zip.cpp

这是CODEPROJECT上的zip压缩解压代码

 * Allocate the match buffer, initialize the various tables and save the
 * location of the internal file attribute (ascii/binary) and method
 * (DEFLATE/STORE).
 */
void ct_init(TState &state, ush *attr)
{
    int n;        /* iterates over tree elements */
    int bits;     /* bit counter */
    int length;   /* length value */
    int code;     /* code value */
    int dist;     /* distance index */

    state.ts.file_type = attr;
    //state.ts.file_method = method;
    state.ts.cmpr_bytelen = state.ts.cmpr_len_bits = 0L;
    state.ts.input_len = 0L;

    if (state.ts.static_dtree[0].dl.len != 0) return; /* ct_init already called */

    /* Initialize the mapping length (0..255) -> length code (0..28) */
    length = 0;
    for (code = 0; code < LENGTH_CODES-1; code++) {
        state.ts.base_length[code] = length;
        for (n = 0; n < (1<<extra_lbits[code]); n++) {
            state.ts.length_code[length++] = (uch)code;
        }
    }
    Assert(state,length == 256, "ct_init: length != 256");
    /* Note that the length 255 (match length 258) can be represented
     * in two different ways: code 284 + 5 bits or code 285, so we
     * overwrite length_code[255] to use the best encoding:
     */
    state.ts.length_code[length-1] = (uch)code;

    /* Initialize the mapping dist (0..32K) -> dist code (0..29) */
    dist = 0;
    for (code = 0 ; code < 16; code++) {
        state.ts.base_dist[code] = dist;
        for (n = 0; n < (1<<extra_dbits[code]); n++) {
            state.ts.dist_code[dist++] = (uch)code;
        }
    }
    Assert(state,dist == 256, "ct_init: dist != 256");
    dist >>= 7; /* from now on, all distances are divided by 128 */
    for ( ; code < D_CODES; code++) {
        state.ts.base_dist[code] = dist << 7;
        for (n = 0; n < (1<<(extra_dbits[code]-7)); n++) {
            state.ts.dist_code[256 + dist++] = (uch)code;
        }
    }
    Assert(state,dist == 256, "ct_init: 256+dist != 512");

    /* Construct the codes of the static literal tree */
    for (bits = 0; bits <= MAX_BITS; bits++) state.ts.bl_count[bits] = 0;
    n = 0;
    while (n <= 143) state.ts.static_ltree[n++].dl.len = 8, state.ts.bl_count[8]++;
    while (n <= 255) state.ts.static_ltree[n++].dl.len = 9, state.ts.bl_count[9]++;
    while (n <= 279) state.ts.static_ltree[n++].dl.len = 7, state.ts.bl_count[7]++;
    while (n <= 287) state.ts.static_ltree[n++].dl.len = 8, state.ts.bl_count[8]++;
    /* fc.codes 286 and 287 do not exist, but we must include them in the
     * tree construction to get a canonical Huffman tree (longest code
     * all ones)
     */
    gen_codes(state,(ct_data *)state.ts.static_ltree, L_CODES+1);

    /* The static distance tree is trivial: */
    for (n = 0; n < D_CODES; n++) {
        state.ts.static_dtree[n].dl.len = 5;
        state.ts.static_dtree[n].fc.code = (ush)bi_reverse(n, 5);
    }

    /* Initialize the first block of the first file: */
    init_block(state);
}

/* ===========================================================================
 * Initialize a new block.
 */
void init_block(TState &state)
{
    int n; /* iterates over tree elements */

    /* Initialize the trees. */
    for (n = 0; n < L_CODES;  n++) state.ts.dyn_ltree[n].fc.freq = 0;
    for (n = 0; n < D_CODES;  n++) state.ts.dyn_dtree[n].fc.freq = 0;
    for (n = 0; n < BL_CODES; n++) state.ts.bl_tree[n].fc.freq = 0;

    state.ts.dyn_ltree[END_BLOCK].fc.freq = 1;
    state.ts.opt_len = state.ts.static_len = 0L;
    state.ts.last_lit = state.ts.last_dist = state.ts.last_flags = 0;
    state.ts.flags = 0; state.ts.flag_bit = 1;
}

#define SMALLEST 1
/* Index within the heap array of least frequent node in the Huffman tree */


/* ===========================================================================
 * Remove the smallest element from the heap and recreate the heap with
 * one less element. Updates heap and heap_len.
 */
#define pqremove(tree, top) \
{\
    top = state.ts.heap[SMALLEST]; \
    state.ts.heap[SMALLEST] = state.ts.heap[state.ts.heap_len--]; \
    pqdownheap(state,tree, SMALLEST); \
}

