java String构造函数逻辑

Question

我试图了解 java String 是如何实现的。下面的 jdk7 source 代码显示了对 originalValue.length > size 的检查.我无法弄清楚它是如何/何时实现的。我试图在一些 java String 创建语句上使用 eclipse 调试器，但这个检查从来没有为真。是否有可能设计一个 String 参数来使这个检查为真？

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;
    }
...
}

Answer 1

看看这段代码:

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

第二个构造函数将匹配条件。诀窍在于 substring 方法。它不会生成真正的子字符串，而是复制底层数组并为其设置新的边界。构造新字符串的思想是复制一个字符串，而不是仅仅分配相同的数组。这就是为什么从大字符串中提取小子字符串可能会导致 OOM 异常的原因。因为用大数组来表示一小段信息。