I'm a Java-beginner, so please bear with this one.
I have a class:
class Point {
public int x;
public int y;
public Point (int x, int y) {
this.x = x;
this.y = y;
}
}
I create two instances:
Point a = new Point(1, 1);
Point b = new Point(1, 1);
I want to check if these two points are at the same place. The obvious way, if (a == b) { ... }
, does not work since this seems to be an "are the objects equal?" kind of test, which is not what I want.
I can do if ( (ax == bx) && (ay == by) ) { ... }
, but that solution does not feel good.
How can I take two Point-objects and test them for equality, coordinate wise, in an elegant way?
The standard protocol is to implement the equals()
method:
class Point {
...
@Override
public boolean equals(Object obj) {
if (!(obj instanceof Point)) return false;
Point rhs = (Point)obj;
return x == rhs.x && y == rhs.y;
}
Then you can use a.equals(b)
.
Note that once you've done this, you also need to implement the hashCode()
method.
For classes like yours, I often use Apache Commons Lang 's EqualsBuilder
and HashCodeBuilder
:
class Point {
...
@Override
public boolean equals(Object obj) {
return EqualsBuilder.reflectionEquals(this, obj);
}
@Override
public int hashCode() {
return HashCodeBuilder.reflectionHashCode(this);
}
}
You'll want to override the hashCode()
and equals()
method. If you're using Eclipse, you can have Eclipse do this for you by going to Source -> Generate hashCode() and equals()..
. After you override these methods, you can call:
if(a.equals(b)) { ... }
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