java String constructor logic

Question

I was trying to understand how a java String was implemented.The jdk7 source code below shows a check for originalValue.length > size .I cant figure out how/when it would come true.I tried to use eclipse debugger on some java String creation statements,but this check was never true.Is it possible to devise a String argument which would make this check true?

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

    /** The offset is the first index of the storage that is used. */
    private final int offset;

    /** The count is the number of characters in the String. */
    private final int count;

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

     /**
     * Initializes a newly created {@code String} object so that it represents
     * the same sequence of characters as the argument; in other words, the
     * newly created string is a copy of the argument string. Unless an
     * explicit copy of {@code original} is needed, use of this constructor is
     * unnecessary since Strings are immutable.
     *
     * @param  original
     *         A {@code String}
     */
    public String(String original) {
        int size = original.count;
        char[] originalValue = original.value;
        char[] v;
        if (originalValue.length > size) {
            // The array representing the String is bigger than the new
            // String itself.  Perhaps this constructor is being called
            // in order to trim the baggage, so make a copy of the array.
            int off = original.offset;
            v = Arrays.copyOfRange(originalValue, off, off+size);
        } else {
            // The array representing the String is the same
            // size as the String, so no point in making a copy.
            v = originalValue;
        }
        this.offset = 0;
        this.count = size;
        this.value = v;
    }
...
}

Mikhail · Accepted Answer · 2013-08-17 06:53:48Z

3

Have a look at this piece of code:

String s1 = "123456789";
String s2 = new String(s1.substring(0, 2));

The second constructor will match condition. The trick is in substring method. It does not make a real substring, but rather copies underlying array and just sets new boundaries to it. The idea of constructing a new string is to make a copy of a string, not just assign the same array. That`s actually why taking small substring from a big string might lead to OOM exception. Because to represent a small piece of information big array is used.

edited Aug 17, 2013 at 6:53

answered Aug 17, 2013 at 6:04

Mikhail

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3 Comments

Karthikeyan Over a year ago

is s2.length > 3 here ?

oddparity Over a year ago

Yes, as it is the same char array underneath (not just an equal one).

Mikhail Over a year ago

In this case s2`s underling array will be of size two, not nine.

rocketboy · Accepted Answer · 2013-08-17 06:04:55Z

3

You can debug this. Value represents the underlying char[]. count represents the view

 String s = new String("Hello   "); //value= [H, e, l, l, o, , , ]  count=8

 String os = s.trim();  //value= [H, e, l, l, o, , , ]  count=5

answered Aug 17, 2013 at 6:04

rocketboy

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