将包含ascii字符串的byte[]快速转换为int/double/date等，无需new String

Question

我得到 FIX 消息字符串 (ASCII) 作为 ByteBuffer。我解析标签值对并将值存储为树形图中的原始对象，并以标签作为键。所以我需要根据其类型将 byte[] 值转换为 int/double/date 等。

最简单的方法是创建新的字符串并将其传递给标准转换器函数。例如，

int convertToInt(byte[] buffer, int offset, int length)
{
  String valueStr = new String(buffer, offset, length);
  return Integer.parseInt(valueStr);
}

我知道在 Java 中，创建新对象非常便宜，仍然有任何方法可以直接将此 ascii byte[] 转换为原始类型。我尝试了手写函数来执行此操作，但发现它非常耗时并且没有带来更好的性能。

是否有任何第三方库可以这样做，最重要的是它值得这样做吗？

Answer 1

最重要的是它值得做吗？

几乎肯定不是 - 在采取重大措施缓解它之前，您应该采取措施检查这是否是一个性能瓶颈。

你现在的表现怎么样？它需要是什么？ （“尽可能快”不是一个好的目标，否则你永远不会停止 - 弄清楚什么时候你可以说你“完成”了。）

分析代码 - 真正的问题是在字符串创建中？检查垃圾收集等的频率（再次使用分析器）。

每种解析类型可能具有不同的特征。例如，对于解析整数，如果您发现在很长一段时间内您得到的是单个数字，您可能想要特殊情况：

if (length == 1)
{
    char c = buffer[index];
    if (c >= '0' && c <= '9')
    {
        return c - '0';
    }
    // Invalid - throw an exception or whatever
}

...但是检查这种情况的频率发生在你走那条路之前。对实际上从未出现过的特定优化进行大量检查会适得其反。

Answer 2

同意 Jon 的观点，但是在处理许多 FIX 消息时，这种情况很快就会增加。
下面的方法将允许使用空格填充数字。如果您需要处理小数，那么代码会略有不同。两种方法之间的速度差异为 11 倍。ConvertToLong 导致 0 次 GC。下面的代码是c#中的：

///<summary>
///Converts a byte[] of characters that represent a number into a .net long type. Numbers can be padded from left
/// with spaces.
///</summary>
///<param name="buffer">The buffer containing the number as characters</param>
///<param name="startIndex">The startIndex of the number component</param>
///<param name="endIndex">The EndIndex of the number component</param>
///<returns>The price will be returned as a long from the ASCII characters</returns>
public static long ConvertToLong(this byte[] buffer, int startIndex, int endIndex)
{
    long result = 0;
    for (int i = startIndex; i <= endIndex; i++)
    {
        if (buffer[i] != 0x20)
        {
            // 48 is the decimal value of the '0' character. So to convert the char value
            // of an int to a number we subtract 48. e.g '1' = 49 -48 = 1
            result = result * 10 + (buffer[i] - 48);
        }
    }
    return result;
}

/// <summary>
/// Same as above but converting to string then to long
/// </summary>
public static long ConvertToLong2(this byte[] buffer, int startIndex, int endIndex)
{
    for (int i = startIndex; i <= endIndex; i++)
    {
        if (buffer[i] != SpaceChar)
        {
            return long.Parse(System.Text.Encoding.UTF8.GetString(buffer, i, (endIndex - i) + 1));
        }
    }
    return 0;
}

[Test]
public void TestPerformance(){
    const int iterations = 200 * 1000;
    const int testRuns = 10;
    const int warmUp = 10000;
    const string number = "    123400";
    byte[] buffer = System.Text.Encoding.UTF8.GetBytes(number);

