of two things: a variable called iter (which may be undeclared or may be a name of a global) and the static variable T::iterator.

Uh oh! The same series of tokens can be parsed in two entirely different ways, and there’s no way to disambiguate them until instantiation. C++ frowns on this situation. Rather than delaying interpretation of the tokens until instantiation, they change the language:

Before a qualified dependent type, you need typename

To be legal, assuming the programmer intended line 3 as a declaration, they would have to write

template <class T>
void foo() {
   typename T::iterator * iter;

Without typename, there is a C++ parsing rule that says that qualified dependent names should be parsed as non-types even if it leads to a syntax error. Thus if there was a variable called iter in scope, the example would be legal; it would just be interpreted as multiplication. Then when the programmer instantiated foo with ContainsAType, there would be an error because you can’t multiply something by a type.

typename states that the name that follows should be treated as a type. Otherwise, names are interpreted to refer to non-types.

This rule even holds if it doesn’t make sense even if it doesn’t make sense to refer to a non-type. For instance, suppose we were to do something more typical and declare an iterator instead of a pointer to an iterator:

template <class T>
void foo() {
   typename T::iterator iter;

Even in this case, typename is required, and omitting it will cause compile error. As another example, typedefs also require use:

template <class T>
void foo() {
   typedef typename T::iterator iterator_type;

The rules

Here, in excruciating detail, are the rules for the use of typename. Unfortunately, due to something which is hopefully not-contagious apparently affecting the standards committee, they are pretty complicated.

  1. typename is prohibited in each of the following scenarios:
    • Outside of a template definition. (Be aware: an explicit template specialization (more commonly called a total specialization, to contrast with partial specializations) is not itself a template, because there are no missing template parameters! Thus typename is always prohibited in a total specialization.)
    • Before an unqualified type, like int or my_thingy_t.
    • When naming a base class. For example, template class my_class : C::some_base_type { ... }; may not have a typename before C::some_base_type.
    • In a constructor initialization list.
  2. typename is mandatory before a qualified, dependent name which refers to a type (unless that name is naming a base class, or in an initialization list).
  3. typename is optional in other scenarios. (In other words, it is optional before a qualified but non-dependent name used within a template, except again when naming a base class or in an initialization list.)
    Again, these rules are for standard C++98/03. C++11 loosens the restrictions. I will update this page after I figure out what they are.