网络Java优化将值作为位写入bytebuffer

Question

我目前正在编写一些网络代码（这是我的第一台服务器），并有一个关于优化特定函数的快速问题，该函数将值写入位，然后将它们打包成字节。优化此函数的原因是，在将数据打包发送到多个客户端时，每个服务器都会使用它数千次

为了更好地解释函数试图实现的功能，可以使用一个示例： 值3可由两位表示。 在二进制中，它看起来像00000011。函数将把这个二进制值转换成11000000。再次调用该函数时，它将知道从第三个最高有效位（从右/十进制32开始的第三个）开始，最多向当前字节写入6位。如果还有剩余的位要写，它将从一个新字节开始

如果有多个小于字节的值，则这样做的目的是节省空间

我当前的函数如下所示：

 private ByteBuffer out = ByteBuffer.allocate(1024);
 private int bitIndex = 0;
 /*
  * Value: The value to write
  * Amount: The number of bits to represent the value in.
  */
     public OutputBuffer writeBits(long value, int amount) {
    if (bitIndex != 0) {
        int remainingBits = 8 - bitIndex;
        int bytePos = out.position() - 1;
        byte current = out.get(bytePos);
        int shiftAmount = amount - remainingBits;
        int bitsWritten = amount < remainingBits ? amount : remainingBits;
        int clearShiftAmount = 8 - bitsWritten + 56;

        byte b;
        if (shiftAmount < 0) {
            b = (byte) (current | (value << remainingBits - amount));
        } else {
            //deal with negative values
            long temp = (value >> shiftAmount);
            temp =  (temp << clearShiftAmount);
            temp = (byte) (temp >>> clearShiftAmount);
            b = (byte) (current | temp);
        }
        out.put(bytePos,b);
        bitIndex = (bitIndex + bitsWritten) % 8;
        amount -= bitsWritten;
    }
    if (amount <= 0) {
        return this;
    }
    bitIndex = amount & 7;
    int newAmount = amount - bitIndex;
    //newValue should not equal 2047
    for (int i = 0; i != newAmount; i += 8) {
        writeByte((byte) ((value >> i)), false);
    }
    if (bitIndex > 0)
        writeByte((byte) (value << (8 - bitIndex)), false);
    return this;
}

由于我是新手，我认为可能有更有效的方法，可能使用位屏蔽或某种查找表？任何正确方向的想法或方向都会很好。干杯

Answer 1

像这样的怎么样

private ByteBuffer out = ByteBuffer.allocate(1024);
private int position = 0;
private int dataBits = 0;
private long data = 0;

/**
 * value: The value to write
 * amount: The number of bits to represent the value in.
 */
public void writeBits(long value, int amount) {
    if (amount <= 0) {
        // need to flush what's left
        if (dataBits > 0) {
            dataBits = Byte.SIZE;
        }
    } else {
        int totalBits = dataBits + amount;

        // need to handle overflow?
        if (totalBits > Long.SIZE) {
            // the new data is to big for the room that remains;  by how much?
            int excess = totalBits - Long.SIZE;

            // drop off the excess and write what's left
            writeBits(value >> excess, amount - excess);

            // now we can continue processing just the rightmost excess bits
            amount = excess;
        }

        // push the bits we're interested in all the way to the left of the long
        long temp = value << (Long.SIZE - amount);

        // make room for any existing (leftover) data bits, filling with zeros to the left (important)
        temp = temp >> dataBits;

        // append the new data to the existing
        data |= temp;

        // account for new bits of data
        dataBits += amount;
    }

    while (dataBits >= Byte.SIZE) {
        // shift one byte left, rotating the byte that falls off into the rightmost byte
        data = Long.rotateLeft(data, Byte.SIZE);

        // add the rightmost byte to the buffer
        out.put(position++, (byte)(data & 0xff));

        // drop off the rightmost byte
        data &= 0xffffffffffffff00L;

        // setup for next byte
        dataBits -= Byte.SIZE;
    }
}

Answer 2

这里有一个使用递归的更好的解决方案（比我以前的解决方案），它非常适合这个问题

private static final long[] mask = { 0, 0x1, 0x3, 0x7, 0xf, 0x1f, 0x3f, 0x7f, 0xff };

private ByteBuffer out = ByteBuffer.allocate(1024);
private int position = 0;
private int dataBits = 0;
private byte remainder = 0;

/**
 * value: The value to write
 * amount: The number of bits to represent the value in.
 */
public void writeBits(long value, int amount) {
    if (amount <= Long.SIZE) {
        if (amount > 0) {
            // left align the bits in value
            writeBitsLeft(value << (Long.SIZE - amount), amount);
        } else {
            // flush what's left
            out.put(position++, remainder);
        }
    } else {
        // the data provided is invalid
        throw new IllegalArgumentException("the amount of bits to write is out of range");
    }
}

/**
 * write amount bits from the given value
 * 
 * @param value represents bits aligned to the left of a long
 * @param amount bits left to be written from value
 */
private void writeBitsLeft(long value, int amount) {
    if (amount > 0) {
        // how many bits are left to be filled in the current byte?
        int room = Byte.SIZE - dataBits;

        // how many bits are we going to add to the current byte?
        int taken = Math.min(amount, room);

        // rotate those number of bits into the rightmost position
        long temp = Long.rotateLeft(value, taken);

        // add count taken to the count of bits in the current byte
        dataBits += taken;

        // add in that number of data bits
        remainder &= temp & mask[taken];

        // have we filled the byte yet?
        if (Byte.SIZE == dataBits) {
            out.put(position++, remainder);

            // reset the current byte
            remainder = 0;
            dataBits = 0;

            // process any bits left over
            writeBitsLeft(temp, amount - taken);
        }
    } 
} // writeBitsLeft()

此解决方案具有较少的数学运算、移位运算和if语句，因此应该比原始解决方案更有效，更不用说它可能更容易理解

Answer 3

好的，我调整了你原来的算法，去掉了一些多余的数学运算，我减少了大约10%（在我的机器上从0.016毫秒减少到了大约0.014毫秒）。我还修改了测试，使每个算法运行1000次

在最后一个for循环中似乎也有一些节省，因为相同的位被反复移位。如果你能以某种方式保留上一次换班的结果，那可能会有所帮助。但这会改变字节输出的顺序，因此需要更多的思考

public void writeBits3(long value, int amount) {
    if (bitIndex != 0) {
        int remainingBits = 8 - bitIndex;
        int bytePos = out.position() - 1;
        byte current = out.get(bytePos);
        int shiftAmount = amount - remainingBits;

        int bitsWritten = 0;
        if (shiftAmount < 0) {
            bitsWritten = amount;
            out.put(bytePos, (byte) (current | (value << -shiftAmount)));
        } else {
            bitsWritten = remainingBits;
            out.put(bytePos, (byte) (current | (value >> shiftAmount)));
        }

        bitIndex += bitsWritten;
        amount -= bitsWritten;
        if (bitIndex >= 8) {
            bitIndex = 0;
        }
    }
    if (amount <= 0) {
        return;
    }
    bitIndex = amount & 7;
    int newAmount = amount - bitIndex;
    long newValue = (value >> bitIndex);
    for (; newAmount >= 8; newAmount -= 8) {
        out.put((byte) (newValue >> newAmount));
    }
    out.put((byte) (value << (8 - bitIndex)));
}