测试是否只设置了一位... [英] testing if just one bit is set...
问题描述
。我现在正在寻找一些*超快*的方法，以确定是否一个int有多个位设置
。任何想法？
" .rhavin grobert" < cl *** @ yahoo.dewrote in message
news：ec *************************** ******* @ u29g2000 pro.googlegroups.com ...
猜你有一个处理器可以原生处理32位你有
a 32-bit-int。我现在正在寻找一些*超快*的方法，以确定是否一个int有多个位设置
。有任何想法吗？
如果n有无符号类型（即unsigned int或unsigned long），则（n& -n）
等于n除非n设置了多个位。
所以你要找的表达式是n！=（n& -n）
.. rhavin grobert写道：
猜你有一个可以原生处理32位的处理器，你有32美元的b $ ba $ bit-int。我现在正在寻找一些*超快*的方法，以确定是否一个int有多个位设置
。有任何想法吗？
我认为二元搜索是指方法很快，但你需要
来测量它。类似
inline bool moreThanOneBitSet（无符号值，
unsigned mask1，unsigned mask2）
{
return（value& mask1）&& （值& mask2）;
}
bool moreThanOneBitSet（无符号值）
{
static const unsigned masks [] = {0x0000FFFF，0xFFFF0000，
0x00FF00FF，0xFF00FF00，
0x0F0F0F0F，0xF0F0F0F0，
0x33333333,0xCCCCCCCC，
0x55555555,0xAAAAAAAA};
if（value 0）{
返回moreThanOneBitSet（value，掩码[0]，掩码[1]）| |
moreThanOneBitSet（value，masks [2]，mask [3]）||
moreThanOneBitSet（value，masks [4]，mask [5]）||
moreThanOneBitSet（value，masks [6]，mask [7]）||
moreThanOneBitSet（value，masks [8]，mask [9]）;
}
其他
返回false;
}
最多10按位AND，5逻辑AND，5逻辑OR，并且不确定
许多测试对0和跳跃...
V
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Andrew Koenig写道：
" .rhavin grobert" < cl *** @ yahoo.dewrote in message
news：ec *************************** ******* @ u29g2000 pro.googlegroups.com ...
>猜测你有一个可以处理32位本地处理器的处理器你有
一个32位的int。我现在正在寻找一些*超快*的方法来确定一个int是否有多个位设置。有任何想法吗？
如果n有无符号类型（即unsigned int或unsigned long），则（n& -n）
等于n除非n设置了多个位。
所以你要找的表达式是n！=（n& -n）
哇...它是否适用于任何表示（两个补码，一个'
补码，签名幅度）？
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guess you have a processor that can handle 32bit natively and you have
a 32-bit-int. im now looking for some *ultrafast* way to determine if
an int has more than one bit set. any ideas?
".rhavin grobert" <cl***@yahoo.dewrote in message
news:ec**********************************@u29g2000 pro.googlegroups.com...
guess you have a processor that can handle 32bit natively and you have
a 32-bit-int. im now looking for some *ultrafast* way to determine if
an int has more than one bit set. any ideas?If n has an unsigned type (i.e. unsigned int or unsigned long), then (n&-n)
is equal to n unless n has more than one bit set.
So the expression you''re looking for is n!=(n&-n)
..rhavin grobert wrote:guess you have a processor that can handle 32bit natively and you have
a 32-bit-int. im now looking for some *ultrafast* way to determine if
an int has more than one bit set. any ideas?I think the "binary search" method is quick enough, but you will need to
measure it. Something like
inline bool moreThanOneBitSet(unsigned value,
unsigned mask1, unsigned mask2)
{
return (value & mask1) && (value & mask2);
}
bool moreThanOneBitSet(unsigned value)
{
static const unsigned masks[] = { 0x0000FFFF,0xFFFF0000,
0x00FF00FF,0xFF00FF00,
0x0F0F0F0F,0xF0F0F0F0,
0x33333333,0xCCCCCCCC,
0x55555555,0xAAAAAAAA };
if (value 0) {
return moreThanOneBitSet(value, masks[0], masks[1]) ||
moreThanOneBitSet(value, masks[2], masks[3]) ||
moreThanOneBitSet(value, masks[4], masks[5]) ||
moreThanOneBitSet(value, masks[6], masks[7]) ||
moreThanOneBitSet(value, masks[8], masks[9]);
}
else
return false;
}
At most, 10 bitwise AND, 5 logical AND, 5 logical OR, and not sure how
many tests against 0 and jumps...
V
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Andrew Koenig wrote:".rhavin grobert" <cl***@yahoo.dewrote in message
news:ec**********************************@u29g2000 pro.googlegroups.com...
>guess you have a processor that can handle 32bit natively and you have
a 32-bit-int. im now looking for some *ultrafast* way to determine if
an int has more than one bit set. any ideas?
If n has an unsigned type (i.e. unsigned int or unsigned long), then (n&-n)
is equal to n unless n has more than one bit set.
So the expression you''re looking for is n!=(n&-n)Wow... Does it work for any representation (two''s complement, one''s
complement, signed magnitude)?
V
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I do not respond to top-posted replies, please don''t ask
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