[英]The easiest way to achieve a responsive orientation enum? java
我将以“最清晰”的方式快速展示我想要实现的目标:
public enum Orientation {
NORTH, WEST, SOUTH, EAST }
public enum Turn {
LEFT, RIGHT }
因此,我希望这两个枚举能够安全有效地根据移动查找更改的方向:
Orientation orient = Orientation.NORTH;
// orient points to NORTH now
orient = orient.turn(Turn.LEFT);
// orient points to WEST now
我尝试实现这一目标的第一种方法是创建一个地图:
EnumMap<Orientation, EnumMap<Turn, Orientation>>
并静态映射所有方向,但这是一大块map.get.put.get ....并且可能有点过量,导致这种期望的效果:
directionMap.get(NORTH).get(LEFT)
// the value would then be WEST
我接下来的方式是通过Compass类将所有方向链接成一个圆圈。像LinkedCircuitList一样...... < - > NORTH < - > EAST < - > SOUTH < - > WEST < - > NORTH < - >。 ..因此Compass可以有一个静态函数,可以调用该链表的任何成员并向左或向右移动,从而导致方向的正确更改。 但是代码并没有按照我想要的方式运行。
所以我的问题到底是什么,有没有人有这种代码的经验,或者知道如何以一种美好的方式实现理想的结果?
我认为地图解决方案没有错。 如果你想要更简洁的东西:
public enum Orientation {
NORTH, EAST, SOUTH, WEST;
private static Orientation[] vals = values();
Orientation turnTo(Turn t) {
return vals[(4 + this.ordinal() + (t == Turn.RIGHT ? 1 : -1)) % 4];
}
}
然而,这不太干净和可维护(如果有人改变了枚举的顺序,它会破裂)。
一点点清洁(但不太明确):
public enum Orientation {
NORTH(0), EAST(1), SOUTH(2), WEST(3);
private final int p;
Orientation(int p) {
this.p = p;
}
private static Orientation[] vals = new Orientation[4];
static {
for( Orientation o : Orientation.values() )
vals[o.p] = o;
}
Orientation turnTo(Turn t) {
return vals[(4 + this.p + (t == Turn.RIGHT ? 1 : -1)) % 4];
}
}
欺骗java接受这个的另一种方法是使用方法而不是字段:
enum Orientation {
NORTH {
Orientation left(){return WEST;}
Orientation right(){return EAST;}
},
EAST {
Orientation left(){return NORTH;}
Orientation right(){return SOUTH;}
},
SOUTH {
Orientation left(){return EAST;}
Orientation right(){return WEST;}
},
WEST {
Orientation left(){return SOUTH;}
Orientation right(){return NORTH;}
};
abstract Orientation left();
abstract Orientation right();
public Orientation turn(Turn where){
return where == Turn.LEFT ? this.left() : this.right();
}
}
如果你愿意,你可以保存自己的turn()
并只写像Orientation.North.left()
这样的东西。 为您提供非常简洁的语法。
我认为你对循环列表的想法非常好:
public enum Turn {
LEFT(-1), RIGHT(1);
private final int offset;
private Turn(int offset) {
this.offset = offset;
}
public int offset() {
return this.offset;
}
}
public enum Orientation {
NORTH, EAST, SOUTH, WEST;
private static final List<Orientation> orientations =
Arrays.asList(NORTH, EAST, SOUTH, WEST); // to not depend on ordinal
public Orientation turn(Turn to) {
int size = orientations.size();
int myIndex = orientations.indexOf(this);
int remainder = (myIndex + to.offset()) % size;
int index = remainder < 0 ? size + remainder : remainder;
return orientations.get(index);
}
}
这似乎是非常可扩展的,即HARD_LEFT将具有-2
偏移,并且循环的方向列表应该从左到右排序。
不使用序数,易于理解:
public enum Orientation { NORTH, WEST, EAST, SOUTH;
static {
NORTH.left = WEST;
NORTH.right = EAST;
WEST.left = SOUTH;
WEST.right = NORTH;
EAST.left = NORTH;
EAST.right = SOUTH;
SOUTH.left = EAST;
SOUTH.right = WEST;
}
private Orientation left;
private Orientation right;
public Orientation turnTo(Turn t) { return t == Turn.LEFT ? left : right; }
}
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