In Java, String str = new String ("ABC") creates several objects __java

The answer is two, and now let's be specific:
string s = new String ("abc");
First we need to understand two concepts, referencing variables and objects, and objects are generally created in the heap by new, and S is just a reference variable.

All strings are string objects, because of the large use of string literals, in Java in order to save time, in the compilation phase, the string will be placed in a constant pool, the advantage of a constant pool is that you can combine the same string, occupy a space, we can use = = Determine if the two reference variable points to an address that is an object.

The ' = = ' in 1.Java:

1 if it is 8 basic types, = = Returns true;

String S1 = "abc";
 String s2 = "abc";
System.out.println (S1==S2);/return true</span>

2 If it is a reference type, = = Compares two variables to point to the same memory.

string S1 = new String ("abc");
String s2 = new String ("abc");
System.out.println (S1==S2);/return false</span>

2.String s = new string ("abc") is actually "ABC" itself is an object in a constant pool, and when you run the new string (), copy the constant pool, the string "abc" in pools, to the heap and give the application of this object to S. So we created two string objects, one in the pool and one in the heap. Look at the program below and create a few objects.

public class Demo03 {public
    static void Main (string[] args) {
        //TODO auto-generated method stub
        String S1 = New String ("abc");
        String s2 = new String ("abc");
        if (S1 = = s2) {//will not execute the statement}
       System.out.println ("created an object in the heap");

    }
        else{
        System.out.println ("two objects created in the heap");}}

Two objects were created in the heap, but there was one object in the text pool, so three objects were created