Java Array.contains方法代碼示例

import com.badlogic.gdx.utils.Array; //導入方法依賴的package包/類
boolean checkMatch(TileType[] tiles, Array<TileType> black, Array<TileType> white, boolean nullMatches) {
	if(tiles.length != map.length)
		throw new IllegalArgumentException("tile type array must be of equal size to overlap map array; "+tiles.length+"!="+map.length);
	
	// if the white or black tiles are null, then it means it works if it doesn't match the other color.
	
	for(int i = 0; i < tiles.length; i++) {
		int matchRule = map[i];
		TileType type = tiles[i];
		if(type == null) {
			if(nullMatches) continue;
			else return false;
		}
		
		if(matchRule == WHITE && (white == null && black.contains(type, true) || white != null && !white.contains(type, true)))
			return false;
		if(matchRule == BLACK && (black == null && white.contains(type, true) || black != null && !black.contains(type, true)))
			return false;
		
		/*if(black == null && type != white && (matchRule == BLACK || matchRule == WHITE)) {
			// here, make sure that if the tile is adjacent to a white tile, that the current type is not greater than the white tile in z order.
			if(i >= height && map[i-height] == WHITE && tiles[i-height] == white || i+height < tiles.length && map[i+height] == WHITE && tiles[i+height] == white || i%height > 0 && map[i-1] == WHITE && tiles[i-1] == white || i%height < height-1 && map[i+1] == WHITE && tiles[i+1] == white)
				if(TileType.tileSorter.compare(type, white) > 0)
					return false;
		}*/
	}
	
	return true;
}
 

import com.badlogic.gdx.utils.Array; //導入方法依賴的package包/類
/** @noinspection unchecked*/
@NotNull
public static <S, T extends S> Class<? extends S> getDirectSubclass(@NotNull Class<S> superClass, @NotNull Class<T> target) {
	if (superClass.equals(target))
		return target; // this returns the super class, not it's direct subclass, but in this situation that's that best we're gonna get.
	
	// because we only ever use super classes from the original target, it is safe at any time to assume that any class found is super T.
	
	Class<? super T> targetSuper = target.getSuperclass(); // due to the equality check above, this class will always extend S.
	
	boolean hasSuper = targetSuper != null; // if target is the Object class, then there won't a super class.
	if (hasSuper && targetSuper.equals(superClass))
		return target;
	
	Array<Class<? super T>> parents = new Array<>((Class<? super T>[]) target.getInterfaces());
	if (parents.contains(superClass, false))
		return target;
	
	// target is not a direct subclass of superClass, so recurse first into the super class, and then into the interfaces.
	
	if (hasSuper && superClass.isAssignableFrom(targetSuper))
		return getDirectSubclass(superClass, (Class<? extends S>) targetSuper);
	else {
		// it HAS to be one of the interfaces that extend the superClass.
		for(Class<? super T> parent: parents)
			if(superClass.isAssignableFrom(parent))
				return getDirectSubclass(superClass, (Class<? extends S>) parent);
		
		// if the code reaches here, then something went terribly wrong, because target is required to extend superClass, but at this point we've gone through both the interfaces and the super class, and haven't found it.
		
		throw new IncompatibleClassChangeError("Class " + target + " is required by generics to extend " + superClass + ", but neither the super class nor any implemented interfaces do.");
	}
}