我得到FIX消息字符串（ASCII）作为ByteBuffer。我解析标记值对并将值存储为树图中的基本对象，标记为键。所以我需要将byte []值转换为int / double / date等，具体取决于它的类型。
最简单的方法是创建新的String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);
}
据我所知，在Java中，创建新对象非常便宜，仍然有办法将此ascii byte []直接转换为原始类型。我尝试过手写功能，但发现这很耗时，并没有带来更好的性能。
是否有任何第三方图书馆这样做，最重要的是值得做什么？
答案 0 :(得分：2)
最重要的是值得做什么？
几乎肯定不是 - 你应该测量一下这个 是否是性能瓶颈，然后再做出重大努力来缓解它。
你现在的表现如何？它需要什么？ （“尽可能快”不是一个好目标，或者你永远不会停止 - 当你可以说你“完成”时锻炼出去。）
配置代码 - 字符串创建中的真正问题是什么？检查你的垃圾收集频率等（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，如果您在很长一段时间内发现了一个数字，那么可能想要特殊情况：
if (length == 1)
{
char c = buffer[index];
if (c >= '0' && c <= '9')
{
return c - '0';
}
// Invalid - throw an exception or whatever
}
...但是在你沿着这条路走下去之前，请检查这种情况。对从未真正实现的特定优化应用大量检查会产生适得其反的效果。
答案 1 :(得分：2)
与Jon同意，但是当处理许多FIX消息时，这很快就会增加。 下面的方法将允许空格填充数字。如果你需要处理小数，那么代码会略有不同。两种方法之间的速度差异为11. ConvertToLong产生0个GC。以下代码位于c＃：
///<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
答案 2 :(得分：1)
看看ByteBuffer。它具有执行此操作的功能，包括处理字节顺序（字节顺序）。
答案 3 :(得分：1)
一般来说，我没有偏好粘贴这样的代码，但无论如何，100行怎么做（生产代码） 我不建议使用它，但有一些参考代码它很好（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;
}
}
}