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  • Java 字符串拼接 五种方法的性能比较分析 从执行100次到90万次

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    > 字符串拼接一般使用“+”,但是“+”不能满足大批量数据的处理,Java中有以下五种方法处理字符串拼接,各有优缺点,程序开发应选择合适的方法实现。

    1. 加号 “+”

    2. String contact() 方法

    3. StringUtils.join() 方法

    4. StringBuffer append() 方法

    5. StringBuilder append() 方法

    > 经过简单的程序测试,从执行100次到90万次的时间开销如下表:

     

     由此可以看出:

    1. 方法1 加号 “+” 拼接 和 方法2 String contact() 方法 适用于小数据量的操作,代码简洁方便,加号“+” 更符合我们的编码和阅读习惯;

    2. 方法3 StringUtils.join() 方法 适用于将ArrayList转换成字符串,就算90万条数据也只需68ms,可以省掉循环读取ArrayList的代码;

    3. 方法4 StringBuffer append() 方法 和 方法5 StringBuilder append() 方法 其实他们的本质是一样的,都是继承自AbstractStringBuilder,效率最高,大批量的数据处理最好选择这两种方法。

    4. 方法1 加号 “+” 拼接 和 方法2 String contact() 方法 的时间和空间成本都很高(分析在本文末尾),不能用来做批量数据的处理。

    > 源代码,供参考

    package cnblogs.twzheng.lab2;
    
    /**
     * @author Tan Wenzheng
     *
     */
    import java.util.ArrayList;
    import java.util.List;
    
    import org.apache.commons.lang3.StringUtils;
    
    public class TestString {
    
        private static final int max = 100;
    
        public void testPlus() {
            System.out.println(">>> testPlus() <<<");
    
            String str = "";
    
            long start = System.currentTimeMillis();
    
            for (int i = 0; i < max; i++) {
                str = str + "a";
            }
    
            long end = System.currentTimeMillis();
    
            long cost = end - start;
    
            System.out.println("   {str + "a"} cost=" + cost + " ms");
        }
    
        public void testConcat() {
            System.out.println(">>> testConcat() <<<");
    
            String str = "";
    
            long start = System.currentTimeMillis();
    
            for (int i = 0; i < max; i++) {
                str = str.concat("a");
            }
    
            long end = System.currentTimeMillis();
    
            long cost = end - start;
    
            System.out.println("   {str.concat("a")} cost=" + cost + " ms");
        }
    
        public void testJoin() {
            System.out.println(">>> testJoin() <<<");
    
            long start = System.currentTimeMillis();
    
            List<String> list = new ArrayList<String>();
    
            for (int i = 0; i < max; i++) {
                list.add("a");
            }
    
            long end1 = System.currentTimeMillis();
            long cost1 = end1 - start;
    
            StringUtils.join(list, "");
    
            long end = System.currentTimeMillis();
            long cost = end - end1;
    
            System.out.println("   {list.add("a")} cost1=" + cost1 + " ms");
            System.out.println("   {StringUtils.join(list, "")} cost=" + cost
                    + " ms");
        }
    
        public void testStringBuffer() {
            System.out.println(">>> testStringBuffer() <<<");
    
            long start = System.currentTimeMillis();
    
            StringBuffer strBuffer = new StringBuffer();
    
            for (int i = 0; i < max; i++) {
                strBuffer.append("a");
            }
            strBuffer.toString();
    
            long end = System.currentTimeMillis();
    
            long cost = end - start;
    
            System.out.println("   {strBuffer.append("a")} cost=" + cost + " ms");
        }
    
        public void testStringBuilder() {
            System.out.println(">>> testStringBuilder() <<<");
    
            long start = System.currentTimeMillis();
    
            StringBuilder strBuilder = new StringBuilder();
    
            for (int i = 0; i < max; i++) {
                strBuilder.append("a");
            }
            strBuilder.toString();
    
            long end = System.currentTimeMillis();
    
            long cost = end - start;
    
            System.out
                    .println("   {strBuilder.append("a")} cost=" + cost + " ms");
        }
    }

    > 测试结果:

    1. 执行100次, private static final int max = 100;

    >>> testPlus() <<<
       {str + "a"} cost=0 ms
    >>> testConcat() <<<
       {str.concat("a")} cost=0 ms
    >>> testJoin() <<<
       {list.add("a")} cost1=0 ms
       {StringUtils.join(list, "")} cost=20 ms
    >>> testStringBuffer() <<<
       {strBuffer.append("a")} cost=0 ms
    >>> testStringBuilder() <<<
       {strBuilder.append("a")} cost=0 ms

    2. 执行1000次, private static final int max = 1000;

    >>> testPlus() <<<
       {str + "a"} cost=10 ms
    >>> testConcat() <<<
       {str.concat("a")} cost=0 ms
    >>> testJoin() <<<
       {list.add("a")} cost1=0 ms
       {StringUtils.join(list, "")} cost=20 ms
    >>> testStringBuffer() <<<
       {strBuffer.append("a")} cost=0 ms
    >>> testStringBuilder() <<<
       {strBuilder.append("a")} cost=0 ms

    3. 执行1万次, private static final int max = 10000;

    >>> testPlus() <<<
       {str + "a"} cost=150 ms
    >>> testConcat() <<<
       {str.concat("a")} cost=70 ms
    >>> testJoin() <<<
       {list.add("a")} cost1=0 ms
       {StringUtils.join(list, "")} cost=30 ms
    >>> testStringBuffer() <<<
       {strBuffer.append("a")} cost=0 ms
    >>> testStringBuilder() <<<
       {strBuilder.append("a")} cost=0 ms

