位掩码 [英] Bit mask
问题描述
什么是一种有效的方法来掩盖8位字符的最后3位并使它们
全部为零？
位 - 转移是可能的，但不确定它是否足够有效。
示例：
01011 [010] - > 01011 [000]
谢谢！
What''s an efficient way to mask a last 3 bits of a 8-bit char and make them
all zero?
Bit-shifting is possible but not sure if it is efficient enough.
Example:
01011[010] --> 01011[000]
Thanks!
推荐答案
Als< no **** @ nowhere。净>潦草地写道：
Als <no****@nowhere.net> scribbled the following:
什么是一种有效的方法来掩盖8位字符的最后3位并使它们全部为零？
可以进行位移，但不确定它是否足够有效。
示例：
01011 [010] - > 01011 [000]
谢谢！
What''s an efficient way to mask a last 3 bits of a 8-bit char and make them
all zero? Bit-shifting is possible but not sure if it is efficient enough. Example: 01011[010] --> 01011[000] Thanks!
好吧，一个非常简单的方法是和〜7进行AND运算。
例如：
unsigned char x = 0x5A; / *与二进制01011010相同* /
x = x& 〜7; / *清除最后3位* /
效率高吗？这取决于你的实施。
-
/ - Joona Palaste（pa*****@cc.helsinki.fi）--- ----------芬兰-------- \
\-- http://www.helsinki.fi/~palaste --------------------- rules ！ -------- /
C ++看起来像线路噪音。
- Fred L. Baube III
Well, a really simple way would be ANDing with ~7.
For example:
unsigned char x = 0x5A; /* the same as binary 01011010 */
x = x & ~7; /* clear last 3 bits */
Is it efficient? It depends on your implementation.
--
/-- Joona Palaste (pa*****@cc.helsinki.fi) ------------- Finland --------\
\-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
"C++ looks like line noise."
- Fred L. Baube III
Mark A. Odell写道：
Mark A. Odell wrote:
Joona I Palaste< pa ***** @ cc.helsinki.fi>在
新闻中写道：bt ********** @ oravannahka.helsinki.fi：
Joona I Palaste <pa*****@cc.helsinki.fi> wrote in
news:bt**********@oravannahka.helsinki.fi:
unsigned char x = 0x5A;
unsigned char x = 0x5A;
x& =（unsigned char）~0x07;
x &= (unsigned char) ~0x07;
这几乎是一回事，无论效率差异如何，编译器优化都可能会丢失。我只是认为我的版本更清晰易读。适合自己。
It''s practically the same thing, and whatever efficiency difference
there would be will probably get lost in compiler optimisation. I just
thought my version was clearer to read. Suit yourself.
我的意思是没有WRT效率，只是因为我认为& =是更常见的习惯用法，你需要将~0x07转换为unsigned char。
I meant nothing WRT efficiency, just that I thought that &= is the more
common idiom and that you needed to cast ~0x07 to unsigned char.
为什么需要演员？这对我来说似乎毫无用处，因为结果
将被提升回int（无论如何，可能是unsigned int理论上也是
）。
-Kevin
-
我的电子邮件地址有效，但会定期更改。
要联系我，请使用地址来自最近的帖子。
Why would the cast be needed? It seems useless to me, since the result
will be promoted back to int (probably - unsigned int is theoretically
possible also) anyway.
-Kevin
--
My email address is valid, but changes periodically.
To contact me please use the address from a recent posting.
Eric Sosman< Er ********* @ sun.com>在消息中写道
新闻：3F *************** @sun.com ...
Eric Sosman <Er*********@sun.com> wrote in message
news:3F***************@sun.com...
Als写道：
什么是掩盖8位字符的最后3位并使
全部为零的有效方法？
位移是可能但不确定它是否足够有效。
示例：
01011 [010] - > 01011 [000]
What''s an efficient way to mask a last 3 bits of a 8-bit char and make them all zero?
Bit-shifting is possible but not sure if it is efficient enough.
Example:
01011[010] --> 01011[000]
许多甚至大多数C实现都使用了一个8位`char''，但实际上并没有保证语言的语言并且已知存在使用更宽的`char''
的实现。仍然：
unsigned char byte = 0x5A; / * 00 ... 01011010 * /
Many or perhaps even most C implementations use an
eight-bit `char'', but that is not actually guaranteed
by the language, and implementations using wider `char''
are known to exist. Still:
unsigned char byte = 0x5A; /* 00...01011010 */
您是否有任何理由使用unsigned char？而不是char以上？
谢谢！
Is there any reason that you use "unsigned char" instead of "char" above?
Thanks!
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