aureooms/js-bit

/**
 * Determine if a word has a byte between m and n
 * When m < n, this technique
 * tests if a word x contains an unsigned byte value, such that m < value < n. It
 * uses 7 arithmetic/logical operations when n and m are constant.
 *
 * Note: Bytes that equal n can be reported by likelyhasbetween as false
 * positives, so this should be checked by character if a certain result is
 * needed.
 *
 * Requirements: x>=0; 0<=m<=127; 0<=n<=128
 *
 * #define likelyhasbetween(x,m,n) \
 * ((((x)-~0UL/255*(n))&~(x)&((x)&~0UL/255*127)+~0UL/255*(127-(m)))&~0UL/255*128)
 *
 * This technique would be suitable for a fast pretest. A variation that takes
 * one more operation (8 total for constant m and n) but provides the exact
 * answer is:
 *
 * #define hasbetween(x,m,n) \
 * ((~0UL/255*(127+(n))-((x)&~0UL/255*127)&~(x)&((x)&~0UL/255*127)+~0UL/255*(127-(m)))&~0UL/255*128)
 *
 * To count the number of bytes in x that are between m and n (exclusive) in 10
 * operations, use:
 *
 * #define countbetween(x,m,n) (hasbetween(x,m,n)/128%255)
 *
 * Juha Järvi suggested likelyhasbetween on April 6, 2005. From there, Sean
 * Anderson created hasbetween and countbetween on April 10, 2005.
 */

const likelyhasbetween = ( x , m , n ) => ( ( ( (x) - ( (~0 >>> 0 ) / 255 )*(n))&~(x)&((x)&( (~0 >>> 0 ) / 255 )*127)+( (~0 >>> 0 ) / 255 )*(127-(m)))&( (~0 >>> 0 ) / 255 )*128) ;

const hasbetween = ( x , m , n ) =>
((( (~0 >>> 0 ) / 255 )*(127+(n))-((x)&( (~0 >>> 0 ) / 255 )*127)&~(x)&((x)&( (~0 >>> 0 ) / 255 )*127)+( (~0 >>> 0 ) / 255 )*(127-(m)))&( (~0 >>> 0 ) / 255 )*128) ;

const countbetween = ( x , m , n ) => ( hasbetween( x , m , n ) >>> 7 ) % 255 ;

exports.likelyhasbetween = likelyhasbetween ;
exports.hasbetween = hasbetween ;
exports.countbetween = countbetween ;