problem with contains in ArrayList

clankill3r

problem with contains in ArrayList

in Library and Tool Development • 10 months ago

I can't post all code so i hope someone still have a clue what it might be.
I have a path and a selection tool to select anchors.
If i hold shift i want a toggle selection.

For debugging i print A, B, C etc. at serveral steps.

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public void handleReleased(float x, float y) {
if(startX != x || startY != y) {
// rectangular selection
float x1 = PApplet.min(x, startX);
float y1 = PApplet.min(y, startY);
float x2 = PApplet.max(x, startX);
float y2 = PApplet.max(y, startY);
boolean toggle = checkToggle();
if(toggle == false) {
toolHandler.layerStack.clearAnchorSelection();
}
lastAnchorSelection = new ArrayList<CVector>(toolHandler.layerStack.anchorSelection);
newAnchorSelection.clear();
System.out.println(lastAnchorSelection.size());
for (Layer layer : toolHandler.layerStack.layers) {
rectangularSelection(x1, y1, x2, y2, layer, toggle);
}
toolHandler.layerStack.anchorSelection.clear();
toolHandler.layerStack.anchorSelection.addAll(newAnchorSelection);
}
}

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public void rectangularSelection(float x1, float y1, float x2, float y2, Layer layer, boolean toggle) {
for (BezierVertexPath p : layer.paths) {
if(p.locked()) continue;
for(CVector v : p.getAnchors()) {
if(x1 <= v.x() && y1 <= v.y() && x2 >= v.x() && y2 >= v.y()) {
// CVectors can be shared, there for with toggle we don't want to turn them on and off
// every time we meat them. So we look if they where present in the old selection.
// And based on we add them to the new selection.
System.out.println("A");
if(toggle) {
System.out.println("B");
// ->
if(lastAnchorSelection.contains(v) == false) {
System.out.println("lastAnchorSelection.size(): "+lastAnchorSelection.size());
System.out.println("C");
// we only want to add it once
if(newAnchorSelection.contains(v) == false) {
System.out.println("D");
newAnchorSelection.add(v);
}
}
else {
System.out.println("E");
if(newAnchorSelection.contains(v) == false) {
System.out.println("F");
newAnchorSelection.add(v);
}
}
}
else {
System.out.println("G");
if(newAnchorSelection.contains(v) == false) {
System.out.println("H");
newAnchorSelection.add(v);
}
}
}
}
}
// now the subLayers
if(layer.subLayers != null) {
for(Layer l : layer.subLayers) {
rectangularSelection(x1, y1, x2, y2, l, toggle);
}
}
}

If i select a anchor, and then select it again while holding shift then it prints

A
B
lastAnchorSelection.size(): 1
C
D

A and B are correct.
howver, there is this line:

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if(lastAnchorSelection.contains(v) == false) {

lastAnchorSelection holds 1 anchor, the one i select again, but still lastAnchorSelection.contains(v) is false.

I only don't understand how that can be.
An ArrayList holds a reference right...

anyone a clue?

Replies(6)

clankill3r

Re: problem with contains in ArrayList

10 months ago

I added 2 new lines to show how strange this is.
It prints 1 for the second line, so lastAnchorSelection contains 1 item.
But since it passed the if value, you would assume that item can not be v.
Then if i test the item in the array against v it equals true :S

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if(lastAnchorSelection.contains(v) == false) {
System.out.println("lastAnchorSelection.size(): "+lastAnchorSelection.size());
CVector test = lastAnchorSelection.get(0);
System.out.println(test == v);
System.out.println("C");

If someone wants to look into this then please drop me a message and i will parse all the files needed.

clankill3r

Re: problem with contains in ArrayList

10 months ago

got it!
Posted the problem on stackOverFlow and someone said v has probably it's own equal method.

my v is basicly a copy of PVector and PVector has it's own equal:

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PVector a = new PVector();
PVector b = new PVector();
println(a.equals(b));
PGraphics pg = createGraphics(100, 100);
PGraphics pg2 = createGraphics(100, 100);
println(pg.equals(pg2));

the above will print true and then false.
The equal method of PVector just checks if xyz are matching.

Wouldn't it be better if the method was called matches() ?

quarks

Re: problem with contains in ArrayList

10 months ago

We have

a class called CVector
an ArrayList of CVector object references

Now consider the following code

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CVector v1, v2, v3;
v1 = new CVector();
v2 = new CVector();
v3 = v2;

Although we have 3 CVector references we only have 2 CVector objects. The references v2 and v3 refer to the same object. SInce we have used the default constructor then the attributes of the 2 objects created will be the same.

If you have not implemented the Comparable interface in the CVector class then comparisons are based on the object reference so we get
In the Java collections (ArrayList, LinkedList, HashMap etc) comparisons are done with the equals() method so if we use it with our objects we get

v1.equals(v2) is false
v1.equals(v3) is false
v2.equals(v3) is true

even though the internal states of v1 and v2 are the same.

The contains method compares the object with each one in the ArrayList using the equals() method so if you want don't want to compare references but rather compare objects based on their internal state your class must implement the Comparable interface and it must have its implementation of the equals() method.

HTH

clankill3r

Re: problem with contains in ArrayList

10 months ago

Well just like PVector, i didn't suspect to have it's own equal method, it's also confusing

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PVector a = new PVector();
PVector b = new PVector();
ArrayList<PVector> list = new ArrayList<PVector>();
list.add(a);
println(list.contains(b));

this will print true

rbrauer

Re: problem with contains in ArrayList

10 months ago

It looks like the PVector equals(...) method (which gets called by ArrayList's contains(...)) returns true if both represent the same coordinate. When you use the no-args constructor they are both 0,0,0 so a.equals(b) is true.

timpulver

Re: problem with contains in ArrayList

10 months ago

You can overwrite the equals method, to make sure comparison is done as you want it to be:

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public class Thing { public int value; public Thing (int x) { this.value = x; }  public boolean equals(Object object) { boolean sameSame = false; if (object != null && object instanceof Thing) { sameSame = this.value == ((Thing) object).value; } return sameSame; } }