Lines Matching full:old
101 struct range old;
116 * reverse order. Its first element must be all zero, the "old" and "new"
284 (*dd)[i].old.from = a;
285 (*dd)[i].old.to = b;
289 if ((*dd)[i].old.from < (*dd)[i - 1].old.to ||
295 (*dd)[i].old.from = (*dd)[i].old.to = (*dd)[i - 1].old.to;
381 change(1, &d1->old, false);
396 change(2, &d2->old, false);
410 d1[1].old.from = d1->old.from;
418 d2[1].old.from = d2->old.from;
425 dup = duplicate(&d1->old, &d2->old);
432 change(1, &d1->old, dup);
433 change(2, &d2->old, false);
434 d3 = d1->old.to > d1->old.from ? d1 : d2;
449 d2->old.from -= d2->new.from - d1->new.from;
452 d1->old.from -= d1->new.from - d2->new.from;
456 d2->old.to += d1->new.to - d2->new.to;
459 d1->old.to += d2->new.to - d1->new.to;
556 * Return 1 or 0 according as the old range (in file 1) contains exactly
620 de[j].old.from = diff->old.from;
621 de[j].old.to = diff->old.to;
661 struct range *new, *old;
665 old = &de[n].old;
671 else if (old->from == new->from && old->to == new->to) {
672 printf("%dc\n", old->from);
673 printrange(fp[2], old);
679 prange(old, delete);
681 printf("%da\n", old->to - 1);
690 printf("%da\n%s %s\n.\n", old->from - 1,
714 struct range *new, *old;
718 old = &de[n].old;
720 deleteold = (old->from == old->to);
724 prange(old, deletenew);
738 printrange(fp[1], old);
743 startmark = old->to - 1;
746 prange(old, deletenew);
754 r.from = old->from-1;
758 printrange(fp[1], old);
775 startmark - (old->to - old->from),
794 struct range r, *new, *old;
803 old = &de[n].old;
812 else if (de[n].type == DIFF_TYPE3 && (old->from == old->to)) {
813 r.from = old->from - 1;
816 r.to = old->from;
827 printrange(fp[1], old);
838 printrange(fp[0], old);
842 if (old->from == old->to) {
844 or.from = old->from - 1;
848 printrange(fp[1], old);
863 if (old->from == old->to)
866 r.from = old->to;
874 * If the new range is 0 length (from == to), we need to use the old
878 old = &de[n-1].old;
880 if (old->from == new->from && old->to == new->to)
883 r.from = old->from;