Grammar
- 0 $axiom: exp $
+ 0 $accept: exp $end
1 exp: exp OP exp
2 | NUM
Terminals, with rules where they appear
-$ (0) 0
+$end (0) 0
error (256)
NUM (258) 2
OP (259) 1
Nonterminals, with rules where they appear
-$axiom (5)
+$accept (5)
on left: 0
exp (6)
on left: 1 2, on right: 0 1
state 0
- 0 $axiom: . exp $
+ 0 $accept: . exp $end
1 exp: . exp OP exp
2 | . NUM
state 2
- 0 $axiom: exp . $
+ 0 $accept: exp . $end
1 exp: exp . OP exp
- $ shift, and go to state 3
- OP shift, and go to state 4
+ $end shift, and go to state 3
+ OP shift, and go to state 4
state 3
- 0 $axiom: exp $ .
+ 0 $accept: exp $end .
$default accept
state 5
- 1 exp: exp . OP exp [$, OP]
- 1 | exp OP exp . [$, OP]
+ 1 exp: exp . OP exp [$end, OP]
+ 1 | exp OP exp . [$end, OP]
OP shift, and go to state 4
AT_CHECK([cat input.output], [],
[[Grammar
- 0 $axiom: exp $
+ 0 $accept: exp $end
1 exp: exp OP exp
2 | NUM
Terminals, with rules where they appear
-$ (0) 0
+$end (0) 0
error (256)
NUM (258) 2
OP (259) 1
Nonterminals, with rules where they appear
-$axiom (5)
+$accept (5)
on left: 0
exp (6)
on left: 1 2, on right: 0 1
state 0
- 0 $axiom: . exp $
+ 0 $accept: . exp $end
1 exp: . exp OP exp
2 | . NUM
state 2
- 0 $axiom: exp . $
+ 0 $accept: exp . $end
1 exp: exp . OP exp
- $ shift, and go to state 3
- OP shift, and go to state 4
+ $end shift, and go to state 3
+ OP shift, and go to state 4
state 3
- 0 $axiom: exp $ .
+ 0 $accept: exp $end .
$default accept
state 5
- 1 exp: exp . OP exp [$, OP]
- 1 | exp OP exp . [$, OP]
+ 1 exp: exp . OP exp [$end, OP]
+ 1 | exp OP exp . [$end, OP]
$default reduce using rule 1 (exp)
Conflict between rule 1 and token OP resolved as reduce (%left OP).
# When there are RR conflicts, some rules are disabled. Usually it is
# simply displayed as:
#
-# $ reduce using rule 3 (num)
-# $ [reduce using rule 4 (id)]
+# $end reduce using rule 3 (num)
+# $end [reduce using rule 4 (id)]
#
# But when `reduce 3' is the default action, we'd produce:
#
-# $ [reduce using rule 4 (id)]
+# $end [reduce using rule 4 (id)]
# $default reduce using rule 3 (num)
#
# In this precise case (a reduction is masked by the default
# reduction), we make the `reduce 3' explicit:
#
-# $ reduce using rule 3 (num)
-# $ [reduce using rule 4 (id)]
+# $end reduce using rule 3 (num)
+# $end [reduce using rule 4 (id)]
# $default reduce using rule 3 (num)
#
# Maybe that's not the best display, but then, please propose something
Grammar
- 0 $axiom: exp $
+ 0 $accept: exp $end
1 exp: num
2 | id
Terminals, with rules where they appear
-$ (0) 0
+$end (0) 0
'0' (48) 3 4
error (256)
Nonterminals, with rules where they appear
-$axiom (4)
+$accept (4)
on left: 0
exp (5)
on left: 1 2, on right: 0
state 0
- 0 $axiom: . exp $
+ 0 $accept: . exp $end
1 exp: . num
2 | . id
3 num: . '0'
state 1
- 3 num: '0' . [$]
- 4 id: '0' . [$]
+ 3 num: '0' . [$end]
+ 4 id: '0' . [$end]
- $ reduce using rule 3 (num)
- $ [reduce using rule 4 (id)]
+ $end reduce using rule 3 (num)
+ $end [reduce using rule 4 (id)]
$default reduce using rule 3 (num)
state 2
- 0 $axiom: exp . $
+ 0 $accept: exp . $end
- $ shift, and go to state 5
+ $end shift, and go to state 5
state 3
state 5
- 0 $axiom: exp $ .
+ 0 $accept: exp $end .
$default accept
]])