How to get subclass array reference from UGameplaystatistics::GetAllActorsOfClass

Development Programming & Scripting C++
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1

I’m sorry if this is badly put. Here is my issue

I use GetAllActorsOfClass to get all actors in the scene of the ASomeActor class
It outputs to an array of AActor
I want it to output to an array of ASomeActor

I hope this makes sense. Any help would be greatly appreciated :slight_smile:

2

I don’t think there is a way to cast an entire array’s contents easily. You could try this method, though:

TArray<AActor*> ActorArray;
TArray<ASomeActor*> SomeActorArray;
UGameplayStatics::GetAllActorsOfClass(GetWorld(), ASomeActor::StaticClass(), ActorArray);
for (AActor* UncastedActor : ActorArray) {
    SomeActorArray.Add(Cast<ASomeActor>(UncastedActor));
}

// SomeActorArray is now an array of ASomeActor.

Beware, don’t run this every frame as it is somewhat expensive to run.

3

If you only need this functionality from C++ and you plan to do this for more than 1 actor class, you can create a templated version of GetAllActorsOfClass. The implementation below is similar to UGameplayStatics::GetAllActorsOfClass but omits the ActorClass==nullptr check since the compiler will make sure ActorClass is valid.

#include "EngineUtils.h"

template <typename ActorClass>
void TGetAllActorsOfClass(UWorld* World, TArray<ActorClass*>& OutActors)
{
	OutActors.Reset();

	if (World)
	{
		for (TActorIterator<ActorClass> It(World); It; ++It)
		{
			OutActors.Add(*It);
		}
	}
}

void Test(UWorld* World)
{
	TArray<APawn*> MyPawns;
	TGetAllActorsOfClass(World, MyPawns);

	TArray<ACharacter*> MyCharacters;
	TGetAllActorsOfClass(World, MyCharacters);

	/*
	Won't compile because UObject not a sub-class of AActor.
	TArray<UObject*> MyObjects;
	TGetAllActorsOfClass(World, MyObjects);
	*/
}

Note this function cannot be used from Blueprints. If you need to create typed arrays accessible from blueprints, you’ll need to make a blueprint library function for each array type you want to get actors for.

2 Likes
4

Wow! That’s really neat!