Exemplo n.º 1
0
 public int deflateEnd(){
     if(dstate==null) return Z_STREAM_ERROR;
     int ret=dstate.deflateEnd();
     dstate=null;
     return ret;
 }
Exemplo n.º 2
0
        // 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.
        internal void gen_bitlen(Deflate s)
        {
            short[] tree = dyn_tree;
            short[] stree = stat_desc.static_tree;
            int[] extra = stat_desc.extra_bits;
            int based = stat_desc.extra_base;
            int max_length = stat_desc.max_length;
            int h;              // heap index
            int n, m;           // iterate over the tree elements
            int bits;           // bit length
            int xbits;          // extra bits
            short f;            // frequency
            int overflow = 0;   // number of elements with bit length too large

            for (bits = 0; bits <= MAX_BITS; bits++) s.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[s.heap[s.heap_max]*2+1] = 0; // root of the heap

            for(h=s.heap_max+1; h<HEAP_SIZE; h++){
                n = s.heap[h];
                bits = tree[tree[n*2+1]*2+1] + 1;
                if (bits > max_length){ bits = max_length; overflow++; }
                tree[n*2+1] = (short)bits;
                // We overwrite tree[n*2+1] which is no longer needed

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

                s.bl_count[bits]++;
                xbits = 0;
                if (n >= based) xbits = extra[n-based];
                f = tree[n*2];
                s.opt_len += f * (bits + xbits);
                if (stree!=null) s.static_len += f * (stree[n*2+1] + xbits);
            }
            if (overflow == 0) return;

            // 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(s.bl_count[bits]==0) bits--;
                s.bl_count[bits]--;      // move one leaf down the tree
                s.bl_count[bits+1]+=2;   // move one overflow item as its brother
                s.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);

            for (bits = max_length; bits != 0; bits--) {
                n = s.bl_count[bits];
                while (n != 0) {
                    m = s.heap[--h];
                    if (m > max_code) continue;
                    if (tree[m*2+1] != bits) {
                        s.opt_len += (int)(((long)bits - (long)tree[m*2+1])*(long)tree[m*2]);
                        tree[m*2+1] = (short)bits;
                    }
                    n--;
                }
            }
        }
Exemplo n.º 3
0
 public int deflateInit(int level, int bits, bool nowrap){
     dstate=new Deflate();
     return dstate.deflateInit(this, level, nowrap?-bits:bits);
 }
Exemplo n.º 4
0
        // 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.
        internal void build_tree(Deflate s)
        {
            short[] tree=dyn_tree;
            short[] stree=stat_desc.static_tree;
            int elems=stat_desc.elems;
            int n, m;          // iterate over heap elements
            int max_code=-1;   // largest code with non zero frequency
            int node;          // new node being created

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

            for(n=0; n<elems; n++) {
                if(tree[n*2] != 0) {
                    s.heap[++s.heap_len] = max_code = n;
                    s.depth[n] = 0;
                }
                else{
                    tree[n*2+1] = 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 (s.heap_len < 2) {
                node = s.heap[++s.heap_len] = (max_code < 2 ? ++max_code : 0);
                tree[node*2] = 1;
                s.depth[node] = 0;
                s.opt_len--; if (stree!=null) s.static_len -= stree[node*2+1];
                // node is 0 or 1 so it does not have extra bits
            }
            this.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=s.heap_len/2;n>=1; n--)
                s.pqdownheap(tree, n);

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

            node=elems;                 // next internal node of the tree
            do{
                // n = node of least frequency
                n=s.heap[1];
                s.heap[1]=s.heap[s.heap_len--];
                s.pqdownheap(tree, 1);
                m=s.heap[1];                // m = node of next least frequency

                s.heap[--s.heap_max] = n; // keep the nodes sorted by frequency
                s.heap[--s.heap_max] = m;

                // Create a new node father of n and m
                tree[node*2] = (short)(tree[n*2] + tree[m*2]);
                s.depth[node] = (byte)(System.Math.Max(s.depth[n],s.depth[m])+1);
                tree[n*2+1] = tree[m*2+1] = (short)node;

                // and insert the new node in the heap
                s.heap[1] = node++;
                s.pqdownheap(tree, 1);
            }
            while(s.heap_len>=2);

            s.heap[--s.heap_max] = s.heap[1];

            // At this point, the fields freq and dad are set. We can now
            // generate the bit lengths.

            gen_bitlen(s);

            // The field len is now set, we can generate the bit codes
            gen_codes(tree, max_code, s.bl_count);
        }