Tại sao không xem xét cách thức GZIP source thực hiện điều đó, cụ thể là mã giải nén Huffman trong unpack.c cụ thể? Nó làm chính xác những gì bạn đang có, ngoại trừ nó làm nó nhiều, nhanh hơn nhiều.
Từ những gì tôi có thể biết, nó sử dụng mảng tra cứu và hoạt động thay đổi/mặt nạ hoạt động trên toàn bộ từ để chạy nhanh hơn. Mặc dù mã khá dày đặc.
EDIT: đây là nguồn gốc hoàn toàn
/* unpack.c -- decompress files in pack format.
* Copyright (C) 1992-1993 Jean-loup Gailly
* This is free software; you can redistribute it and/or modify it under the
* terms of the GNU General Public License, see the file COPYING.
*/
#ifdef RCSID
static char rcsid[] = "$Id: unpack.c,v 1.4 1993/06/11 19:25:36 jloup Exp $";
#endif
#include "tailor.h"
#include "gzip.h"
#include "crypt.h"
#define MIN(a,b) ((a) <= (b) ? (a) : (b))
/* The arguments must not have side effects. */
#define MAX_BITLEN 25
/* Maximum length of Huffman codes. (Minor modifications to the code
* would be needed to support 32 bits codes, but pack never generates
* more than 24 bits anyway.)
*/
#define LITERALS 256
/* Number of literals, excluding the End of Block (EOB) code */
#define MAX_PEEK 12
/* Maximum number of 'peek' bits used to optimize traversal of the
* Huffman tree.
*/
local ulg orig_len; /* original uncompressed length */
local int max_len; /* maximum bit length of Huffman codes */
local uch literal[LITERALS];
/* The literal bytes present in the Huffman tree. The EOB code is not
* represented.
*/
local int lit_base[MAX_BITLEN+1];
/* All literals of a given bit length are contiguous in literal[] and
* have contiguous codes. literal[code+lit_base[len]] is the literal
* for a code of len bits.
*/
local int leaves [MAX_BITLEN+1]; /* Number of leaves for each bit length */
local int parents[MAX_BITLEN+1]; /* Number of parents for each bit length */
local int peek_bits; /* Number of peek bits currently used */
/* local uch prefix_len[1 << MAX_PEEK]; */
#define prefix_len outbuf
/* For each bit pattern b of peek_bits bits, prefix_len[b] is the length
* of the Huffman code starting with a prefix of b (upper bits), or 0
* if all codes of prefix b have more than peek_bits bits. It is not
* necessary to have a huge table (large MAX_PEEK) because most of the
* codes encountered in the input stream are short codes (by construction).
* So for most codes a single lookup will be necessary.
*/
#if (1<<MAX_PEEK) > OUTBUFSIZ
error cannot overlay prefix_len and outbuf
#endif
local ulg bitbuf;
/* Bits are added on the low part of bitbuf and read from the high part. */
local int valid; /* number of valid bits in bitbuf */
/* all bits above the last valid bit are always zero */
/* Set code to the next 'bits' input bits without skipping them. code
* must be the name of a simple variable and bits must not have side effects.
* IN assertions: bits <= 25 (so that we still have room for an extra byte
* when valid is only 24), and mask = (1<<bits)-1.
*/
#define look_bits(code,bits,mask) \
{ \
while (valid < (bits)) bitbuf = (bitbuf<<8) | (ulg)get_byte(), valid += 8; \
code = (bitbuf >> (valid-(bits))) & (mask); \
}
/* Skip the given number of bits (after having peeked at them): */
#define skip_bits(bits) (valid -= (bits))
#define clear_bitbuf() (valid = 0, bitbuf = 0)
/* Local functions */
local void read_tree OF((void));
local void build_tree OF((void));
/* ===========================================================================
* Read the Huffman tree.
*/
local void read_tree()
{
int len; /* bit length */
int base; /* base offset for a sequence of leaves */
int n;
/* Read the original input size, MSB first */
orig_len = 0;
for (n = 1; n <= 4; n++) orig_len = (orig_len << 8) | (ulg)get_byte();
max_len = (int)get_byte(); /* maximum bit length of Huffman codes */
if (max_len > MAX_BITLEN) {
error("invalid compressed data -- Huffman code > 32 bits");
}
/* Get the number of leaves at each bit length */
n = 0;
for (len = 1; len <= max_len; len++) {
leaves[len] = (int)get_byte();
n += leaves[len];
}
if (n > LITERALS) {
error("too many leaves in Huffman tree");
}
Trace((stderr, "orig_len %ld, max_len %d, leaves %d\n",
orig_len, max_len, n));
/* There are at least 2 and at most 256 leaves of length max_len.
