按位旋转 [英] bitwise rotation

问题描述

问候，


我在C＃中模拟汇编语言位轮换并且工作原理

奇妙


---------

....

public uint价值;

....

public uint RotateRight（字节数）{

Value =（（Value>> count）|（Value<<（32-count）））;

返回值;

}

---------


只要是值是32位还是64位（由于

移位运算符的限制）。我需要它来处理8位或16位值。我b $ b假设我必须知道价值的预期大小并且确定转移并移动该位和0或7，或0或者15位的位置并清除

如果有的额外位。这似乎是很多工作。


有没有办法模拟8位或16位旋转32位或64位

价值？

谢谢，

Shawn

Greetings,

I am simulating an assembly language bit rotation in C# and it works
wonderfully

---------
....
public uint Value;
....
public uint RotateRight(byte count) {
Value = ((Value >> count) | (Value << (32-count)));
return Value;
}
---------

as long is the value is 32-bit or 64-bit (because of a limitation of the
shift operator). I need it to work on an 8-bit or 16-bit value, as well. I
presume I''d have to have knowledge of the intended size of the value and do
the shift and move the bit and the 0 or 7, or 0 or 15 bit position and clear
the extra bits if there are any. That seems like a lot of work.

Is there a way to simulate an 8 or 16-bit rotate with a 32-bit or 64-bit
value?
Thanks,
Shawn

推荐答案

：


public uint RotateRight16（uint值，字节数）

{

return（（值>>计数） ）|（值<<（16-Count）））& 0xffff;

}

public uint RotateRight8（uint值，字节数）

{

return（（值>>计数）|（值<<（8-Count）））& 0xff;

}


public uint RotateLeft16（uint值，字节数）

{

return（（Value<< Count）| （值>>（16-Count）））& 0xffff;

}

public uint RotateLeft8（uint值，字节数）

{

return（（Value<< Count）|（Value>>（8-Count）））& 0xff;

}


-

Robert Jeppesen

robert.jeppesen％at％durius-dot-se

" Shawn B." <乐**** @ html.com>在留言新闻中写道：O9 *************** @ TK2MSFTNGP11.phx.gbl ...

How about:

public uint RotateRight16(uint Value, byte Count)
{
return ((Value >> Count) | (Value << (16-Count)))&0xffff;
}
public uint RotateRight8(uint Value, byte Count)
{
return ((Value >> Count) | (Value << (8-Count)))&0xff;
}

public uint RotateLeft16(uint Value, byte Count)
{
return ((Value << Count) | (Value >> (16-Count)))&0xffff;
}
public uint RotateLeft8(uint Value, byte Count)
{
return ((Value << Count) | (Value >> (8-Count)))&0xff;
}

--
Robert Jeppesen
robert.jeppesen%at%durius-dot-se
"Shawn B." <le****@html.com> wrote in message news:O9***************@TK2MSFTNGP11.phx.gbl...

问候，

我是在C＃中模拟汇编语言位旋转并且它的工作原理非常好

---------
...
公共uint值;
...
public uint RotateRight（字节数）{
Value =（（Value>> count）|（Value<<（32-count）））;
返回值;
}
---------

只要值为32位或64位（由于受限制
班次操作员）。我需要它来处理8位或16位值。我认为我必须知道该值的预期大小，并且确定移位并移动位和0或7，或0或15位位置并清除
如果有的额外位。这看起来很多工作。

有没有办法模拟具有32位或64位
值的8位或16位旋转？

谢谢，
Shawn

Greetings,

I am simulating an assembly language bit rotation in C# and it works
wonderfully

---------
...
public uint Value;
...
public uint RotateRight(byte count) {
Value = ((Value >> count) | (Value << (32-count)));
return Value;
}
---------

as long is the value is 32-bit or 64-bit (because of a limitation of the
shift operator). I need it to work on an 8-bit or 16-bit value, as well. I
presume I''d have to have knowledge of the intended size of the value and do
the shift and move the bit and the 0 or 7, or 0 or 15 bit position and clear
the extra bits if there are any. That seems like a lot of work.

