- /* LINTED */
- for (n = m->last; n; n = n->parent) {
- c = rew_dohalt(tok, t, n);
- if (REWIND_HALT == c) {
- if (MDOC_BLOCK != t)
- return(1);
- if ( ! (MDOC_EXPLICIT & mdoc_macros[tok].flags))
- return(1);
- return(mdoc_perr(m, line, ppos, ENOCTX));
+ /*
+ * Iterate backwards, searching for the block matching tok,
+ * that is, the block breaking the *broken block.
+ */
+ for (breaker = broken->parent; breaker; breaker = breaker->parent) {
+
+ /*
+ * If the *broken block had already been broken before
+ * and we encounter its breaker, make the tok block
+ * pending on the inner breaker.
+ * Graphically, "[A breaker=[B broken=[C->B B] tok=A] C]"
+ * becomes "[A broken=[B [C->B B] tok=A] C]"
+ * and finally "[A [B->A [C->B B] A] C]".
+ */
+ if (breaker == broken->pending) {
+ broken = breaker;
+ continue;
+ }
+
+ if (REWIND_THIS != rew_dohalt(tok, MDOC_BLOCK, breaker))
+ continue;
+ if (MDOC_BODY == broken->type)
+ broken = broken->parent;
+
+ /*
+ * Found the breaker.
+ * If another, outer breaker is already pending on
+ * the *broken block, we must not clobber the link
+ * to the outer breaker, but make it pending on the
+ * new, now inner breaker.
+ * Graphically, "[A breaker=[B broken=[C->A A] tok=B] C]"
+ * becomes "[A breaker=[B->A broken=[C A] tok=B] C]"
+ * and finally "[A [B->A [C->B A] B] C]".
+ */
+ if (broken->pending) {
+ struct mdoc_node *taker;
+
+ /*
+ * If the breaker had also been broken before,
+ * it cannot take on the outer breaker itself,
+ * but must hand it on to its own breakers.
+ * Graphically, this is the following situation:
+ * "[A [B breaker=[C->B B] broken=[D->A A] tok=C] D]"
+ * "[A taker=[B->A breaker=[C->B B] [D->C A] C] D]"
+ */
+ taker = breaker;
+ while (taker->pending)
+ taker = taker->pending;
+ taker->pending = broken->pending;