    4. 执行10万次, private static final int max = 100000;

    >>> testPlus() <<<
       {str + "a"} cost=4198 ms
    >>> testConcat() <<<
       {str.concat("a")} cost=1862 ms
    >>> testJoin() <<<
       {list.add("a")} cost1=21 ms
       {StringUtils.join(list, "")} cost=49 ms
    >>> testStringBuffer() <<<
       {strBuffer.append("a")} cost=10 ms
    >>> testStringBuilder() <<<
       {strBuilder.append("a")} cost=10 ms

    5. 执行20万次, private static final int max = 200000;

    >>> testPlus() <<<
       {str + "a"} cost=17196 ms
    >>> testConcat() <<<
       {str.concat("a")} cost=7653 ms
    >>> testJoin() <<<
       {list.add("a")} cost1=20 ms
       {StringUtils.join(list, "")} cost=51 ms
    >>> testStringBuffer() <<<
       {strBuffer.append("a")} cost=20 ms
    >>> testStringBuilder() <<<
       {strBuilder.append("a")} cost=16 ms

    6. 执行50万次, private static final int max = 500000;

    >>> testPlus() <<<
       {str + "a"} cost=124693 ms
    >>> testConcat() <<<
       {str.concat("a")} cost=49439 ms
    >>> testJoin() <<<
       {list.add("a")} cost1=21 ms
       {StringUtils.join(list, "")} cost=50 ms
    >>> testStringBuffer() <<<
       {strBuffer.append("a")} cost=20 ms
    >>> testStringBuilder() <<<
       {strBuilder.append("a")} cost=10 ms

    7. 执行90万次, private static final int max = 900000;

    >>> testPlus() <<<
       {str + "a"} cost=456739 ms
    >>> testConcat() <<<
       {str.concat("a")} cost=186252 ms
    >>> testJoin() <<<
       {list.add("a")} cost1=20 ms
       {StringUtils.join(list, "")} cost=68 ms
    >>> testStringBuffer() <<<
       {strBuffer.append("a")} cost=30 ms
    >>> testStringBuilder() <<<
       {strBuilder.append("a")} cost=24 ms


    > 查看源代码,以及简单分析

    String contact 和 StringBuffer,StringBuilder 的源代码都可以在Java库里找到,有空可以研究研究。

    1. 其实每次调用contact()方法就是一次数组的拷贝,虽然在内存中是处理都是原子性操作,速度非常快,但是,最后的return语句会创建一个新String对象,限制了concat方法的速度。

        public String concat(String str) {
            int otherLen = str.length();
            if (otherLen == 0) {
                return this;
            }
            int len = value.length;
            char buf[] = Arrays.copyOf(value, len + otherLen);
            str.getChars(buf, len);
            return new String(buf, true);
        }

    2. StringBuffer 和 StringBuilder 的append方法都继承自AbstractStringBuilder,整个逻辑都只做字符数组的加长,拷贝,到最后也不会创建新的String对象,所以速度很快,完成拼接处理后在程序中用strBuffer.toString()来得到最终的字符串。

        /**
         * Appends the specified string to this character sequence.
         * <p>
         * The characters of the {@code String} argument are appended, in
         * order, increasing the length of this sequence by the length of the
         * argument. If {@code str} is {@code null}, then the four
         * characters {@code "null"} are appended.
         * <p>
         * Let <i>n</i> be the length of this character sequence just prior to
         * execution of the {@code append} method. Then the character at
         * index <i>k</i> in the new character sequence is equal to the character
         * at index <i>k</i> in the old character sequence, if <i>k</i> is less
         * than <i>n</i>; otherwise, it is equal to the character at index
         * <i>k-n</i> in the argument {@code str}.
         *
         * @param   str   a string.
         * @return  a reference to this object.
         */
        public AbstractStringBuilder append(String str) {
            if (str == null) str = "null";
            int len = str.length();
            ensureCapacityInternal(count + len);
            str.getChars(0, len, value, count);
            count += len;
            return this;
        }
        /**
         * This method has the same contract as ensureCapacity, but is
         * never synchronized.
         */
        private void ensureCapacityInternal(int minimumCapacity) {
            // overflow-conscious code
            if (minimumCapacity - value.length > 0)
                expandCapacity(minimumCapacity);
        }
    
        /**
         * This implements the expansion semantics of ensureCapacity with no
         * size check or synchronization.
         */
        void expandCapacity(int minimumCapacity) {
            int newCapacity = value.length * 2 + 2;
            if (newCapacity - minimumCapacity < 0)
                newCapacity = minimumCapacity;
            if (newCapacity < 0) {
                if (minimumCapacity < 0) // overflow
                    throw new OutOfMemoryError();
                newCapacity = Integer.MAX_VALUE;
            }
            value = Arrays.copyOf(value, newCapacity);
        }

    3. 字符串的加号“+” 方法, 虽然编译器对其做了优化,使用StringBuilder的append方法进行追加,但是每循环一次都会创建一个StringBuilder对象,且都会调用toString方法转换成字符串,所以开销很大。

      注:执行一次字符串“+”,相当于 str = new StringBuilder(str).append("a").toString();

    4. 本文开头的地方统计了时间开销,根据上述分析再想想空间的开销。常说拿空间换时间,反过来是不是拿时间换到了空间呢,但是在这里,其实时间是消耗在了重复的不必要的工作上(生成新的对象,toString方法),所以对大批量数据做处理时,加号“+” 和 contact 方法绝对不能用,时间和空间成本都很高。

     【请尊重原创版权,如需引用,请注明来源及地址】

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  • 原文地址:https://www.cnblogs.com/twzheng/p/5923642.html
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