Is there a way to simulate an 8 or 16-bit rotate with a 32-bit or 64-bit
value?
Thanks,
Shawn

糟糕，忘了确保价值在限制范围内：


public static uint RotateRight16（uint Value，byte Count）

{

return（（（Value& 0xffff）>> Count）|（值<<（16-Count）））& 0xffff;

}

public static uint RotateLeft16（uint Value，byte Count）

{

return（（Value<< Count）|（（Value& 0xffff）>>（16-Count）））& 0xffff;

}


等8位。


-

Robert Jeppesen

robert.jeppesen％at％durius-dot-se

" Robert Jeppesen" < robert.jeppesen（＃）durius.se>在留言新闻中写道：eE ************** @ TK2MSFTNGP11.phx.gbl ...

Oops, forgot to make sure that value is within limits:

public static uint RotateRight16(uint Value, byte Count)
{
return (((Value&0xffff) >> Count) | (Value << (16-Count)))&0xffff;
}
public static uint RotateLeft16(uint Value, byte Count)
{
return ((Value << Count) | ((Value&0xffff) >> (16-Count)))&0xffff;
}

etc for 8-bit.

--
Robert Jeppesen
robert.jeppesen%at%durius-dot-se
"Robert Jeppesen" <robert.jeppesen(#)durius.se> wrote in message news:eE**************@TK2MSFTNGP11.phx.gbl...

怎么样：

public uint RotateRight16（uint Value，byte Count）
返回（（Value>> Count）|（Value<<（16-Count）））& 0xffff;
}
public uint RotateRight8（uint Value，byte Count）
{
return（（Value>> Count）|（Value<<（8- Count）））& 0xff;
}
公共uint RotateLeft16（uint值，字节数）
返回（（值<< Count） |（值>>（16-Count）））& 0xffff;
}
public uint RotateLeft8（uint Value，byte Count）
{
return（（Value） << Count）|（Value>>（8-Count）））& 0xff;
}

-
Robert Jeppesen
罗伯特.jeppesen％at％durius-dot-se

" Shawn B." <乐**** @ html.com>在留言新闻中写道：O9 *************** @ TK2MSFTNGP11.phx.gbl ...

How about:

public uint RotateRight16(uint Value, byte Count)
{
return ((Value >> Count) | (Value << (16-Count)))&0xffff;
}
public uint RotateRight8(uint Value, byte Count)
{
return ((Value >> Count) | (Value << (8-Count)))&0xff;
}

public uint RotateLeft16(uint Value, byte Count)
{
return ((Value << Count) | (Value >> (16-Count)))&0xffff;
}
public uint RotateLeft8(uint Value, byte Count)
{
return ((Value << Count) | (Value >> (8-Count)))&0xff;
}

--
Robert Jeppesen
robert.jeppesen%at%durius-dot-se
"Shawn B." <le****@html.com> wrote in message news:O9***************@TK2MSFTNGP11.phx.gbl...

问候，

我是在C＃中模拟汇编语言位旋转并且它的工作原理非常好

---------
...
公共uint值;
...
public uint RotateRight（字节数）{
Value =（（Value>> count）|（Value<<（32-count）））;
返回值;
}
---------

只要值为32位或64位（由于受限制
班次操作员）。我需要它来处理8位或16位值。我认为我必须知道该值的预期大小，并且确定移位并移动位和0或7，或0或15位位置并清除
如果有的额外位。这看起来很多工作。

有没有办法模拟具有32位或64位
值的8位或16位旋转？

谢谢，
Shawn

Greetings,

I am simulating an assembly language bit rotation in C# and it works
wonderfully

---------
...
public uint Value;
...
public uint RotateRight(byte count) {
Value = ((Value >> count) | (Value << (32-count)));
return Value;
}
---------

as long is the value is 32-bit or 64-bit (because of a limitation of the
shift operator). I need it to work on an 8-bit or 16-bit value, as well. I
presume I''d have to have knowledge of the intended size of the value and do
the shift and move the bit and the 0 or 7, or 0 or 15 bit position and clear
the extra bits if there are any. That seems like a lot of work.

