When Java overwrites equals always overwrite hashcode
- 2020-05-19 04:54:10
- OfStack
When Java overwrites equals, always overwrite hashcode
Recently, I have learned the basic knowledge of java. When I encounter Java covering equals, I always need to cover hashcode. After direct discussion with my colleagues and online inquiry, I will sort out the information here to help you understand.
Specific implementation code:
package cn.xf.cp.ch02.item9;
import java.util.HashMap;
import java.util.Map;
public class PhoneNumber
{
private final short areaCode;
private final short prefix;
private final short lineNumber;
public PhoneNumber(int areaCode, int prefix, int lineNumber)
{
rangeCheck(areaCode, 999, "area code");
rangeCheck(prefix, 999, "prefix");
rangeCheck(lineNumber, 9999, "line number");
this.areaCode = (short) areaCode;
this.prefix = (short) prefix;
this.lineNumber = (short) lineNumber;
}
private static void rangeCheck(int arg, int max, String name)
{
if (arg < 0 || arg > max)
throw new IllegalArgumentException(name + ": " + arg);
}
@Override
public boolean equals(Object o)
{
if (o == this)
return true;
if (!(o instanceof PhoneNumber))
return false;
PhoneNumber pn = (PhoneNumber) o;
return pn.lineNumber == lineNumber && pn.prefix == prefix && pn.areaCode == areaCode;
}
/*
@Override
// As to why it is used 31 , this is the recommended value, and studies have shown that this number performs better
public int hashCode()
{
int result = 17;
result = 31 * result + areaCode;
result = 31 * result + prefix;
result = 31 * result + lineNumber;
return result;
}
*/
// if 1 The hash code is cached inside the object if the object is not changed very often and is expensive
// with volatile The thread reads the most modified value of the variable every time it USES the variable.
private volatile int hashcode;
@Override
public int hashCode()
{
int result = hashcode;
if (result == 0)
{
result = 17;
result = 31 * result + areaCode;
result = 31 * result + prefix;
result = 31 * result + lineNumber;
hashcode = result;
}
return result;
}
public static void main(String[] args)
{
Map<PhoneNumber, String> m = new HashMap<PhoneNumber, String>();
m.put(new PhoneNumber(707, 867, 5309), "Jenny");
// There is no return jenny Oh, it will null This is because put Objects put them into different hash buckets
System.out.println(m.get(new PhoneNumber(707, 867, 5309)));
}
}
Thank you for reading, I hope to help you, thank you for your support of this site!