How new String() and normal String created ? String class in java (Confusion) [duplicate]

In java exist the heap and the stack, Heap is where all Objects are saved stack is where vars are saved Now also exist another kind of list for Strings and Integers (numbers)

As you know a String can be created in some ways like

like new String("word") or just = "word" when you use the first way you create a new object (heap) when you use the other you save the word in a stack of words (Java engenniers thought it would be good if you don't create manny objects or words are repeated so they created an special stack for words, same for Integers from 0 to 127) So as I said You have to know that there is an stack and a Heap look at this example

String wordOne ="hola";
        String wordTwo = "hola";
        String wordTres = "hola";
        System.out.println(wordOne == wordTwo);
        System.out.println(wordTres == wordTwo);
        System.out.println(wordOne == wordTres);

        String wordFour = new String("hola");
        System.out.println(wordOne == wordFour);

        Integer uno = 127;
        Integer dos = 127;
        System.out.println(uno == uno);
        Integer tres = 128;
        Integer cuatro = 128;
        System.out.println(tres == cuatro);

String x = "word"; is saved in an special Stack String y = new String("it is not");

But tbh I don't remeber so well the rules for tha stack, but in any case i recomend you to compare all words using wordX.equals(wordY)

An also numbers in objects could be compared using == from 0 to 127 but the same if you use objects use equals, although using numbers there is a better do to do it in spite of use equals, convert one number to a primitive value (the memory will be better)

When you are making string with new keyword,JVM will create a new string object in normal(non pool) heap memory and the literal will be placed in the string constant pool. In your case, The variable s1 will refer to the object in heap(non pool).

String s1 = new String(new String("2")+new String("2"));

But in the next line your are calling intern() method.

When the intern method is invoked, if the pool already contains a string equal to this String object as determined by the equals(Object) method, then the string from the pool is returned. Otherwise, this String object is added to the pool and a reference to this String object is returned.

Check Javadocs. As "22" is not in string pool, a new string literal "22" will be created and a reference of it will be returned. When you are writing:

String s2="22";

it simply refers "22" in string pool. But calling s3.intern() will not create a new string literal as "22" exists in the pool. Check the Javadocs for intern() again. It says if exists in pool, then string from the pool is returned not reference. So, this time s3 references to a different object. But s4 is referred to same object as s1,s2. You can print the objects hashcode for checking if the are same or not. Like:

System.out.println(System.identityHashCode(s1));

Notice that the type String is capitalized and is not one of Java's 8 primitive types (int, boolean, double, char, etc.). This indicates that any instance of a String is an object that was built using the 'blueprint' of the class String. Because variables in Java that refer to objects only store the memory address where the actual object is stored, when you compare Strings with == it compares memory location.

String str1 = new String("hello");
String str2 = str1; //sets str1 and str2 pointing to same memory loc
if (str1 == str2){
    //do stuff; the code will enter this if-statement in this case
}

The way to compare the values within objects in Java is with equals(), such as:

String str1 = new String("hello");
String str2 = new String("hello"); //str2 not same memory loc as str1
if (str1.equals(str2)){
    //do stuff; the code will enter this if-statement in this case
}

This is a common error for beginners, since the primitive types are not objects and you CAN compare two ints for equality like:

int one = 1; //primitive types are NOT objects
int two = 2; //notice when I make an int, I don't have to say "new"
             //which means a new **object**
if (int1 == int2) {
    //do stuff; in this case the program will not enter this if-statement
}

It seems that you understand everything but the meaning of the very last line. See my comment on the last line.

String s1 = new String(new String("2")+new String("2")); //declare AND initialize s1 as a new String object
s1.intern();
String s2="22"; //declare a new variable s2 and point it to the same object that s1 is pointing to
System.out.print(s1==s2);
String s3 =new String (new String("2")+new String("2"));
s3.intern();
String s4="22";
System.out.print(s3==s4); //check if s3 and s4 are stored in the same memory location = FALSE

In java object1 == object2 means

that do object1 and object2 have the same address in memory?

object1.equals(object2)

means are they equal, for example do they have the same values of all fields?

So, For two Strings S1 and S2, string1.equals(S2) means, do they have the same characters in the same sequence?
S1 == S1 means are string1 and string2 stored at the same address in memory?