Is there a way to simulate an 8 or 16-bit rotate with a 32-bit or 64-bit
value?
Thanks,
Shawn

很难相信它是如此简单，我试过那...除非你在哪里
16-count是，我把它留在64位的8位掩码，这就解释了为什么它b / b
错了。


非常感谢，

Shawn

" Robert Jeppesen" < robert.jeppesen（＃）durius.se>在消息中写道

新闻：eE ************** @ TK2MSFTNGP11.phx.gbl ...

Hard to believe it was so simple, I tried that... except where you''re
16-count is, I left it at 64 bits with the 8-bit mask, that explains why it
was wrong.

Thanks very much,
Shawn
"Robert Jeppesen" <robert.jeppesen(#)durius.se> wrote in message
news:eE**************@TK2MSFTNGP11.phx.gbl...

怎么样 ：

public uint RotateRight16（uint Value，byte Count）
{
return（（Value>> Count）|（Value<<（ 16-Count）））& 0xffff;
}
public uint RotateRight8（uint Value，byte Count）
{
return（（Value>> Count）|（值<<（8-Count）））& 0xff;
}
公共uint RotateLeft16（uint值，字节数）
{
返回（ （值<< Count）|（值>>（16-Count）））& 0xffff;
}
public uint RotateLeft8（uint Value，byte Count）
{
return（（值<< Count）|（Value>>（8-Count）））& 0xff;
}

-
Robert Jeppesen
robert.jeppesen％at％durius-dot-se

Shawn B. <乐**** @ html.com>在消息中写道

How about:

public uint RotateRight16(uint Value, byte Count)
{
return ((Value >> Count) | (Value << (16-Count)))&0xffff;
}
public uint RotateRight8(uint Value, byte Count)
{
return ((Value >> Count) | (Value << (8-Count)))&0xff;
}

public uint RotateLeft16(uint Value, byte Count)
{
return ((Value << Count) | (Value >> (16-Count)))&0xffff;
}
public uint RotateLeft8(uint Value, byte Count)
{
return ((Value << Count) | (Value >> (8-Count)))&0xff;
}

--
Robert Jeppesen
robert.jeppesen%at%durius-dot-se
"Shawn B." <le****@html.com> wrote in message

新闻：O9 *************** @ TK2MSFTNGP11.phx.gbl ...

news:O9***************@TK2MSFTNGP11.phx.gbl...

问候，

我在C＃中模拟汇编语言位旋转，它的工作原理很好

------ ---
...
公共uint值;
...
公共uint RotateRight（字节数）{
值=（（值>> count）|（值<<（32-count）））;
返回值;
}
---------

只要该值为32位或64位（由于
移位运算符的限制）。我需要它来处理8位或16位值，因为
很好。我认为我必须知道值的预期大小，并且
执行移位并移动位和0或7，或0或15位位置和
清除额外位如果有的话。这看起来很多工作。

有没有办法模拟具有32位或64位
值的8位或16位旋转？

谢谢，
Shawn

Greetings,

I am simulating an assembly language bit rotation in C# and it works
wonderfully

---------
...
public uint Value;
...
public uint RotateRight(byte count) {
Value = ((Value >> count) | (Value << (32-count)));
return Value;
}
---------

as long is the value is 32-bit or 64-bit (because of a limitation of the
shift operator). I need it to work on an 8-bit or 16-bit value, as well. I presume I''d have to have knowledge of the intended size of the value and do the shift and move the bit and the 0 or 7, or 0 or 15 bit position and clear the extra bits if there are any. That seems like a lot of work.

Is there a way to simulate an 8 or 16-bit rotate with a 32-bit or 64-bit
value?
Thanks,
Shawn

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