想对问题1的结果作出解释。
*** 1。以下方法的输出是什么?
public static void main(String[] args) {
Integer i1=new Integer(1);
Integer i2=new Integer(1);
String s1=new String("Today");
String s2=new String("Today");
System.out.println(i1==i2);
System.out.println(s1==s2);
System.out.println(s1.equals(s2));
System.out.println(s1!=s2);
System.out.println( (s1!=s2) || s1.equals(s2));
System.out.println( (s1==s2) && s1.equals(s2));
System.out.println( ! (s1.equals(s2)));
}
答案:
false
false
true
true
true
false
false
答案 0 :(得分:5)
我认为重点是==
比较两个对象引用以查看它们是否引用同一个实例,而equals
比较这些值。
例如,s1和s2是两个不同的字符串实例,因此==
返回false,但它们都包含值"Today"
,因此equals
返回true。
答案 1 :(得分:5)
Integer i1=new Integer(1);
Integer i2=new Integer(1);
String s1=new String("Today");
String s2=new String("Today");
// do i1 and 12 point at the same location in memory? No - they used "new"
System.out.println(i1==i2);
// do s1 and s2 point at the same location in memory? No - the used "new"
System.out.println(s1==s2);
// do s1 and s2 contain the same sequence of characters ("Today")? Yes.
System.out.println(s1.equals(s2));
// do s1 and s2 point at different locations in memory? Yes - they used "new"
System.out.println(s1!=s2);
// do s1 and s2 point to different locations in memory? Yes - they used "new".
// Do not check s1.equals(s2) because the first part of the || was true.
System.out.println( (s1!=s2) || s1.equals(s2));
// do s1 and s2 point at the same location in memory? No - they used "new".
// do not check s1.equals(s2) because the first part of the && was false.
System.out.println( (s1==s2) && s1.equals(s2));
// do s1 and s2 not contain the same sequence of characters ("Today")? No.
System.out.println( ! (s1.equals(s2)));
答案 2 :(得分:1)
请记住Integer
和String
是对象,==
运算符会比较这两个指针的内存地址,而不是实际的对象本身。因此,前{2} ==
将为假,因为i1
与i2
不是同一个对象。如果初始化是:
Integer i1=new Integer(1);
Integer i2=i1;
然后第一个println()就是真的。
s1.equals(s2)
是比较对象中的相等性的正确方法。 String.equals()方法将检查字符串是否相等,因此“Today”和“Today”是相等的字符串。
s1!=s2
为真,因为s1和s2是不同的对象,类似于==
其余的应该是非常简单的布尔操作。
答案 3 :(得分:0)
特别是哪些结果? this有帮助吗?