haxe - 如何使用具有多个枚举值的开关？

Question

我正在尝试这段代码:

enum SideType
{
    Vex;
    Cav;
    Plano;
}

function drawLense(aLeftType:SideType){

switch (aLeftType) 
        {
            case Cav:
                leftCenter = -aRadius - d * 0.5;

            case Vex:
                leftCenter = -Math.cos(( -90 + offset) * Math.PI / 180) * aRadius-d*0.5;

            case Plano:return ;
            case Cav, Vex:
                points1= drawCurve(1, -90 + offset + trim, 180 - offset * 2 - (trim * 2), leftCenter, aRadius);
                _LB = points1[0];
                _LA = points1[1];

        }
}

但是编译的时候报错:

characters 8-16 : This pattern is unused

因此，它指向案例 Cav，Vex:

在上述情况下如何检查 Cav 或 Vex？

编辑

我发现如果我删除 case Cav & Case Vex，那么 case Cav, Vex 会起作用，但这不是我想要的，我不能在 or experision 中重复模式使用吗？像(案例 Cav||Vex)？

case (Cav || Vex) 会导致:

src/com/optics/components/Lense.hx:343: characters 8-38 : Case expression must be a constant value or a pattern, not an arbitrary expression

Answer 1

aLeftType 的值只有 3 个选择，Vex、Cav 或 Plano。

var aLeftType = Vex;
switch (aLeftType) 
{
    case Cav:
        // If aLeftType is `Cav`, run this line.
    case Vex:
        // If aLeftType is `Vex`, run this line.
    case Plano:
        // If aLeftType is `Plano`, run this line.
    case Cav, Vex:
        // If aLeftType is `Vex` or `Plano`, run this line...
        // But the first 2 cases already covered `Vex` and `Plano`,
        // so it will never be reached.
}

所以真的，第 4 种情况的代码永远不会运行。它类似于:

if (a == 1) {
    trace("a is 1");
} else if (a == 1) {
    trace("a is really 1"); // This can never be reached.
}

这意味着，你必须重新考虑你是否真的想做。