概要
- 类继承关系
java.lang.Object
java.lang.String
- 定义
public final class String
extends Object
implements Serializable, Comparable<String>, CharSequence
- 要点
一旦创建就不可改变
实现
- storage
/** The value is used for character storage. */
private final char value[];
可以看出 String 中的数据是如何存储的。
- 初始化
public String(String original) {
this.value = original.value;
this.hash = original.hash;
}
可以看出使用 String 类型初始化，新 String 实际上与原来的 String 指向同一块内存。
public String(char value[]) {
this.value = Arrays.copyOf(value, value.length);
}
如果用 char[] 初始化，可以看出，新分配了内存，并复制，保证了两者相互独立，只是内容相同。
public String(StringBuffer buffer) {
synchronized(buffer) {
this.value = Arrays.copyOf(buffer.getValue(), buffer.length());
}
}
注意用 StringBuffer 初始化时，对同一 buffer 是线程安全的，即初始化 String 的过程中，其它线程不会改变 buffer 的内容。
另外，能告诉我下面这段代码是怎么回事么？
public String(StringBuilder builder) {
this.value = Arrays.copyOf(builder.getValue(), builder.length());
}
为啥这次不同步了呢？
- equals
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;
i++;
}
return true;
}
}
return false;
}
注意：
1) 检查类型 2) value 直接通过点访问了，value 是 private 的啊，怎么能这样？
- hashCode
public int hashCode() {
int h = hash;
if (h == 0 && value.length > 0) {
char val[] = value;
for (int i = 0; i < value.length; i++) {
h = 31 * h + val[i];
}
hash = h;
}
return h;
}
String 的 hashCode 公式：
s[0]*31^(n-1) + s[1]*31^(n-2) + ... + s[n-1]
- replace
public String replace(char oldChar, char newChar) {
if (oldChar != newChar) {
int len = value.length;
int i = -1;
char[] val = value; /* avoid getfield opcode */
while (++i < len) {
if (val[i] == oldChar) {
break;
}
}
if (i < len) {
char buf[] = new char[len];
for (int j = 0; j < i; j++) {
buf[j] = val[j];
}
while (i < len) {
char c = val[i];
buf[i] = (c == oldChar) ? newChar : c;
i++;
}
return new String(buf, true);
}
}
return this;
}
从中可以看出，虽然说是 replace，但是实际上还是新生成了 buf ，然后再生成新的 String，而不是在原来的 value 上修改。如果有大量的替换，还是自己实现比较好诶～
- indexOf
/**
* Code shared by String and StringBuffer to do searches. The
* source is the character array being searched, and the target
* is the string being searched for.
*
* @param source the characters being searched.
* @param sourceOffset offset of the source string.
* @param sourceCount count of the source string.
* @param target the characters being searched for.
* @param targetOffset offset of the target string.
* @param targetCount count of the target string.
* @param fromIndex the index to begin searching from.
*/
static int indexOf(char[] source, int sourceOffset, int sourceCount,
char[] target, int targetOffset, int targetCount,
int fromIndex) {
if (fromIndex >= sourceCount) {
return (targetCount == 0 ? sourceCount : -1);
}
if (fromIndex < 0) {
fromIndex = 0;
}
if (targetCount == 0) {
return fromIndex;
}
char first = target[targetOffset];
int max = sourceOffset + (sourceCount - targetCount);
for (int i = sourceOffset + fromIndex; i <= max; i++) {
/* Look for first character. */
if (source[i] != first) {
while (++i <= max && source[i] != first);
}
/* Found first character, now look at the rest of v2 */
if (i <= max) {
int j = i + 1;
int end = j + targetCount - 1;
for (int k = targetOffset + 1; j < end && source[j]
== target[k]; j++, k++);
if (j == end) {
/* Found whole string. */
return i - sourceOffset;
}
}
}
return -1;
}
这段代码从 source 中寻找 target 第一次出现的位置，for 循环每次都先让 i 停留在一个位置，此位置上内容与 target 首字符相同，然后开始遍历。可以看出这是一个 O(n^2) 的算法，所以，标准库也不一定是最高效的，要是要高效，还是需要自己实现，或者找其它库的。
- matches
public boolean matches(String regex) {
return Pattern.matches(regex, this);
}
正则表达式匹配函数。可以看出，是直接调用了 Pattern 中的相应函数。
public String[] split(String regex, int limit) {
/* fastpath if the regex is a
(1)one-char String and this character is not one of the
RegEx's meta characters ".$|()[{^?*+\\", or
(2)two-char String and the first char is the backslash and
the second is not the ascii digit or ascii letter.
*/
char ch = 0;
if (((regex.value.length == 1 &&
".$|()[{^?*+\\".indexOf(ch = regex.charAt(0)) == -1) ||
(regex.length() == 2 &&
regex.charAt(0) == '\\' &&
(((ch = regex.charAt(1))-'0')|('9'-ch)) < 0 &&
((ch-'a')|('z'-ch)) < 0 &&
((ch-'A')|('Z'-ch)) < 0)) &&
(ch < Character.MIN_HIGH_SURROGATE ||
ch > Character.MAX_LOW_SURROGATE))
{
int off = 0;
int next = 0;
boolean limited = limit > 0;
ArrayList<String> list = new ArrayList<>();
while ((next = indexOf(ch, off)) != -1) {
if (!limited || list.size() < limit - 1) {
list.add(substring(off, next));
off = next + 1;
} else { // last one
//assert (list.size() == limit - 1);
list.add(substring(off, value.length));
off = value.length;
break;
}
}
// If no match was found, return this
if (off == 0)
return new String[]{this};
// Add remaining segment
if (!limited || list.size() < limit)
list.add(substring(off, value.length));
// Construct result
int resultSize = list.size();
if (limit == 0)
while (resultSize > 0 && list.get(resultSize - 1).length() == 0)
resultSize--;
String[] result = new String[resultSize];
return list.subList(0, resultSize).toArray(result);
}
return Pattern.compile(regex).split(this, limit);
}
按 regex 将字符串分割，思路是如果是单个字符，或者转义字符，就手工分割，否则就直接调用 Pattern.comile(regex).split函数。手工分割时每次都将 [off, next] 之间的内容加入 list，最后将 剩余的 [off, ] 加入。另外注意 limit 对分割次数的限制。