Dive into String


String is a final class inheritanced from java.lang.Object, it represents for character strings. All string literals we see in Java, such as “abc”, are implemented as instances of this class.

According to the document, Strings are defined as constants, that means the value of them cannot be changed after they are created. If you want mutable strings, you should use StringBuilder(or StringBuffer).

As a result that String objects are immutable, the can be shared. For instance:

String str = "abc";

is equivalent to:

char data[] = {'a', 'b', 'c'};
String str = new String(data);

String class has two attributes, value and hash.

 /** The value is used for character storage. */
  private final char value[];

  /** Cache the hash code for the string */
   private int hash; // Default to 0

and one of the constructors of String class is:

public String(char value[]) {
    this.value = Arrays.copyOf(value, value.length);

Since String is a final class, it cannot be inheritanced.


String class has 15 constructors, in which there are two deprecated. Usually we use:

String str1 = "abc";


String str2 = new String("abc");

to create String object. The two ways above are different. In the former one, characters are stored in stack memory when the latter are stored in heap memory. Enventually, if we use str1==str2 to compare them, it would return false cause java compares objects use their address for operation: ==.

In the former one, Java would create a reference object of String class, and then searching the stack whether there is an address storing value “abc”, if yes, let the reference str1 point to it, otherwise, create a new String type object with value “abc”, and let str1 point to it.

So, if we create another String object with value “abc”:

String str3 = "abc";

it’s obvious that str1==str3 would return true cause they are point to the same address. When, to

String str4 = new String("abc");

str2==str4 would return false.


In this section, I will start with below example.

public static void main(String[] args) {
	String a = "123";
	String b = "123";

	String c = new String("123");
	String d = new String("123");

	System.out.println(a == b);
	System.out.println(c == d);
	System.out.println(a == c);


The output is:


Because java compares two objects by their address when use ==. So if you want to check two String objects whether they have the same content, you should use equals method.

Let’s have a look at equals method’s code:

    public boolean equals(Object anObject) {
        if (this == anObject) {
            return true;
        if (anObject instanceof String) {
            String anotherString = (String) anObject;
            int n = value.length;
            if (n == anotherString.value.length) {
                char v1[] = value;
                char v2[] = anotherString.value;
                int i = 0;
                while (n-- != 0) {
                    if (v1[i] != v2[i])
                            return false;
                return true;
        return false;

Here we can see that String class compute the equal by check the characters one by one.


Sometimes you may want to use:

	str1 = str1 + "def";

to concatenate two strings. In this process, str1 is not pointing to the address of abc anymore, according to the document, concatenation operator “+” of String is implemented through StringBuilder(or StringBuffer) class and its append method.

First, java will create a new object StringBuilder in heap memory, then copy the value of str1 and value “def” to the StringBuilder object, finally let the str1 point to the object.

So that’s why:

	String a = "abc";
	String b = "abcdef";

	a = a + "def";

	System.out.println(a == b);

would return false.

Another method to concatenate two String objects is use concat method. Here is the code:

    public String concat(String str) {
        int otherLen = str.length();
        if (otherLen == 0) {
            return this;
        int len = value.length;
        char buf[] = Arrays.copyOf(value, len + otherLen);
        str.getChars(buf, len);
        return new String(buf, true);

When the length of the argument str is 0 (means it’s an empty String), the original String will return, otherwise, a new String object would be created. As a result, it would take more time and cost more memory to concatenate the strings. (Remind yourself that String is immutable object.)

StringBuilder(or StringBuffer) is recommended to use to concatenate strings, cause they can change the length through certain method calls. StringBuilder is added since version 1.5, regarded as a replacement of StringBuffer. The difference between them is that StringBuilder is not thread-safe but StringBuffer is. So it’s recommended that StringBuilder be used in preference to StringBuffer when possible as it will be faster. Otherwise, if in multi-thread environment, you should use StringBuffer.

Here is an example to test the performance of these three classes.

	public static void main(String[] args) {
		String str1 = "string";
		String str2 = "1";
		StringBuilder sbd = new StringBuilder("string");
		StringBuffer sbf = new StringBuffer("string");
		long start = System.currentTimeMillis();
		for (int i = 0; i < 50000; i++) {
			str1 = str1 + str2;
		long end = System.currentTimeMillis();
		System.out.println("Time of operator + :" + (end - start) + " ms");
		start = System.currentTimeMillis();
		for (int i = 0; i < 50000; i++) {
			str1 = str1.concat(str2);
		end = System.currentTimeMillis();
		System.out.println("Time of concat method :" + (end - start) + " ms");
		start = System.currentTimeMillis();
		for (int i = 0; i < 50000; i++) {
		end = System.currentTimeMillis();
		System.out.println("Time of StringBuilder :" + (end - start) + " ms");
		start = System.currentTimeMillis();
		for (int i = 0; i < 50000; i++) {
		end = System.currentTimeMillis();
		System.out.println("Time of StringBuffer :" + (end - start) + " ms");


Time of operator + :3931 ms
Time of concat method :6211 ms
Time of StringBuilder :2 ms
Time of StringBuffer :4 ms

I’ve tested some cases, I found that operator + is most like :

	StringBuilder sb = null ;
	for (int i = 0; i < 50000; i++) {
		sb = new StringBuilder(str1);
	str1 = sb.toString();

concat method is slower than operator + is because that operator + concatenate strings in stack when concat in heap.

( If you set str2 to an empty String(String str2 = ""), the output is:

Time of operator + :27 ms
Time of concat method :1 ms
Time of StringBuilder :1 ms
Time of StringBuffer :3 ms


I’m sure you already know why.

See Also

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