java中为什么重写equals时必须重写hashCode方法？

/** * Returns a hash code value for the object. This method is * supported for the benefit of hash tables such as those provided by * {@link java.util.HashMap}. * <p> * The general contract of {@code hashCode} is: * <ul> * <li>Whenever it is invoked on the same object more than once during * an execution of a Java application, the {@code hashCode} method * must consistently return the same integer, provided no information * used in {@code equals} comparisons on the object is modified. * This integer need not remain consistent from one execution of an * application to another execution of the same application. * <li>If two objects are equal according to the {@code equals(Object)} * method, then calling the {@code hashCode} method on each of * the two objects must produce the same integer result. * <li>It is <em>not</em> required that if two objects are unequal * according to the {@link java.lang.Object#equals(java.lang.Object)} * method, then calling the {@code hashCode} method on each of the * two objects must produce distinct integer results. However, the * programmer should be aware that producing distinct integer results * for unequal objects may improve the performance of hash tables. * </ul> * <p> * As much as is reasonably practical, the hashCode method defined by * class {@code Object} does return distinct integers for distinct * objects. (This is typically implemented by converting the internal * address of the object into an integer, but this implementation * technique is not required by the * Java&trade; programming language.) * * @return a hash code value for this object. * @see java.lang.Object#equals(java.lang.Object) * @see java.lang.System#identityHashCode */ public native int hashCode();

/** * Indicates whether some other object is "equal to" this one. * <p> * The {@code equals} method implements an equivalence relation * on non-null object references: * <ul> * <li>It is <i>reflexive</i>: for any non-null reference value * {@code x}, {@code x.equals(x)} should return * {@code true}. * <li>It is <i>symmetric</i>: for any non-null reference values * {@code x} and {@code y}, {@code x.equals(y)} * should return {@code true} if and only if * {@code y.equals(x)} returns {@code true}. * <li>It is <i>transitive</i>: for any non-null reference values * {@code x}, {@code y}, and {@code z}, if * {@code x.equals(y)} returns {@code true} and * {@code y.equals(z)} returns {@code true}, then * {@code x.equals(z)} should return {@code true}. * <li>It is <i>consistent</i>: for any non-null reference values * {@code x} and {@code y}, multiple invocations of * {@code x.equals(y)} consistently return {@code true} * or consistently return {@code false}, provided no * information used in {@code equals} comparisons on the * objects is modified. * <li>For any non-null reference value {@code x}, * {@code x.equals(null)} should return {@code false}. * </ul> * <p> * The {@code equals} method for class {@code Object} implements * the most discriminating possible equivalence relation on objects; * that is, for any non-null reference values {@code x} and * {@code y}, this method returns {@code true} if and only * if {@code x} and {@code y} refer to the same object * ({@code x == y} has the value {@code true}). * <p> * Note that it is generally necessary to override the {@code hashCode} * method whenever this method is overridden, so as to maintain the * general contract for the {@code hashCode} method, which states * that equal objects must have equal hash codes. * * @param obj the reference object with which to compare. * @return {@code true} if this object is the same as the obj * argument; {@code false} otherwise. * @see #hashCode() * @see java.util.HashMap */ public boolean equals(Object obj) { return (this == obj); }

1.在 Java 应用程序执行期间，在对同一对象多次调用 hashCode 方法时，必须一致地返回相同的整数，前提是将对象进行 equals 比较时所用的信息没有被修改。
 从某一应用程序的一次执行到同一应用程序的另一次执行，该整数无需保持一致。 2.如果根据 equals(Object) 方法，两个对象是相等的，那么对这两个对象中的每个对象调用 hashCode 方法都必须生成相同的整数结果。 3.如果根据 equals(java.lang.Object) 方法，两个对象不相等，那么两个对象不一定必须产生不同的整数结果。
 但是，程序员应该意识到，为不相等的对象生成不同整数结果可以提高哈希表的性能。

/** * @see Person * @param args */ public static void main(String[] args) { HashMap<Person, Integer> map = new HashMap<Person, Integer>(); Person p = new Person("jack",22,"男"); Person p1 = new Person("jack",22,"男"); System.out.println("p的hashCode:"+p.hashCode()); System.out.println("p1的hashCode:"+p1.hashCode()); System.out.println(p.equals(p1)); System.out.println(p == p1); map.put(p,888); map.put(p1,888); map.forEach((key,val)->{ System.out.println(key); System.out.println(val); }); }

public class Person { private String name; private int age; private String sex; Person(String name,int age,String sex){ this.name = name; this.age = age; this.sex = sex; } }

p的hashCode:356573597 p1的hashCode:1735600054 false false com.blueskyli.练习.Person@677327b6 888 com.blueskyli.练习.Person@1540e19d 888

public class Person { private String name; private int age; private String sex; Person(String name,int age,String sex){ this.name = name; this.age = age; this.sex = sex; } @Override public boolean equals(Object obj) { if(obj instanceof Person){ Person person = (Person)obj; return name.equals(person.name); } return super.equals(obj); } }

p的hashCode:356573597 p1的hashCode:1735600054 true false com.blueskyli.练习.Person@677327b6 888 com.blueskyli.练习.Person@1540e19d 888

public class Person { private String name; private int age; private String sex; Person(String name,int age,String sex){ this.name = name; this.age = age; this.sex = sex; } @Override public boolean equals(Object obj) { if(obj instanceof Person){ Person person = (Person)obj; return name.equals(person.name); } return super.equals(obj); } @Override public int hashCode() { return name.hashCode(); } }

p的hashCode:3254239 p1的hashCode:3254239 true false com.blueskyli.练习.Person@31a7df 888