"digraph %s\n"
"{\n",
quote (grammar_file));
- fprintf (fout, "node [shape=box]\n");
+ fprintf (fout,
+ " node [fontname = courier, shape = box, colorscheme = paired6]\n"
+ " edge [fontname = courier]\n"
+ "\n");
}
void
}
static void
-conclude_red (struct obstack *out, int source, int ruleno, bool enabled,
- bool first, FILE *fout)
+conclude_red (struct obstack *out, int source, rule_number ruleno,
+ bool enabled, bool first, FILE *fout)
{
/* If no lookahead tokens were valid transitions, this reduction is
actually hidden, so cancel everything. */
return (void) obstack_finish0 (out);
else
{
- char const *ed = enabled ? "e" : "d";
+ char const *ed = enabled ? "" : "d";
+ char const *color = enabled ? ruleno ? "3" : "1" : "5";
- /* First, build the edge's head. */
- if (! first)
- fprintf (fout, " %1$d -> \"%1$dR%2$d%3$s\" [label = \"",
- source, ruleno, ed);
+ /* First, build the edge's head. The name of reduction nodes is "nRm",
+ with n the source state and m the rule number. This is because we
+ don't want all the reductions bearing a same rule number to point to
+ the same state, since that is not the desired format. */
+ fprintf (fout, " %1$d -> \"%1$dR%2$d%3$s\" [",
+ source, ruleno, ed);
/* (The lookahead tokens have been added to the beginning of the
obstack, in the caller function.) */
+ if (! obstack_empty_p (out))
+ {
+ char *label = obstack_finish0 (out);
+ fprintf (fout, "label=\"[%s]\", ", label);
+ obstack_free (out, label);
+ }
/* Then, the edge's tail. */
- obstack_sgrow (out, "\" style = solid]\n");
-
- /* Build the associated diamond representation or the target rule. */
- obstack_printf (out, " \"%dR%d%s\" "
- "[style = filled shape = diamond fillcolor = %s "
- "label = \"R%d\"]\n",
- source, ruleno, ed,
- enabled ? "yellowgreen" : "firebrick1",
- ruleno);
- fprintf (fout, obstack_finish0 (out));
+ fprintf (fout, "style=solid]\n");
+
+ /* Build the associated diamond representation of the target rule. */
+ fprintf (fout, " \"%dR%d%s\" [label=\"",
+ source, ruleno, ed);
+ if (ruleno)
+ fprintf (fout, "R%d", ruleno);
+ else
+ fprintf (fout, "Acc");
+
+ fprintf (fout, "\", fillcolor=%s, shape=diamond, style=filled]\n",
+ color);
}
}
char const *q = escape (tok);
if (! first)
- obstack_sgrow (out, ",");
+ obstack_sgrow (out, ", ");
obstack_sgrow (out, q);
return false;
}
bitset no_reduce_set;
int j;
int source = s->number;
- struct obstack dout, eout;
+
+ /* Two obstacks are needed: one for the enabled reductions, and one
+ for the disabled reductions, because in the end we want two
+ separate edges, even though in most cases only one will actually
+ be printed. */
+ struct obstack dout;
+ struct obstack eout;
no_reduce_bitset_init (s, &no_reduce_set);
obstack_init (&dout);
for (j = 0; j < reds->num; ++j)
{
bool defaulted = false;
- bool firstd = true, firste = true; // first{en,dis}abled
- int ruleno = reds->rules[j]->user_number;
+ bool firstd = true;
+ bool firste = true;
+ rule_number ruleno = reds->rules[j]->number;
rule *default_reduction = NULL;
if (yydefact[s->number] != 0)
/* Build the lookahead tokens lists, one for enabled transitions and one
for disabled transistions. */
if (default_reduction && default_reduction == reds->rules[j])
- {
- firste = print_token (&eout, true, "$default");
- defaulted = true;
- }
+ defaulted = true;
if (reds->lookahead_tokens)
{
int i;
}
/* Do the actual output. */
- conclude_red (&eout, source, ruleno, true, firste, fout);
conclude_red (&dout, source, ruleno, false, firstd, fout);
+ conclude_red (&eout, source, ruleno, true, firste && !defaulted, fout);
}
- obstack_free (&eout, 0);
obstack_free (&dout, 0);
+ obstack_free (&eout, 0);
+ bitset_free (no_reduce_set);
}
void