英文:
Is there a way to simplify this code comparing two Points with fields x and y for equality?
问题
下面的代码旨在比较具有x和y字段的两个Point对象是否相等。有没有简化这段代码的方法？
public class Point {
private int x;
private int y;
public Point(int x, int y) {
this.x = x;
this.y = y;
}
public boolean equals(Object o) {
if (o instanceof Point) {
Point c = (Point) o;
if (!(c.x == this.x && c.y == this.y)) {
return false;
} else {
return true;
}
}
return false;
}
}
英文:
Is there a way to simplify the following code which aims to compare two Points with fields x and y for equality?
public class Point {
private int x;
private int y;
public Point(int x, int y) {
this.x = x;
this.y = y;
}
public boolean equals(Object o) {
if (o instanceof Point) {
Point c = (Point) o;
if (!(c.x == this.x && c.y == this.y)) {
return false;
} else {
return true;
}
}
return false;
}
}
答案1
得分: 1
这可以简化为：
@Override
public boolean equals(Object o) {
if (!(o instanceof Point)) {
return false;
}
Point c = (Point) o;
return x == c.x && y == c.y;
}
解释
将条件中返回true/false的if/then布尔条件替换为布尔本身的结果（或取反的结果）通常会使代码变得更短、更清晰。
例如：
if (someBooleanCondition()) { return true; } else { return false; }
可以简化为：
return someBooleanCondition();
将这个思想应用到你的例子中：
if (!(c.x == this.x && c.y == this.y)) {
return false;
} else {
return true;
}
可以简化为：
return c.x == this.x && c.y == this.y;
我将instanceof检查放在前面，因为我认为在异常情况下使用短路返回会使代码更易读，不过这是主观的。
英文:
This can be simplified to:
@Override
public boolean equals(Object o) {
if (!(o instanceof Point)) {
return false;
}
Point c = (Point) o;
return x == c.x && y == c.y;
}
Explanation
It is shorter and often clearer to replace an if/then boolean conditional that returns true/false in the condition with the results (or negated results) of the boolean itself.
For instance:
if (someBooleanCondition()) { return true; } else { return false; }
can be simplified to
return someBooleanCondition();
Applying this to your example:
if (!(c.x == this.x && c.y == this.y)) {
return false;
} else {
return true;
}
becomes:
return c.x == this.x && c.y == this.y;
I've put the instanceof check first, since I feel short-circuit returning the exceptional case leads to easier to read code, though that is subjective.
答案2
得分: 0
以下是最简单的代码示例：
public boolean equals(Object o) {
if (o == null || o.getClass() == getClass()) return false;
return x == ((Point) o).x && y == ((Point) o).y;
}
希望对您有帮助！
英文:
I believe the following is the simplest
public boolean equals(Object o) {
if (o == null || o.getClass() == getClass()) return false;
return x == ((Point) o).x && y == ((Point) o).y;
}
Hope it helped you!
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