String池(又名实习池)和Integer池进一步模糊了区别，并允许您在某些情况下对对象使用==而不是.equals

这可以为您提供更好的性能(?)，但代价是更大的复杂性。

例如:

assert "ab" == "a" + "b";

Integer i = 1;
Integer j = i;
assert i == j;

复杂性权衡:以下内容可能会让你大吃一惊:

assert new String("a") != new String("a");

Integer i = 128;
Integer j = 128;
assert i != j;

我建议你远离这样的微观优化，总是用.equals表示对象，用==表示原语:

assert (new String("a")).equals(new String("a"));

Integer i = 128;
Integer j = 128;
assert i.equals(j);

可能值得添加的是，对于基本类型的包装器对象，例如Int, Long, Double - ==如果两个值相等将返回true。

Long a = 10L;
Long b = 10L;

if (a == b) {
    System.out.println("Wrapped primitives behave like values");
}

相比之下，将上述两个long放入两个单独的数组列表中，equals将它们视为相同，而==则不同。

ArrayList<Long> c = new ArrayList<>();
ArrayList<Long> d = new ArrayList<>();

c.add(a);
d.add(b);
if (c == d) System.out.println("No way!");
if (c.equals(d)) System.out.println("Yes, this is true.");

==运算符测试两个变量是否有相同的引用 (又名指向内存地址的指针)。

String foo = new String("abc");
String bar = new String("abc");

if(foo==bar)
// False (The objects are not the same)

bar = foo;

if(foo==bar)
// True (Now the objects are the same)

而equals()方法测试两个变量是否引用对象 具有相同的状态(值)。

String foo = new String("abc");
String bar = new String("abc");

if(foo.equals(bar))
// True (The objects are identical but not same)

欢呼:-)

简而言之，答案是肯定的。

在Java中，==操作符比较两个对象，以确定它们是否指向相同的内存位置;而.equals()方法实际上是比较两个对象，看看它们是否具有相同的对象值。

一般来说，你问题的答案是“是的”，但是……

.equals(...) will only compare what it is written to compare, no more, no less. If a class does not override the equals method, then it defaults to the equals(Object o) method of the closest parent class that has overridden this method. If no parent classes have provided an override, then it defaults to the method from the ultimate parent class, Object, and so you're left with the Object#equals(Object o) method. Per the Object API this is the same as ==; that is, it returns true if and only if both variables refer to the same object, if their references are one and the same. Thus you will be testing for object equality and not functional equality. Always remember to override hashCode if you override equals so as not to "break the contract". As per the API, the result returned from the hashCode() method for two objects must be the same if their equals methods show that they are equivalent. The converse is not necessarily true.

==操作符总是比较引用。但是如果

equals()方法

这取决于我们的实现，如果我们是overridden equals method，而不是在overridden method中给出的实现的基础上比较对象。

 class A
 {
   int id;
   String str;

     public A(int id,String str)
     {
       this.id=id;
       this.str=str;
     }

    public static void main(String arg[])
    {
      A obj=new A(101,"sam");
      A obj1=new A(101,"sam");

      obj.equals(obj1)//fasle
      obj==obj1 // fasle
    }
 }

在上面的代码中，obj和obj1对象包含相同的数据，但引用不相同，所以=也返回false和==。 但如果我们重写equals method than

 class A
 {
   int id;
   String str;

     public A(int id,String str)
     {
       this.id=id;
       this.str=str;
     }
    public boolean equals(Object obj)
    {
       A a1=(A)obj;
      return this.id==a1.id;
    }

    public static void main(String arg[])
    {
      A obj=new A(101,"sam");
      A obj1=new A(101,"sam");

      obj.equals(obj1)//true
      obj==obj1 // fasle
    }
 }

知道签出它将返回true和false对于相同的情况，只是我们覆盖了

等于方法。

它以对象的内容(id)为基础对对象进行比较

但= =

还比较对象的引用。