如何在Java中将字符串转换为32位对齐的字节数组？

Question

The title itself might not be self-explaining. Let me show an example:

_test is 5f 74 65 73 74 (totally 5 bytes in ascii)

However for some reason I need to pack it into a byte array like:

5f 74 65 73 74 00 00 00

(the last 3 bytes of 0 are padding bytes to do the alignment)

In other words, I need to calculate the number of padding bytes before I use write() to write those bytes one by one.

I can use this way to get the number I want: (4 - str.length() % 4) % 4 but I'm curious if there is another efficient way to get that (eg bit-wise operation).

Answer 1

是的，尽管仅当使用定点CPU时效率更高，而定点CPU会在软件中实现/和%等除运算： ((str.length()-1)|3)+1 。

Answer 2

To figure out the amount of padding to get to a multiple of 4 all we need to look at is the last 2 bits of the length.

Last 2 bits   Padding   Padding
 of Length    Needed    in binary
    00          0          00
    01          3          11
    10          2          10
    11          1          01

This suggests a pattern. If the length is even (last bit 0), the padding is equal to the last 2 bits of the length; otherwise it's the last 2 bits with the first bit flipped.

This can be expressed as

(len & 0x03) ^ ((len & 0x01) << 1)

Ie take the last 2 bits of the length and xor with the last bit shifted 1 bit to the left.

NOTE: If you do this, be sure to comment it well in the code so the next person who has to read this doesn't have to spend 10 minutes figuring it out.

Answer 3

str.length() is not the same as the number of bytes. This is only true if you use ASCII strings. If oyu have non-ASCII string (German, French, ...), then the correct way to determine the number of bytes is following:

int nBytes = str.getBytes("UTF8").length();

Determining the 32-bit aligned length is trivial.