    double result = 0;
    for (int i = 0; i < warmUp; i++){
        result = buffer.ConvertToLong(0, buffer.Length - 1);
    }
    for (int testRun = 0; testRun < testRuns; testRun++){
        Stopwatch sw = new Stopwatch();
        sw.Start();
        for (int i = 0; i < iterations; i++){
            result = buffer.ConvertToLong(0, buffer.Length - 1);
        }
        sw.Stop();
        Console.WriteLine("Test {4}: {0} ticks, {1}ms, 1 conversion takes = {2}μs or {3}ns. GCs: {5}", sw.ElapsedTicks,
            sw.ElapsedMilliseconds, (((decimal) sw.ElapsedMilliseconds)/((decimal) iterations))*1000,
            (((decimal) sw.ElapsedMilliseconds)/((decimal) iterations))*1000*1000, testRun,
            GC.CollectionCount(0) + GC.CollectionCount(1) + GC.CollectionCount(2));
    }
}
RESULTS
ConvertToLong:
Test 0: 9243 ticks, 4ms, 1 conversion takes = 0.02000μs or 20.00000ns. GCs: 2
Test 1: 8339 ticks, 4ms, 1 conversion takes = 0.02000μs or 20.00000ns. GCs: 2
Test 2: 8425 ticks, 4ms, 1 conversion takes = 0.02000μs or 20.00000ns. GCs: 2
Test 3: 8333 ticks, 4ms, 1 conversion takes = 0.02000μs or 20.00000ns. GCs: 2
Test 4: 8332 ticks, 4ms, 1 conversion takes = 0.02000μs or 20.00000ns. GCs: 2
Test 5: 8331 ticks, 4ms, 1 conversion takes = 0.02000μs or 20.00000ns. GCs: 2
Test 6: 8409 ticks, 4ms, 1 conversion takes = 0.02000μs or 20.00000ns. GCs: 2
Test 7: 8334 ticks, 4ms, 1 conversion takes = 0.02000μs or 20.00000ns. GCs: 2
Test 8: 8335 ticks, 4ms, 1 conversion takes = 0.02000μs or 20.00000ns. GCs: 2
Test 9: 8331 ticks, 4ms, 1 conversion takes = 0.02000μs or 20.00000ns. GCs: 2
ConvertToLong2:
Test 0: 109067 ticks, 55ms, 1 conversion takes = 0.275000μs or 275.000000ns. GCs: 4
Test 1: 109861 ticks, 56ms, 1 conversion takes = 0.28000μs or 280.00000ns. GCs: 8
Test 2: 102888 ticks, 52ms, 1 conversion takes = 0.26000μs or 260.00000ns. GCs: 9
Test 3: 105164 ticks, 53ms, 1 conversion takes = 0.265000μs or 265.000000ns. GCs: 10
Test 4: 104083 ticks, 53ms, 1 conversion takes = 0.265000μs or 265.000000ns. GCs: 11
Test 5: 102756 ticks, 52ms, 1 conversion takes = 0.26000μs or 260.00000ns. GCs: 13
Test 6: 102219 ticks, 52ms, 1 conversion takes = 0.26000μs or 260.00000ns. GCs: 14
Test 7: 102086 ticks, 52ms, 1 conversion takes = 0.26000μs or 260.00000ns. GCs: 15
Test 8: 102672 ticks, 52ms, 1 conversion takes = 0.26000μs or 260.00000ns. GCs: 17
Test 9: 102025 ticks, 52ms, 1 conversion takes = 0.26000μs or 260.00000ns. GCs: 18

Answer 3

看一下 ByteBuffer。它具有执行此操作的功能，包括处理字节顺序（字节顺序）。

Answer 4

一般来说，我不喜欢粘贴这样的代码，但无论如何，100 行是如何完成的（生产代码）
我不建议使用它，但有一些参考代码很好（通常）

package t1;

import java.io.UnsupportedEncodingException;
import java.nio.ByteBuffer;

public class IntParser {
    final static byte[] digits = {
        '0' , '1' , '2' , '3' , '4' , '5' ,
        '6' , '7' , '8' , '9' , 'a' , 'b' ,
        'c' , 'd' , 'e' , 'f' , 'g' , 'h' ,
        'i' , 'j' , 'k' , 'l' , 'm' , 'n' ,
        'o' , 'p' , 'q' , 'r' , 's' , 't' ,
        'u' , 'v' , 'w' , 'x' , 'y' , 'z'
    };

    static boolean isDigit(byte b) {
    return b>='0' &&  b<='9';
  }

    static int digit(byte b){
        //negative = error

        int result  = b-'0';
        if (result>9)
            result = -1;
        return result;
    }

    static NumberFormatException forInputString(ByteBuffer b){
        byte[] bytes=new byte[b.remaining()];
        b.get(bytes);
        try {
            return new NumberFormatException("bad integer: "+new String(bytes, "8859_1"));
        } catch (UnsupportedEncodingException e) {
            throw new RuntimeException(e);
        }
    }
    public static int parseInt(ByteBuffer b){
        return parseInt(b, 10, b.position(), b.limit());
    }
    public static int parseInt(ByteBuffer b, int radix, int i, int max) throws NumberFormatException{
        int result = 0;
        boolean negative = false;


        int limit;
        int multmin;
        int digit;      

        if (max > i) {
            if (b.get(i) == '-') {
                negative = true;
                limit = Integer.MIN_VALUE;
                i++;
            } else {
                limit = -Integer.MAX_VALUE;
            }
            multmin = limit / radix;
            if (i < max) {
                digit = digit(b.get(i++));
                if (digit < 0) {
                    throw forInputString(b);
                } else {
                    result = -digit;
                }
            }
            while (i < max) {
                // Accumulating negatively avoids surprises near MAX_VALUE
                digit = digit(b.get(i++));
                if (digit < 0) {
                    throw forInputString(b);
                }
                if (result < multmin) {
                    throw forInputString(b);
                }
                result *= radix;
                if (result < limit + digit) {
                    throw forInputString(b);
                }
                result -= digit;
            }
        } else {
            throw forInputString(b);
        }
        if (negative) {
            if (i > b.position()+1) {
                return result;
            } else {    /* Only got "-" */
                throw forInputString(b);
            }
        } else {
            return -result;
        }
    }

}