예제 #1
0
		public int deflateEnd()
		{
			if (dstate == null)
				return Z_STREAM_ERROR;
			int ret = dstate.deflateEnd();
			dstate = null;
			return ret;
		}
예제 #2
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((byte) s.depth[n], (byte) 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);
		}
예제 #3
0
		public int deflateInit(int level, int bits)
		{
			dstate = new Deflate();
			return dstate.deflateInit(this, level, bits);
		}
예제 #4
0
		internal StaticTree stat_desc; // the corresponding static tree
		
		// 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 base_Renamed = 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 >= base_Renamed)
					xbits = extra[n - base_Renamed];
				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] = (short) (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) (s.opt_len + ((long) bits - (long) tree[m * 2 + 1]) * (long) tree[m * 2]);
						tree[m * 2 + 1] = (short) bits;
					}
					n--;
				}
			}
		}