mirror of
https://github.com/mudler/luet.git
synced 2025-06-27 07:50:18 +00:00
Instead of using gox on one side and an action to release, we can merge them together with goreleaser which will build for extra targets (arm, mips if needed in the future) and it also takes care of creating checksums, a source archive, and a changelog and creating a release with all the artifacts. All binaries should respect the old naming convention, so any scripts out there should still work. Signed-off-by: Itxaka <igarcia@suse.com>
377 lines
10 KiB
Go
377 lines
10 KiB
Go
// Copyright 2009 The Go Authors. All rights reserved.
|
|
// Use of this source code is governed by a BSD-style
|
|
// license that can be found in the LICENSE file.
|
|
|
|
package flate
|
|
|
|
import (
|
|
"math"
|
|
"math/bits"
|
|
)
|
|
|
|
const (
|
|
maxBitsLimit = 16
|
|
// number of valid literals
|
|
literalCount = 286
|
|
)
|
|
|
|
// hcode is a huffman code with a bit code and bit length.
|
|
type hcode struct {
|
|
code, len uint16
|
|
}
|
|
|
|
type huffmanEncoder struct {
|
|
codes []hcode
|
|
freqcache []literalNode
|
|
bitCount [17]int32
|
|
}
|
|
|
|
type literalNode struct {
|
|
literal uint16
|
|
freq uint16
|
|
}
|
|
|
|
// A levelInfo describes the state of the constructed tree for a given depth.
|
|
type levelInfo struct {
|
|
// Our level. for better printing
|
|
level int32
|
|
|
|
// The frequency of the last node at this level
|
|
lastFreq int32
|
|
|
|
// The frequency of the next character to add to this level
|
|
nextCharFreq int32
|
|
|
|
// The frequency of the next pair (from level below) to add to this level.
|
|
// Only valid if the "needed" value of the next lower level is 0.
|
|
nextPairFreq int32
|
|
|
|
// The number of chains remaining to generate for this level before moving
|
|
// up to the next level
|
|
needed int32
|
|
}
|
|
|
|
// set sets the code and length of an hcode.
|
|
func (h *hcode) set(code uint16, length uint16) {
|
|
h.len = length
|
|
h.code = code
|
|
}
|
|
|
|
func reverseBits(number uint16, bitLength byte) uint16 {
|
|
return bits.Reverse16(number << ((16 - bitLength) & 15))
|
|
}
|
|
|
|
func maxNode() literalNode { return literalNode{math.MaxUint16, math.MaxUint16} }
|
|
|
|
func newHuffmanEncoder(size int) *huffmanEncoder {
|
|
// Make capacity to next power of two.
|
|
c := uint(bits.Len32(uint32(size - 1)))
|
|
return &huffmanEncoder{codes: make([]hcode, size, 1<<c)}
|
|
}
|
|
|
|
// Generates a HuffmanCode corresponding to the fixed literal table
|
|
func generateFixedLiteralEncoding() *huffmanEncoder {
|
|
h := newHuffmanEncoder(literalCount)
|
|
codes := h.codes
|
|
var ch uint16
|
|
for ch = 0; ch < literalCount; ch++ {
|
|
var bits uint16
|
|
var size uint16
|
|
switch {
|
|
case ch < 144:
|
|
// size 8, 000110000 .. 10111111
|
|
bits = ch + 48
|
|
size = 8
|
|
case ch < 256:
|
|
// size 9, 110010000 .. 111111111
|
|
bits = ch + 400 - 144
|
|
size = 9
|
|
case ch < 280:
|
|
// size 7, 0000000 .. 0010111
|
|
bits = ch - 256
|
|
size = 7
|
|
default:
|
|
// size 8, 11000000 .. 11000111
|
|
bits = ch + 192 - 280
|
|
size = 8
|
|
}
|
|
codes[ch] = hcode{code: reverseBits(bits, byte(size)), len: size}
|
|
}
|
|
return h
|
|
}
|
|
|
|
func generateFixedOffsetEncoding() *huffmanEncoder {
|
|
h := newHuffmanEncoder(30)
|
|
codes := h.codes
|
|
for ch := range codes {
|
|
codes[ch] = hcode{code: reverseBits(uint16(ch), 5), len: 5}
|
|
}
|
|
return h
|
|
}
|
|
|
|
var fixedLiteralEncoding = generateFixedLiteralEncoding()
|
|
var fixedOffsetEncoding = generateFixedOffsetEncoding()
|
|
|
|
func (h *huffmanEncoder) bitLength(freq []uint16) int {
|
|
var total int
|
|
for i, f := range freq {
|
|
if f != 0 {
|
|
total += int(f) * int(h.codes[i].len)
|
|
}
|
|
}
|
|
return total
|
|
}
|
|
|
|
func (h *huffmanEncoder) bitLengthRaw(b []byte) int {
|
|
var total int
|
|
for _, f := range b {
|
|
if f != 0 {
|
|
total += int(h.codes[f].len)
|
|
}
|
|
}
|
|
return total
|
|
}
|
|
|
|
// Return the number of literals assigned to each bit size in the Huffman encoding
|
|
//
|
|
// This method is only called when list.length >= 3
|
|
// The cases of 0, 1, and 2 literals are handled by special case code.