* (Pack arbitrarily rejects empty files and files consisting of
* a single byte even repeated.) To fit the last leaf count in a
* byte, it is offset by 2. However, the last literal is the EOB
* code, and is not transmitted explicitly in the tree, so we must
* adjust here by one only.
*/
leaves[max_len]++;
/* Now read the leaves themselves */
base = 0;
for (len = 1; len <= max_len; len++) {
/* Remember where the literals of this length start in literal[] : */
lit_base[len] = base;
/* And read the literals: */
for (n = leaves[len]; n > 0; n--) {
literal[base++] = (uch)get_byte();
}
}
leaves[max_len]++; /* Now include the EOB code in the Huffman tree */
}
/* ===========================================================================
* Build the Huffman tree and the prefix table.
*/
local void build_tree()
{
int nodes = 0; /* number of nodes (parents+leaves) at current bit length */
int len; /* current bit length */
uch *prefixp; /* pointer in prefix_len */
for (len = max_len; len >= 1; len--) {
/* The number of parent nodes at this level is half the total
* number of nodes at parent level:
*/
nodes >>= 1;
parents[len] = nodes;
/* Update lit_base by the appropriate bias to skip the parent nodes
* (which are not represented in the literal array):
*/
lit_base[len] -= nodes;
/* Restore nodes to be parents+leaves: */
nodes += leaves[len];
}
/* Construct the prefix table, from shortest leaves to longest ones.
* The shortest code is all ones, so we start at the end of the table.
*/
peek_bits = MIN(max_len, MAX_PEEK);
prefixp = &prefix_len[1<<peek_bits];
for (len = 1; len <= peek_bits; len++) {
int prefixes = leaves[len] << (peek_bits-len); /* may be 0 */
while (prefixes--) *--prefixp = (uch)len;
}
/* The length of all other codes is unknown: */
while (prefixp > prefix_len) *--prefixp = 0;
}
/* ===========================================================================
* Unpack in to out. This routine does not support the old pack format
* with magic header \037\037.
*
* IN assertions: the buffer inbuf contains already the beginning of
* the compressed data, from offsets inptr to insize-1 included.
* The magic header has already been checked. The output buffer is cleared.
*/
int unpack(in, out)
int in, out; /* input and output file descriptors */
{
int len; /* Bit length of current code */
unsigned eob; /* End Of Block code */
register unsigned peek; /* lookahead bits */
unsigned peek_mask; /* Mask for peek_bits bits */
ifd = in;
ofd = out;
read_tree(); /* Read the Huffman tree */
build_tree(); /* Build the prefix table */
clear_bitbuf(); /* Initialize bit input */
peek_mask = (1<<peek_bits)-1;
/* The eob code is the largest code among all leaves of maximal length: */
eob = leaves[max_len]-1;
Trace((stderr, "eob %d %x\n", max_len, eob));
/* Decode the input data: */
for (;;) {
/* Since eob is the longest code and not shorter than max_len,
* we can peek at max_len bits without having the risk of reading
* beyond the end of file.
*/
look_bits(peek, peek_bits, peek_mask);
len = prefix_len[peek];
if (len > 0) {
peek >>= peek_bits - len; /* discard the extra bits */
} else {
/* Code of more than peek_bits bits, we must traverse the tree */
ulg mask = peek_mask;
len = peek_bits;
do {
len++, mask = (mask<<1)+1;
look_bits(peek, len, mask);
} while (peek < (unsigned)parents[len]);
/* loop as long as peek is a parent node */
}
/* At this point, peek is the next complete code, of len bits */
if (peek == eob && len == max_len) break; /* end of file? */
put_ubyte(literal[peek+lit_base[len]]);
Tracev((stderr,"%02d %04x %c\n", len, peek,
literal[peek+lit_base[len]]));
skip_bits(len);
} /* for (;;) */
flush_window();
Trace((stderr, "bytes_out %ld\n", bytes_out));
if (orig_len != (ulg)bytes_out) {
error("invalid compressed data--length error");
}
return OK;
}
Xin chúc mừng cho mã được định dạng tốt của bạn. Làm cho mã WinAPI gần như ngon miệng. – Manuel
FYI: bộ giải mã hiện đại có thể làm tốc độ giải nén ~ 1GB/giây (hoặc hơn) trên máy tính trung bình hiện tại: bạn phải giới hạn độ dài của mã huffman đến 11/12, sử dụng bảng giải nén và sử dụng nhiều luồng huffman, vì vậy phụ thuộc dữ liệu được giảm thiểu. – geza