Value v1 = new Value(1);
Value v4 = new Value(4);
map.put(k1, v1);
map.put(k4, v4); // 这里会替换掉v1
// assertEquals(v1, map.get(k1));
assertEquals(v4, map.get(k1));
assertEquals(v4, map.get(k4));
}
}
class Key {
public int m;
public Key(int m) {
this.m = m;
}
@Override
public int hashCode() {
return m%3;
}
}
class Value {
public int m;
public Value(int m) {
this.m = m;
}
}
class Key2 {
public int m;
public Key2(int m) {
this.m = m;
}
@Override
public int hashCode() {
return m%3;
}
@Override
public boolean equals(Object obj) {
if(obj == null || !(obj instanceof Key2)) return false;
Key2 other = (Key2)obj;
return this.hashCode() == other.hashCode();
}
}
package test.junit;
import java.util.HashMap;
import junit.framework.TestCase;
/**
* HashMap中判断两个Key是否相同:先判断hashCode,在判断是否equals
* e.hash == hash && ((k = e.key) == key || key.equals(k))
* @author xuefeng
*
*/
public class HashMapTest extends TestCase {
/**
* v1和v4的hashCode相同,但是equals不符合,所以v1和v4会组成一个链表。
*/
public void testKey1() {
HashMap
Key k1 = new Key(1);
Key k4 = new Key(4);
Value v1 = new Value(1);
Value v4 = new Value(4);
map.put(k1, v1);
map.put(k4, v4);
assertEquals(v1, map.get(k1));
assertEquals(v4, map.get(k4));
}
/**
* v1和v4的hashCode相同,而且equals也符合,所以v1会被v4替换掉,k1和k4指向同一个v4。
*/
public void testKey2() {
HashMap
Key2 k1 = new Key2(1);
Key2 k4 = new Key2(4);
Value v1 = new Value(1);
Value v4 = new Value(4);
map.put(k1, v1);
map.put(k4, v4); // 这里会替换掉v1
// assertEquals(v1, map.get(k1));
assertEquals(v4, map.get(k1));
assertEquals(v4, map.get(k4));
}
}
class Key {
public int m;
public Key(int m) {
this.m = m;
}
@Override
public int hashCode() {
return m%3;
}
}
class Value {
public int m;
public Value(int m) {
this.m = m;
}
}
class Key2 {
public int m;
public Key2(int m) {
this.m = m;
}
@Override
public int hashCode() {
return m%3;
}
@Override
public boolean equals(Object obj) {
if(obj == null || !(obj instanceof Key2)) return false;
Key2 other = (Key2)obj;
return this.hashCode() == other.hashCode();
}
}