问题
I am having a hard time converting this ActionScript code to C#. I don't understand how I can mimic what ByteArray() is doing in C#. If anyone could help me recreate this functionality I would really appreciate it.
ActionScript (seed
is a uint.):
//Start by reversing the byte order of the seed
var ba:ByteArray = new ByteArray();
ba.endian = Endian.BIG_ENDIAN;
ba.writeInt(seed);
ba.position = 0;
ba.endian = Endian.LITTLE_ENDIAN;
seed = ba.readInt();
回答1:
You can use BitConverter.GetBytes(UInt32)
to get your byte[]
, then call Array.Reverse
on the array, then use BitConverter.ToUInt32(byte[])
to get your int back out.
Edit
Here's a more efficient and cryptic way to do it:
public static UInt32 ReverseBytes(UInt32 value)
{
return (value & 0x000000FFU) << 24 | (value & 0x0000FF00U) << 8 |
(value & 0x00FF0000U) >> 8 | (value & 0xFF000000U) >> 24;
}
This is what you need to know to understand what's happening here:
- A UInt32 is 4 bytes.
- In hex, two characters represent one byte. 179 in decimal == B3 in hex == 10110011 in binary.
- A bitwise and (
&
) preserves the bits that are set in both inputs:1011 0011 & 1111 0000 = 1011 0000
; in hex:B3 & F0 = B0
. - A bitwise or (
|
) preserves the bits that are set in either input:1111 0000 | 0000 1111 = 1111 1111
; in hex,F0 | 0F = FF
. - The bitwise shift operators (
<<
and>>
) move the bits left or right in a value. So0011 1100 << 2 = 1111 0000
, and1100 0011 << 4 = 0011 0000
.
So value & 0x000000FFU
returns a UInt32 with all but the 4th byte set to 0. Then << 24
moves that 4th byte to the left 24 places, making it the 1st byte. Then (value & 0x0000FF00U) << 8
zeros out everything but the 3rd byte and shifts it left so it is the second byte. And so on. The four (x & y) << z
create four UInt32s where each of the bytes have been moved to the place they will be in the reversed value. Finally, the |
combines those UIntt32s bytes back together into a single value.
来源:https://stackoverflow.com/questions/18145667/how-can-i-reverse-the-byte-order-of-an-int