如果两个字符串具有相同的引用,为什么输出不被排除(true)？

Question

昨天这个问题由Interviewer询问该线路的输出是什么.

public class StringTest {

  public static void main(String[] args) {
      String s1 = "abc";
      String s2 = "abc";
      String s3 = s1+s2;
      String s4 = s1+s2;

      //check s1 == s2
      System.out.println(s1==s2);

      //check s3 == s4
      System.out.println(s3==s4);
  }
}

当我看着这个问题,然后思考面试官提出的简单问题.我告诉他,输出s1==s2和s3==s4返回true和true我是非常有信心.突然间,他说不是错误的输出然后我认为他可能是在开玩笑或试图失去信心,但直到最后他说错了.当我检查输出是true和false.我也在想如何通过给出适当的答案来解决我的困惑.提前致谢.

Answer 1

s1+s2在编译时不知道,因此它在运行时计算并创建一个新对象,每次运行时都不同.如果将s1和s2更改为final,编译器将内联常量,您将得到真的s3==s4.

  final String s1 = "abc";
  final String s2 = "abc";
  String s3 = s1+s2; // compiler replaces with "abcabc"
  String s4 = s1+s2; // compiler replaces with "abcabc"

  //check s1 == s2
  System.out.println(s1==s2);

  //check s3 == s4
  System.out.println(s3==s4); // is now true.

要么

  String s1 = "abc";
  String s2 = "abc";
  String s3 = (s1+s2).intern(); // use the string literal pool
  String s4 = (s1+s2).intern();

  //check s1 == s2
  System.out.println(s1==s2);

  //check s3 == s4
  System.out.println(s3==s4); // is now true.

但

  String s1 = "abc";
  String s2 = "abc";
  // String s3 = s1+s2;
  String s3 = new StringBuilder().append(s1).append(s2).toString();
  // String s4 = s1+s2;
  String s4 = new StringBuilder().append(s1).append(s2).toString();

  //check s3 == s4
  System.out.println(s3==s4); // different objects.