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Lines Matching defs:Freq

338         ush  freq;       /* frequency count */
347 #define Freq fc.freq
586 s->dyn_ltree[cc].Freq++; \
595 s->dyn_ltree[_length_code[len]+LITERALS+1].Freq++; \
596 s->dyn_dtree[d_code(dist)].Freq++; \
2513 for (n = 0; n < L_CODES; n++) s->dyn_ltree[n].Freq = 0;
2514 for (n = 0; n < D_CODES; n++) s->dyn_dtree[n].Freq = 0;
2515 for (n = 0; n < BL_CODES; n++) s->bl_tree[n].Freq = 0;
2517 s->dyn_ltree[END_BLOCK].Freq = 1;
2542 (tree[n].Freq < tree[m].Freq || \
2543 (tree[n].Freq == tree[m].Freq && depth[n] <= depth[m]))
2578 * IN assertion: the fields freq and dad are set, heap[heap_max] and
2620 f = tree[n].Freq;
2655 *(long)tree[m].Freq;
2707 * IN assertion: the field freq is set for all tree elements.
2729 if (tree[n].Freq != 0) {
2744 tree[node].Freq = 1;
2768 tree[node].Freq = tree[n].Freq + tree[m].Freq;
2774 node, tree[node].Freq, n, tree[n].Freq, m, tree[m].Freq);
2785 /* At this point, the fields freq and dad are set. We can now
2818 s->bl_tree[curlen].Freq += count;
2820 if (curlen != prevlen) s->bl_tree[curlen].Freq++;
2821 s->bl_tree[REP_3_6].Freq++;
2823 s->bl_tree[REPZ_3_10].Freq++;
2825 s->bl_tree[REPZ_11_138].Freq++;
2894 int max_blindex; /* index of last bit length code of non zero freq */
3023 int max_blindex = 0; /* index of last bit length code of non zero freq */
3125 s->dyn_ltree[lc].Freq++;
3134 s->dyn_ltree[_length_code[lc]+LITERALS+1].Freq++;
3135 s->dyn_dtree[d_code(dist)].Freq++;
3146 out_length += (ulg)s->dyn_dtree[dcode].Freq *
3216 * IN assertion: the fields freq of dyn_ltree are set and the total of all
3224 while (n < 7) bin_freq += s->dyn_ltree[n++].Freq;
3225 while (n < 128) ascii_freq += s->dyn_ltree[n++].Freq;
3226 while (n < LITERALS) bin_freq += s->dyn_ltree[n++].Freq;