The difference between == and equals()

1, equals()

What are the execution results of the following codes?

public class Main {public static void main(String[] args) {String s1 = "hello world";String s2 = "hello world";String s3 = new String("hello world");String s4 = "hello"+"world";String s5 = "hello"+new String("world");String s6 = "hello";String s7 = s6+"world";System.out.println(s1.equals(s2));System.out.println(s1.equals(s3));System.out.println(s1.equals(s4));System.out.println(s1.equals(s5));System.out.println(s1.equals(s7));}}

The above output results are true for the following reasons:

equals() can only be used between reference types. It is defined in the Object class. The default is to compare addresses. It is often overridden to compare whether specific property values ​​are the same.The String class overrides equals() to compare the contents of strings for the same.

2, reference type

What is the result of executing the following code?

public class Main {public static void main(String[] args) {String s1 = "hello world";String s2 = "hello world";String s3 = new String("hello world");String s4 = "hello"+"world";String s5 = "hello"+new String("world");String s6 = "hello";String s7 = s6+"world";System.out.println(s1==s2);System.out.println(s1==s3);System.out.println(s1==s4);System.out.println(s1==s5);System.out.println(s1==s7);}}

Running the above code, we can see that the results of s1==s2 and s1==s4 are true, and the other results are false.

The reasons are as follows:

Since the String type is a reference data type rather than a basic data type, == here compares whether the referenced addresses are the same.

Here, s1 creates a new literal object "hello world", which is stored in the heap and cached in the constant pool at the same time, so s2 directly reuses the objects in the constant pool.Therefore, using the same address of the two, the result of s1==s2 is true.

s3 will create another string object when new String() is performed, and refer to the content of the string, so the addresses of s3 and s1 are not the same, so the result of s1==s3 returns false.Similarly, the result of s1==s5 also returns false.

Since s4 uses two literals for splicing, it will directly reuse the objects in the constant pool.Therefore, the address of s4 is the same as that of s1, so the result of s1==s4 returns true.

Because s6 is a variable, it will not be directly connected at compile time, but a new object will be created to store hello world.Therefore, the addresses of s7 and s1 are not the same, and the result returns false.

3, wrapper class

What is the result of executing the following code?

public class Main {public static void main(String[] args) {Integer i1 = 127;Integer i2 = 127;Integer i3 = 128;Integer i4 = 128;Integer i5 = new Integer(127);int i6 = 127;System.out.println(i1==i2);System.out.println(i3==i4);System.out.println(i1==i5);System.out.println(i1==i6);}}

Executing the above code, we can see that the result of i1==i2 is true, the result of i3==i4 is false, the result of i1==i5 is false, and the result of i1==i6 is true.

The reason is as follows: Integer.valueOf() will reuse the data in the range of -128 to 127, so the range between -128 and 127 will return true, that is, the result of i1==i2 is true, i3==i4The result is false.Since i5 creates a new object during new Integer(), the addresses of i1 and i5 are not the same, so the result returns false.When i6 and i1 are == compared, the basic package type will be automatically unboxed to the basic type and then compared, so Integer() will be automatically unboxed to the int type and then compared, so it returns true.