/* ===========================================================================
 * Compares to subtrees, using the tree depth as tie breaker when
 * the subtrees have equal frequency. This minimizes the worst case length.
 */
#define smaller(tree, n, m) \
   (tree[n].fc.freq < tree[m].fc.freq || \
   (tree[n].fc.freq == tree[m].fc.freq && state.ts.depth[n] <= state.ts.depth[m]))

/* ===========================================================================
 * Restore the heap property by moving down the tree starting at node k,
 * exchanging a node with the smallest of its two sons if necessary, stopping
 * when the heap property is re-established (each father smaller than its
 * two sons).
 */
void pqdownheap(TState &state,ct_data *tree, int k)
{
    int v = state.ts.heap[k];
    int j = k << 1;  /* left son of k */
    int htemp;       /* required because of bug in SASC compiler */

    while (j <= state.ts.heap_len) {
        /* Set j to the smallest of the two sons: */
        if (j < state.ts.heap_len && smaller(tree, state.ts.heap[j+1], state.ts.heap[j])) j++;

        /* Exit if v is smaller than both sons */
        htemp = state.ts.heap[j];
        if (smaller(tree, v, htemp)) break;

        /* Exchange v with the smallest son */
        state.ts.heap[k] = htemp;
        k = j;

        /* And continue down the tree, setting j to the left son of k */
        j <<= 1;
    }
    state.ts.heap[k] = v;
}

/* ===========================================================================
 * Compute the optimal bit lengths for a tree and update the total bit length
 * for the current block.
 * IN assertion: the fields freq and dad are set, heap[heap_max] and
 *    above are the tree nodes sorted by increasing frequency.
 * OUT assertions: the field len is set to the optimal bit length, the
 *     array bl_count contains the frequencies for each bit length.
 *     The length opt_len is updated; static_len is also updated if stree is
 *     not null.
 */
void gen_bitlen(TState &state,tree_desc *desc)
{
    ct_data *tree  = desc->dyn_tree;
    const int *extra     = desc->extra_bits;
    int base            = desc->extra_base;
    int max_code        = desc->max_code;
    int max_length      = desc->max_length;
    ct_data *stree = desc->static_tree;
    int h;              /* heap index */
    int n, m;           /* iterate over the tree elements */
    int bits;           /* bit length */
    int xbits;          /* extra bits */
    ush f;              /* frequency */
    int overflow = 0;   /* number of elements with bit length too large */

    for (bits = 0; bits <= MAX_BITS; bits++) state.ts.bl_count[bits] = 0;

    /* In a first pass, compute the optimal bit lengths (which may
     * overflow in the case of the bit length tree).
     */
    tree[state.ts.heap[state.ts.heap_max]].dl.len = 0; /* root of the heap */

    for (h = state.ts.heap_max+1; h < HEAP_SIZE; h++) {
        n = state.ts.heap[h];
        bits = tree[tree[n].dl.dad].dl.len + 1;
        if (bits > max_length) bits = max_length, overflow++;
        tree[n].dl.len = (ush)bits;
        /* We overwrite tree[n].dl.dad which is no longer needed */

        if (n > max_code) continue; /* not a leaf node */

        state.ts.bl_count[bits]++;
        xbits = 0;
        if (n >= base) xbits = extra[n-base];
        f = tree[n].fc.freq;
        state.ts.opt_len += (ulg)f * (bits + xbits);
        if (stree) state.ts.static_len += (ulg)f * (stree[n].dl.len + xbits);
    }
    if (overflow == 0) return;

    Trace("\nbit length overflow\n");
    /* This happens for example on obj2 and pic of the Calgary corpus */

    /* Find the first bit length which could increase: */
    do {
        bits = max_length-1;
        while (state.ts.bl_count[bits] == 0) bits--;
        state.ts.bl_count[bits]--;           /* move one leaf down the tree */
        state.ts.bl_count[bits+1] += (ush)2; /* move one overflow item as its brother */
        state.ts.bl_count[max_length]--;
        /* The brother of the overflow item also moves one step up,
         * but this does not affect bl_count[max_length]
         */
        overflow -= 2;
    } while (overflow > 0);

