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我正在尝试使用 x*x-1 检查整数是否为 2 的幂,然后对其进行计数。

long count_bits(long n) {
unsigned int c;
for c = 0:n
n = n * (n - 1);  %Determines if an integer is a power of two!
c=c+1;
end

disp(c);

在这里找到了我的答案:在matlab中有效地计算汉明权重

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1 回答 1

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使用bitget

% generate a random int number
>> n = uint32( randi( intmax('uint32'), 1, 1 ) )

n = 

   3771981510

>> count = sum(bitget(n,1:32))

count = 

   18

或者,如果您关心性能,您可以使用查找表 (LUT) 来计算位数:

为 8 位整数构建 LUT(仅 256 个条目):

function lut = countBitsLUT()
for ii = 0:255
    lut(ii+1) = sum(bitget(uint8(ii),1:8));
end

您只需构建 LUT 一次。

获得 LUT 后,您可以使用以下方法计算位数:

count = lut( bitand(n,255)+1 ) + ...      % how many set bits in first byte
        lut( bitand( bitshift(n,-8), 255 ) + 1 ) + ... % how many set bits in second byte
        lut( bitand( bitshift(n,-16), 255 ) + 1 ) + ... % how many set bits in third byte
        lut( bitand( bitshift(n,-24), 255 ) + 1 ); % how many set bits in fourth byte

我还做了一个小“基准”:

lutCount = @( n ) lut( bitand(n,255)+1 ) + ...      % how many set bits in first byte
        lut( bitand( bitshift(n,-8), 255 ) + 1 ) + ... % how many set bits in second byte
        lut( bitand( bitshift(n,-16), 255 ) + 1 ) + ... % how many set bits in third byte
        lut( bitand( bitshift(n,-24), 255 ) + 1 ); % how many set bits in fourth byte

t = [ 0 0 ];
for ii=1:1000
    n = uint32( randi( intmax('uint32'), 1, 1 ) );
    tic;
    c1 = sum(bitget(n,1:32));
    t(1) = t(1) + toc;
    tic;
    c2 = lutCount( n );
    t(2) = t(2) + toc;
    assert( c1 == c2 );
end

运行时间是:

t = [0.0115    0.0053]

sum也就是说,LUT 的速度是的两倍bitget

于 2013-01-09T07:14:50.697 回答