Please note, I am asking about writing LALR parser, not writing rules for LALR parser.
What I need is…
…to mimic YACC precedence definitions. I don’t know how it is implemented, and below I describe what I’ve done and read so far.
For now I have basic LALR parser written. Next step — adding precedence, so 2+3*4 could be parsed as 2+(3*4).
I’ve read about precedence parsers, however I don’t see how to fit such model into LALR. I don’t understand two points:
-
how to compute when insert parenthesis generator
-
how to compute how many parenthesis the generator should create
I insert generators when the symbols is taken from input and put at the stack, right? So let’s say I have something like this (| denotes boundary between stack and input):
ID = 5 | + ..., at this point I add open, so it givesID = < 5 | + ..., then I read more inputID = < 5 + | 5 ...and moreID = < 5 + 5 | ; ...and moreID = < 5 + 5 ; | ...
At this point I should have several reduce moves in normal LALR, but the open parenthesis does not match so I continue reading more input. Which does not make sense.
So this was when problem.
And about count, let’s say I have such data < 2 + < 3 * 4 >. As human I can see that the last generator should create 2 parenthesis, but how to compute this? After all there could be two scenarios:
-
( 2 + ( 3 *4 ))— parenthesis is used to show the outcome of generator -
or
(2 + (( 3 * 4 ) ^ 5)because there was more input
Please note that in both cases before 3 was open generator, and after 4 there was close generator. However in both cases, after reading 4 I have to reduce, so I have to know what generator “creates”.
4
Consider the following rules:
[1] E -> E + E
[2] E -> E * E
[3] E -> num
Without precedence, it would produce a shift/reduce error when encountering the second operator. The table looks something like this:
num + * $ E
0 S2 --- --- --- 1
1 --- S3 S4 Accept ---
2 --- R3 R3 R3 ---
3 S2 --- --- --- 5
4 S2 --- --- --- 6
5 --- R1/S3 R1/S4 R1 ---
6 --- R2/S3 R2/S4 R2 ---
With precedence, you are saying which of “shift” and “reduce” you want in the case of conflicts.
- If the first operator has a lower precedence, you shift.
- If the first operator has a higher precedence, you reduce.
- If the operators have equal precedences, and both are
- left associative, you reduce.
- right associative, you shift.
- otherwise, you fail.
Cleaning up the errors in the table, you get something like this:
num + * $ E
0 S2 --- --- --- 1
1 --- S3 S4 Accept ---
2 --- R3 R3 R3 ---
3 S2 --- --- --- 5
4 S2 --- --- --- 6
5 --- R1 S4 R1 ---
6 --- R2 R2 R2 ---
1