    /* Now recompute all bit lengths, scanning in increasing frequency.
     * h is still equal to HEAP_SIZE. (It is simpler to reconstruct all
     * lengths instead of fixing only the wrong ones. This idea is taken
     * from 'ar' written by Haruhiko Okumura.)
     */
    for (bits = max_length; bits != 0; bits--) {
        n = state.ts.bl_count[bits];
        while (n != 0) {
            m = state.ts.heap[--h];
            if (m > max_code) continue;
            if (tree[m].dl.len != (ush)bits) {
                Trace("code %d bits %d->%d\n", m, tree[m].dl.len, bits);
                state.ts.opt_len += ((long)bits-(long)tree[m].dl.len)*(long)tree[m].fc.freq;
                tree[m].dl.len = (ush)bits;
            }
            n--;
        }
    }
}

/* ===========================================================================
 * Generate the codes for a given tree and bit counts (which need not be
 * optimal).
 * IN assertion: the array bl_count contains the bit length statistics for
 * the given tree and the field len is set for all tree elements.
 * OUT assertion: the field code is set for all tree elements of non
 *     zero code length.
 */
void gen_codes (TState &state, ct_data *tree, int max_code)
{
    ush next_code[MAX_BITS+1]; /* next code value for each bit length */
    ush code = 0;              /* running code value */
    int bits;                  /* bit index */
    int n;                     /* code index */

    /* The distribution counts are first used to generate the code values
     * without bit reversal.
     */
    for (bits = 1; bits <= MAX_BITS; bits++) {
        next_code[bits] = code = (ush)((code + state.ts.bl_count[bits-1]) << 1);
    }
    /* Check that the bit counts in bl_count are consistent. The last code
     * must be all ones.
     */
    Assert(state,code + state.ts.bl_count[MAX_BITS]-1 == (1<< ((ush) MAX_BITS)) - 1,
            "inconsistent bit counts");
    Trace("\ngen_codes: max_code %d ", max_code);

    for (n = 0;  n <= max_code; n++) {
        int len = tree[n].dl.len;
        if (len == 0) continue;
        /* Now reverse the bits */
        tree[n].fc.code = (ush)bi_reverse(next_code[len]++, len);

        //Tracec(tree != state.ts.static_ltree, "\nn %3d %c l %2d c %4x (%x) ", n, (isgraph(n) ? n : ' '), len, tree[n].fc.code, next_code[len]-1);
    }
}

/* ===========================================================================
 * Construct one Huffman tree and assigns the code bit strings and lengths.
 * Update the total bit length for the current block.
 * IN assertion: the field freq is set for all tree elements.
 * OUT assertions: the fields len and code are set to the optimal bit length
 *     and corresponding code. The length opt_len is updated; static_len is
 *     also updated if stree is not null. The field max_code is set.
 */
void build_tree(TState &state,tree_desc *desc)
{
    ct_data *tree   = desc->dyn_tree;
    ct_data *stree  = desc->static_tree;
    int elems            = desc->elems;
    int n, m;          /* iterate over heap elements */
    int max_code = -1; /* largest code with non zero frequency */
    int node = elems;  /* next internal node of the tree */

    /* Construct the initial heap, with least frequent element in
     * heap[SMALLEST]. The sons of heap[n] are heap[2*n] and heap[2*n+1].
     * heap[0] is not used.
     */
    state.ts.heap_len = 0, state.ts.heap_max = HEAP_SIZE;

    for (n = 0; n < elems; n++) {
        if (tree[n].fc.freq != 0) {
            state.ts.heap[++state.ts.heap_len] = max_code = n;
            state.ts.depth[n] = 0;
        } else {
            tree[n].dl.len = 0;
        }
    }

    /* The pkzip format requires that at least one distance code exists,
     * and that at least one bit should be sent even if there is only one
     * possible code. So to avoid special checks later on we force at least
     * two codes of non zero frequency.
     */
    while (state.ts.heap_len < 2) {
        int newcp = state.ts.heap[++state.ts.heap_len] = (max_code < 2 ? ++max_code : 0);
        tree[newcp].fc.freq = 1;
        state.ts.depth[newcp] = 0;
        state.ts.opt_len--; if (stree) state.ts.static_len -= stree[newcp].dl.len;
        /* new is 0 or 1 so it does not have extra bits */
    }
    desc->max_code = max_code;

    /* The elements heap[heap_len/2+1 .. heap_len] are leaves of the tree,
     * establish sub-heaps of increasing lengths:
     */
    for (n = state.ts.heap_len/2; n >= 1; n--) pqdownheap(state,tree, n);

    /* Construct the Huffman tree by repeatedly combining the least two
     * frequent nodes.