使用位移位和十六进制代码的 setBit java 方法 - 问题

发布于 2024-08-29 08:32:37 字数 982 浏览 3 评论 0原文

我无法理解 0xFF7F 的两行及其下面的一行发生了什么。这里有一个链接在某种程度上解释了它。 http://www.herongyang.com/ java/Bit-String-Set-Bit-to-Byte-Array.html 我不知道是否 0xFF7F>>posBit) &旧字节）& 0x00FF 应该是 3 个值“AND”在一起或者应该如何读取。如果有人能更好地澄清这里发生的事情，我将不胜感激。

private static void setBit(byte[] data,
                               final int pos,
                               final int val) {
        int posByte = pos/8;
        int posBit = pos%8;
        byte oldByte = data[posByte];
        oldByte = (byte) (((0xFF7F>>posBit) & oldByte) & 0x00FF);
        byte newByte = (byte) ((val<<(8-(posBit+1))) | oldByte);
        data[posByte] = newByte;
    }

作为来自 selectBits 方法的参数传入此方法的是 setBit(out,i,val)； out = is byte[] out = new byte[numOfBytes]; （在这种情况下 numOfBytes 可以是 7） i = 数字 [57]，来自保存 56 个整数的 PC1 int 数组的原始数字。 val = 这是通过 getBit() 方法从字节数组中获取的位。

原文

I am having trouble understanding what is happening in the two lines with the 0xFF7F and the one below it. There is a link here that explains it to some degree.
http://www.herongyang.com/java/Bit-String-Set-Bit-to-Byte-Array.html
I don't know if 0xFF7F>>posBit) & oldByte) & 0x00FF
are supposed to be 3 values 'AND'ed together or how this is supposed to be read. If anyone can clarify what is happening here a little better, I would greatly appreciate it.

private static void setBit(byte[] data,
                               final int pos,
                               final int val) {
        int posByte = pos/8;
        int posBit = pos%8;
        byte oldByte = data[posByte];
        oldByte = (byte) (((0xFF7F>>posBit) & oldByte) & 0x00FF);
        byte newByte = (byte) ((val<<(8-(posBit+1))) | oldByte);
        data[posByte] = newByte;
    }

passed into this method as parameters from a selectBits method was setBit(out,i,val);
out = is byte[] out = new byte[numOfBytes]; (numOfBytes can be 7 in this situation)
i = which is number [57], the original number from the PC1 int array holding the 56-integers.
val = which is the bit taken from the byte array from the getBit() method.

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平生欢 2024-09-05 08:32:37

首先，0xFF7F 是1111 1111 0111 1111。它右移了根据您作为参数传递的位（即您要设置的位）计算出的位数。

如果您指定第三位 posBit = 3 % 8 = 3，那么

0xFF7F 1111 1111 0111 1111
>> 3   0001 1111 1110 1111

该值将与您正在修改的原始字节进行 AND 运算，结果是每个位都保持等于 oldBit 原始位，除了与 0 位进行与运算的位，假设您有例如 oldByte == 0111 1010，您将获得：

    0111 1010
&   1110 1111
-------------
    0110 1010

然后该值与0xFF 只是在进行转换之前丢弃任何不适合字节的位（因为它至少是第九位）。

First of all 0xFF7F is 1111 1111 0111 1111. This is shifted right by an amount of bits calculated from the bit you pass as a parameter (so the one you want to set).

If you specify third bit posBit = 3 % 8 = 3 so

0xFF7F 1111 1111 0111 1111
>> 3   0001 1111 1110 1111

this value is then ANDed with the original byte you are modifying, the result is that every bit is kept equal to oldBit original bit except the one that is anded with the 0 bit, suppose you have for example oldByte == 0111 1010, you'll obtain:

    0111 1010
&   1110 1111
-------------
    0110 1010

Then the value is anded with 0xFF just to discard any bit that doesn't fit a byte (because it's at least the ninth bit) before doing the cast.

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情释 2024-09-05 08:32:37

更好的写法是：

private static void setBit(byte[] data, int index, boolean value)
{
    final int byteIndex = index / 8;
    final int bitIndex = 7 - (index % 8);

    final byte mask = (byte) (1 << bitIndex);
    final byte valueBit = value ? mask : 0;

    data[byteIndex] = (byte) ((data[byteIndex] & ~mask) | valueBit);
}

A better way to write this would be:

private static void setBit(byte[] data, int index, boolean value)
{
    final int byteIndex = index / 8;
    final int bitIndex = 7 - (index % 8);

    final byte mask = (byte) (1 << bitIndex);
    final byte valueBit = value ? mask : 0;

    data[byteIndex] = (byte) ((data[byteIndex] & ~mask) | valueBit);
}

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