ctz_compat.go

//go:build !go1.9
// +build !go1.9

package roaring

// Reuse of portions of go/src/math/big standard lib code
// under this license:
/*
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const deBruijn32 = 0x077CB531

var deBruijn32Lookup = []byte{
	0, 1, 28, 2, 29, 14, 24, 3, 30, 22, 20, 15, 25, 17, 4, 8,
	31, 27, 13, 23, 21, 19, 16, 7, 26, 12, 18, 6, 11, 5, 10, 9,
}

const deBruijn64 = 0x03f79d71b4ca8b09

var deBruijn64Lookup = []byte{
	0, 1, 56, 2, 57, 49, 28, 3, 61, 58, 42, 50, 38, 29, 17, 4,
	62, 47, 59, 36, 45, 43, 51, 22, 53, 39, 33, 30, 24, 18, 12, 5,
	63, 55, 48, 27, 60, 41, 37, 16, 46, 35, 44, 21, 52, 32, 23, 11,
	54, 26, 40, 15, 34, 20, 31, 10, 25, 14, 19, 9, 13, 8, 7, 6,
}

// trailingZeroBits returns the number of consecutive least significant zero
// bits of x.
func countTrailingZeros(x uint64) int {
	// x & -x leaves only the right-most bit set in the word. Let k be the
	// index of that bit. Since only a single bit is set, the value is two
	// to the power of k. Multiplying by a power of two is equivalent to
	// left shifting, in this case by k bits. The de Bruijn constant is
	// such that all six bit, consecutive substrings are distinct.
	// Therefore, if we have a left shifted version of this constant we can
	// find by how many bits it was shifted by looking at which six bit
	// substring ended up at the top of the word.
	// (Knuth, volume 4, section 7.3.1)
	if x == 0 {
		// We have to special case 0; the fomula
		// below doesn't work for 0.
		return 64
	}
	return int(deBruijn64Lookup[((x&-x)*(deBruijn64))>>58])
}