|
|
//
|
|
// list An array of the literals with non-zero frequencies
|
|
// and their associated frequencies. The array is in order of increasing
|
|
// frequency, and has as its last element a special element with frequency
|
|
// MaxInt32
|
|
// maxBits The maximum number of bits that should be used to encode any literal.
|
|
// Must be less than 16.
|
|
// return An integer array in which array[i] indicates the number of literals
|
|
// that should be encoded in i bits.
|
|
func (h *huffmanEncoder) bitCounts(list []literalNode, maxBits int32) []int32 {
|
|
if maxBits >= maxBitsLimit {
|
|
panic("flate: maxBits too large")
|
|
}
|
|
n := int32(len(list))
|
|
list = list[0 : n+1]
|
|
list[n] = maxNode()
|
|
|
|
// The tree can't have greater depth than n - 1, no matter what. This
|
|
// saves a little bit of work in some small cases
|
|
if maxBits > n-1 {
|
|
maxBits = n - 1
|
|
}
|
|
|
|
// Create information about each of the levels.
|
|
// A bogus "Level 0" whose sole purpose is so that
|
|
// level1.prev.needed==0. This makes level1.nextPairFreq
|
|
// be a legitimate value that never gets chosen.
|
|
var levels [maxBitsLimit]levelInfo
|
|
// leafCounts[i] counts the number of literals at the left
|
|
// of ancestors of the rightmost node at level i.
|
|
// leafCounts[i][j] is the number of literals at the left
|
|
// of the level j ancestor.
|
|
var leafCounts [maxBitsLimit][maxBitsLimit]int32
|
|
|
|
for level := int32(1); level <= maxBits; level++ {
|
|
// For every level, the first two items are the first two characters.
|
|
// We initialize the levels as if we had already figured this out.
|
|
levels[level] = levelInfo{
|
|
level: level,
|
|
lastFreq: int32(list[1].freq),
|
|
nextCharFreq: int32(list[2].freq),
|
|
nextPairFreq: int32(list[0].freq) + int32(list[1].freq),
|
|
}
|
|
leafCounts[level][level] = 2
|
|
if level == 1 {
|
|
levels[level].nextPairFreq = math.MaxInt32
|
|
}
|
|
}
|
|
|
|
// We need a total of 2*n - 2 items at top level and have already generated 2.
|
|
levels[maxBits].needed = 2*n - 4
|
|
|
|
level := maxBits
|
|
for {
|
|
l := &levels[level]
|
|
if l.nextPairFreq == math.MaxInt32 && l.nextCharFreq == math.MaxInt32 {
|
|
// We've run out of both leafs and pairs.
|
|
// End all calculations for this level.
|
|
// To make sure we never come back to this level or any lower level,
|
|
// set nextPairFreq impossibly large.
|
|
l.needed = 0
|
|
levels[level+1].nextPairFreq = math.MaxInt32
|
|
level++
|
|
continue
|
|
}
|
|
|
|
prevFreq := l.lastFreq
|
|
if l.nextCharFreq < l.nextPairFreq {
|
|
// The next item on this row is a leaf node.
|
|
n := leafCounts[level][level] + 1
|
|
l.lastFreq = l.nextCharFreq
|
|
// Lower leafCounts are the same of the previous node.
|
|
leafCounts[level][level] = n
|
|
e := list[n]
|
|
if e.literal < math.MaxUint16 {
|
|
l.nextCharFreq = int32(e.freq)
|
|
} else {
|
|
l.nextCharFreq = math.MaxInt32
|
|
}
|
|
} else {
|
|
// The next item on this row is a pair from the previous row.
|
|
// nextPairFreq isn't valid until we generate two
|
|
// more values in the level below
|
|
l.lastFreq = l.nextPairFreq
|
|
// Take leaf counts from the lower level, except counts[level] remains the same.
|
|
copy(leafCounts[level][:level], leafCounts[level-1][:level])
|
|
levels[l.level-1].needed = 2
|
|
}
|
|
|
|
if l.needed--; l.needed == 0 {
|
|
// We've done everything we need to do for this level.
|
|
// Continue calculating one level up. Fill in nextPairFreq
|
|
// of that level with the sum of the two nodes we've just calculated on
|
|
// this level.
|
|
if l.level == maxBits {
|
|
// All done!
|
|
break
|
|
}
|
|
levels[l.level+1].nextPairFreq = prevFreq + l.lastFreq
|
|
level++
|
|
} else {
|
|
// If we stole from below, move down temporarily to replenish it.
|
|
for levels[level-1].needed > 0 {
|
|
level--
|
|
}
|
|
}
|
|
}
|
|
|
|
// Somethings is wrong if at the end, the top level is null or hasn't used
|
|
// all of the leaves.
