二进制问题 [英] Binary question
问题描述
给定任意字节（或者就此而言是int），有没有办法计算
out：
1）一个'套数的*数没有循环？也许是一些公式？
2）从右边看，如果所有1'都是连续的，而不是
干预0'。 0'结束了确定。
TIA
-
William Stacey，MVP
Given any arbitrary byte (or int for that matter), is there a way to figure
out:
1) The *number of one''s set without a loop? Maybe some formula?.
2) Looking from left the right, if all 1''s are continuous with not
intervening 0''s. 0''s on the end ok.
TIA
--
William Stacey, MVP
推荐答案
" William Stacey" < ST *********** @ mvps.org>写道：
"William Stacey" <st***********@mvps.org> wrote:
给定任意字节（或int为此），有没有办法让
弄清楚：
1）没有任何一个'一个循环？也许是一些公式？。
2）从左边看，如果所有1'都是连续的，而不是介入0'。 0'结束了确定。
Given any arbitrary byte (or int for that matter), is there a way to figure out:
1) The *number of one''s set without a loop? Maybe some formula?.
2) Looking from left the right, if all 1''s are continuous with not
intervening 0''s. 0''s on the end ok.
我会在编译时使用一个填充了所需信息的数组
时间，并且字节作为运行时数组的索引。
Jens。
-
http://ManagedXLL.net/ | http://jens-thiel.de/ | http://QuantLib.net/
用我的第一个替换MSDN回复我的电子邮件地址时的名字！
I would use an array that is filled with the required information at compile
time, and the byte as index to the array at run-time.
Jens.
--
http://ManagedXLL.net/ | http://jens-thiel.de/ | http://QuantLib.net/
Replace MSDN with my first name when replying to my email address!
是的我能做到这一点并感谢您的回复。我只是在想这个
可以用数学来完成。
-
-
William Stacey，MVP
" Jens Thiel" < MS ** @ Thiel.de>在消息中写道
新闻：#D ************* @ TK2MSFTNGP10.phx.gbl ...
Yes I could do that and thanks for the reply. I was just thinking this
could be done with math.
--
--
William Stacey, MVP
"Jens Thiel" <MS**@Thiel.de> wrote in message
news:#D*************@TK2MSFTNGP10.phx.gbl...
" William Stacey" < ST *********** @ mvps.org>写道：
"William Stacey" <st***********@mvps.org> wrote:
给定任意字节（或int为此），有没有办法计算
Given any arbitrary byte (or int for that matter), is there a way to figure
out：
1）* number of one'没有循环的设置？也许是一些公式？。
2）从左边看，如果所有1'都是连续的，而不是介入0'。 0'结束了确定。
out:
1) The *number of one''s set without a loop? Maybe some formula?.
2) Looking from left the right, if all 1''s are continuous with not
intervening 0''s. 0''s on the end ok.
我会在
I would use an array that is filled with the required information at
编译时使用一个填充了所需信息的数组，并且字节作为运行时数组的索引。
Jens。
http://ManagedXLL.net/ | http://jens-thiel.de/ | http://QuantLib.net/
在回复时，用我的名字替换MSDN我的电子邮件地址！
compile time, and the byte as index to the array at run-time.
Jens.
--
http://ManagedXLL.net/ | http://jens-thiel.de/ | http://QuantLib.net/
Replace MSDN with my first name when replying to my email address!
仅供参考。做了一些测试和研究，并想出了这个。任何
优化？
///< summary>
///返回1的数字在任何uint中设置的位。
///< / summary>
public static int BitCount（uint x）
{
int bitCount = 0;
if（x！= 0）
{
do
{
bitCount ++;
x = x& （x - 1）;
} while（x！= 0）;
}
返回bitCount;
}
///< summary>
///给定任何uint，如果1位从左边连续到
right。
///此方法也将返回out参数1位的数量
in
/// num，即使这些位不是连续的。
///< / summary>
public static bool ContiguousBits（uint num，out int bitCount）
{
int shift = 0;
int bc = 0;
bc = Bits.BitCount（num）; //上面的BitCount方法。
shift = 32 - bc;
bitCount = bc;
return（（num>> shift ）==（Math.Pow（2，bc） - 1））;
}
-
William Stacey ，MVP
" William Stacey" < ST *********** @ mvps.org>在消息中写道
news：uL ************* @ tk2msftngp13.phx.gbl ...
FYI. Did some more testings and research and came up with this. Any
optimizations?
/// <summary>
/// Returns the number of "1" bits set in any uint.
/// </summary>
public static int BitCount(uint x)
{
int bitCount = 0;
if ( x != 0)
{
do
{
bitCount++;
x = x & (x - 1);
} while (x != 0);
}
return bitCount;
}
/// <summary>
/// Given any uint, returns true if the 1 bits are contiguous from left to
right.
/// This method will also return to the "out" parameter the number of 1 bits
in
/// the num, even if the bits are not contiguous.
/// </summary>
public static bool ContiguousBits(uint num, out int bitCount)
{
int shift = 0;
int bc = 0;
bc = Bits.BitCount(num); //The BitCount Method above.
shift = 32 - bc;
bitCount = bc;
return ( (num >> shift) == (Math.Pow(2, bc) - 1) );
}
--
William Stacey, MVP
"William Stacey" <st***********@mvps.org> wrote in message
news:uL*************@tk2msftngp13.phx.gbl...
给定任意字节（或对于那个问题，有没有办法让
弄明白：
1）没有循环的一个'套数？也许是一些公式？。
2）从左边看，如果所有1'都是连续的，而不是介入0'。 0'结束了确定。
TIA
-
William Stacey，MVP
Given any arbitrary byte (or int for that matter), is there a way to figure out:
1) The *number of one''s set without a loop? Maybe some formula?.
2) Looking from left the right, if all 1''s are continuous with not
intervening 0''s. 0''s on the end ok.
TIA
--
William Stacey, MVP
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