问题总结笔记

1.向量旋转

       问题： 将一个向量旋转90

        方法：旋转矩阵

FVector FrontDir = EndMousePoint - Point;
FrontDir.Normalize();
FVector Left = FVector(-FrontDir.Y, FrontDir.X, 0);
Verties.Add(Point + Left * (WallWedith / 2));
Verties.Add(FVector(Verties[0].X, Verties[0].Y, Verties[0].Z + WallHeight));
Verties.Add(Point - Left * (WallWedith / 2));
Verties.Add(FVector(Verties[2].X, Verties[2].Y, Verties[2].Z + WallHeight));
Verties.Add(EndMousePoint - Left * (WallWedith / 2));
Verties.Add(FVector(Verties[4].X, Verties[4].Y, Verties[4].Z + WallHeight));
Verties.Add(EndMousePoint + Left * (WallWedith / 2));
Verties.Add(FVector(Verties[6].X, Verties[6].Y, Verties[6].Z + WallHeight));

Point：起点坐标， EndMousePoint：终点坐标。

依次加入点的坐标，WallWedith：立方体的宽，WallHeight：立方体的高。

2.ProceduralMeshActor

问题：利用ProceduralMeshActor创建立方体

//侧面
for (int32 i = 0; i < Verties.Num() - 1; i = i + 2) {

	Triangles.Add(i);
	Triangles.Add((i+2)% Verties.Num());
	Triangles.Add(i+1);
	Triangles.Add(i+1);
	Triangles.Add((i+2)% Verties.Num());
	Triangles.Add((i+3) % Verties.Num());
}

//上下面
for (int32 i = 0; i < 2; i++) {
	for (int32 j = 0; j < Verties.Num(); j++) {
		if (j % 2 == i) {
			Triangles.Add(j);
		}
	}
	Triangles.Add(i);
	Triangles.Add(i+4);

}

先构建一个侧面，通过循环构建所有侧面，上下面单独处理

3.鼠标屏幕坐标转换成空间坐标（有偏差，有待改进） 

FVector2D MousePosition;

UWorld* world = GetWorld();
if (world)
{
	APlayerController* pc = world->GetFirstPlayerController();
	if (pc)
	{
		ULocalPlayer* LocalPlayer = pc->GetLocalPlayer();
		if (LocalPlayer && LocalPlayer->ViewportClient)
		{
			LocalPlayer->ViewportClient->GetMousePosition(MousePosition);
		}
	}
}

float Scale = UWidgetLayoutLibrary::GetViewportScale(this);

FVector WorldLocation;
FVector WorldDirection;
APlayerController::DeprojectScreenPositionToWorld(MousePosition.X, MousePosition.Y, WorldLocation,WorldDirection);

MouseLocal = FVector(WorldDirection.X * 1000/ Scale, WorldDirection.Y * 1000/ Scale, 0);

记得在Build.cs文件里添加"UMG"