|
|
if leafCounts[maxBits][maxBits] != n {
|
|
panic("leafCounts[maxBits][maxBits] != n")
|
|
}
|
|
|
|
bitCount := h.bitCount[:maxBits+1]
|
|
bits := 1
|
|
counts := &leafCounts[maxBits]
|
|
for level := maxBits; level > 0; level-- {
|
|
// chain.leafCount gives the number of literals requiring at least "bits"
|
|
// bits to encode.
|
|
bitCount[bits] = counts[level] - counts[level-1]
|
|
bits++
|
|
}
|
|
return bitCount
|
|
}
|
|
|
|
// Look at the leaves and assign them a bit count and an encoding as specified
|
|
// in RFC 1951 3.2.2
|
|
func (h *huffmanEncoder) assignEncodingAndSize(bitCount []int32, list []literalNode) {
|
|
code := uint16(0)
|
|
for n, bits := range bitCount {
|
|
code <<= 1
|
|
if n == 0 || bits == 0 {
|
|
continue
|
|
}
|
|
// The literals list[len(list)-bits] .. list[len(list)-bits]
|
|
// are encoded using "bits" bits, and get the values
|
|
// code, code + 1, .... The code values are
|
|
// assigned in literal order (not frequency order).
|
|
chunk := list[len(list)-int(bits):]
|
|
|
|
sortByLiteral(chunk)
|
|
for _, node := range chunk {
|
|
h.codes[node.literal] = hcode{code: reverseBits(code, uint8(n)), len: uint16(n)}
|
|
code++
|
|
}
|
|
list = list[0 : len(list)-int(bits)]
|
|
}
|
|
}
|
|
|
|
// Update this Huffman Code object to be the minimum code for the specified frequency count.
|
|
//
|
|
// freq An array of frequencies, in which frequency[i] gives the frequency of literal i.
|
|
// maxBits The maximum number of bits to use for any literal.
|
|
func (h *huffmanEncoder) generate(freq []uint16, maxBits int32) {
|
|
if h.freqcache == nil {
|
|
// Allocate a reusable buffer with the longest possible frequency table.
|
|
// Possible lengths are codegenCodeCount, offsetCodeCount and literalCount.
|
|
// The largest of these is literalCount, so we allocate for that case.
|
|
h.freqcache = make([]literalNode, literalCount+1)
|
|
}
|
|
list := h.freqcache[:len(freq)+1]
|
|
// Number of non-zero literals
|
|
count := 0
|
|
// Set list to be the set of all non-zero literals and their frequencies
|
|
for i, f := range freq {
|
|
if f != 0 {
|
|
list[count] = literalNode{uint16(i), f}
|
|
count++
|
|
} else {
|
|
list[count] = literalNode{}
|
|
h.codes[i].len = 0
|
|
}
|
|
}
|
|
list[len(freq)] = literalNode{}
|
|
|
|
list = list[:count]
|
|
if count <= 2 {
|
|
// Handle the small cases here, because they are awkward for the general case code. With
|
|
// two or fewer literals, everything has bit length 1.
|
|
for i, node := range list {
|
|
// "list" is in order of increasing literal value.
|
|
h.codes[node.literal].set(uint16(i), 1)
|
|
}
|
|
return
|
|
}
|
|
sortByFreq(list)
|
|
|
|
// Get the number of literals for each bit count
|
|
bitCount := h.bitCounts(list, maxBits)
|
|
// And do the assignment
|
|
h.assignEncodingAndSize(bitCount, list)
|
|
}
|
|
|
|
func atLeastOne(v float32) float32 {
|
|
if v < 1 {
|
|
return 1
|
|
}
|
|
return v
|
|
}
|
|
|
|
// Unassigned values are assigned '1' in the histogram.
|
|
func fillHist(b []uint16) {
|
|
for i, v := range b {
|
|
if v == 0 {
|
|
b[i] = 1
|
|
}
|
|
}
|
|
}
|
|
|
|
// histogramSize accumulates a histogram of b in h.
|
|
// An estimated size in bits is returned.
|
|
// len(h) must be >= 256, and h's elements must be all zeroes.
|
|
func histogramSize(b []byte, h []uint16, fill bool) (bits int) {
|
|
h = h[:256]
|
|
for _, t := range b {
|
|
h[t]++
|
|
}
|
|
total := len(b)
|
|
if fill {
|
|
for _, v := range h {
|
|
if v == 0 {
|
|
total++
|
|
}
|
|
}
|
|
}
|
|
|
|
invTotal := 1.0 / float32(total)
|
|
shannon := float32(0.0)
|
|
for _, v := range h {
|
|
if v > 0 {
|
|
n := float32(v)
|
|
shannon += atLeastOne(-mFastLog2(n*invTotal)) * n
|
|
}
|
|
}
|
|
|
|
return int(shannon + 0.99)
|
|
}
|