How to use switch with multiple enum values?

Question

I m trying this code:

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];

        }
}

But I get an error when compile:

characters 8-16 : This pattern is unused

So, it pointing at case Cav, Vex:

How can I check Cav or Vex in the case above?

EDIT

I found that if I removed case Cav & Case Vex, then case Cav, Vex will work, but this is not what I want, can't I repeat the pattern usage in an or experision? like (case Cav||Vex)?

case (Cav || Vex) would result into:

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

Short answer: no way currently, you can only match one enum option in one place(not counting guarded options). So, duplicate your code for each enum option and live a happy life(this code will be also easier to read) or use a seconds switch(which might be shorter and easier in some more complex cases).

Answer 2

There are only 3 choices for the value of aLeftType, either Vex, Cav, or 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.
}

So really, the code of the 4th case will never be run. It is similar to:

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

That means, you have to think again want do you really want to do.

Answer 3

Usually when you want to make the same thing in different situation, you make a function for that :)

function drawLense(aLeftType:SideType){

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

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

            case Plano:return ;
        }
}

function functionCalledIfCavOrVex(/*...*/){
        points1= drawCurve(1, -90 + offset + trim, 180 - offset * 2 - (trim * 2), leftCenter, aRadius);
        _LB = points1[0];
        _LA = points1[1];
}

Answer 4

try:

case Cav | Vex:
  trace("cav or vex");

Hope it helps.