深入string的substring方法_JAVA_编程开发_程序员俱乐部

1.应用举例

??????? substring方法，例如s="abcdef"? s.substring(2,5)结果为cde，长度为endindex-beginindex

2.实现原理

class="java" name="code">/**
* Returns a new string that is a substring of this string. The
* substring begins at the specified <code>beginIndex</code> and
* extends to the character at index <code>endIndex - 1</code>.
* Thus the length of the substring is <code>endIndex-beginIndex</code>.
* <p>
* Examples:
* <blockquote><pre>
* "hamburger".substring(4, 8) returns "urge"
* "smiles".substring(1, 5) returns "mile"
* </pre></blockquote>
*
* @param      beginIndex   the beginning index, inclusive.
* @param      endIndex     the ending index, exclusive.
* @return     the specified substring.
* @exception  IndexOutOfBoundsException  if the
*             <code>beginIndex</code> is negative, or
*             <code>endIndex</code> is larger than the length of
*             this <code>String</code> object, or
*             <code>beginIndex</code> is larger than
*             <code>endIndex</code>.
*/
public String substring(int beginIndex, int endIndex) {
if (beginIndex < 0) {
    throw new StringIndexOutOfBoundsException(beginIndex);
}
if (endIndex > count) {
    throw new StringIndexOutOfBoundsException(endIndex);
}
if (beginIndex > endIndex) {
    throw new StringIndexOutOfBoundsException(endIndex - beginIndex);
}
return ((beginIndex == 0) && (endIndex == count)) ? this :
    new String(offset + beginIndex, endIndex - beginIndex, value);
}

??????? 通过源代码可以发现

??????? a).string内部通过char[]数组实现

??????? b).substring方法返回了新的string对象，但是string对象却指向原来得char数组，如果原来的string很大，即使原来的string释放，内存空间也无法回收

????????c).offset值默认为0，第一个参数为偏移，第二个为长度，第三个为char数组

// Package private constructor which shares value array for speed.
String(int offset, int count, char value[]) {
    this.value = value;
    this.offset = offset;
    this.count = count;
}

??????? d).默认string构造方法如下：

/**
  * Initializes a newly created {@code String} object so that it represents
  * an empty character sequence.  Note that use of this constructor is
  * unnecessary since Strings are immutable.
  */
public String() {
    this.offset = 0;
    this.count = 0;
    this.value = new char[0];
}

??????? e).内存泄露测试

package com.bijian.study.string;

import java.io.UnsupportedEncodingException;
import java.util.ArrayList;
import java.util.List;

public class StringTest {

	public static void main(String[] args) throws UnsupportedEncodingException {
		
		String aa=new String(new char[100000]);  
        List<String> list=new ArrayList<String>();  
        for (int i = 0; i < 10000000; i++) {  
            //虽然截取的字符串占据空间很小，但是由于aa巨大的数组空间被共享，没有释放，所以内存溢出  
        	//list.add(aa.substring(2, 3));  
            list.add(new String(aa.substring(2,3)));//加一层构造方法后，构造方法内部进行了数组的拷贝，原有的巨大数组空间被回收，不会内存溢出  
        }  
	}
}

??????? 为了避免内存拷贝、加快速度，Sun JDK直接复用了原String对象的char[]，偏移量和长度来标识不同的字符串内容。也就是说，substring出的来String小对象仍然会指向原String大对象的char[]，split也是同样的情况。这就解释了，为什么HashMap中String对象的char[]都那么大。
原因解释：
??????? a.程序从每个请求中得到一个String大对象，该对象内部char[]的长度达数百K。
??????? b.程序对String大对象做split，将split得到的String小对象放到HashMap中，用作缓存。
??????? c.Sun JDK6对String.split方法做了优化，split出来的Stirng对象直接使用原String对象的char[]
??????? d.HashMap中的每个String对象其实都指向了一个巨大的char[]
??????? e.HashMap的上限是万级的，因此被缓存的Sting对象的总大小=万*百K=G级
??????? f.G级的内存被缓存占用了，大量的内存被浪费，造成内存泄露的迹象
解决方案：
??????? split是要用的，但是我们不要把split出来的String对象直接放到HashMap中，而是调用一下String的拷贝构造函数String(String original)，这个构造函数是安全的，源代码如下：

/**
* 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;
}?