Question

我试图了解java String是如何实施的。下面的jdk7 source代码显示了对originalValue.length > size的检查。我无法弄清楚它将如何/何时实现。我试过了在一些java String创建语句中使用eclipse调试器，但是这个检查从来都不是真的。是否可以设计一个String参数来使这个检查成立？

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

看一下这段代码：

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

第二个构造函数将匹配条件。诀窍在于子串方法。它不是一个真正的子串，而是复制底层数组，并为它设置新的边界。构造新字符串的想法是创建一个字符串的副本，而不仅仅是分配相同的数组。这就是为什么从大字符串中获取小子串可能导致OOM异常的原因。因为代表一小部分信息使用了大数组。

Answer 2

你可以调试它。 Value代表基础char[]。 count代表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

java String构造函数逻辑

2 个答案: