Consider the following grammar:
%token IF EXP ELSE XX %% stmts : stmts stmt ';' | stmt ';' ; stmt : IF '(' EXP ')' stmt ELSE stmt | IF '(' EXP ')' stmt | XX ; %%In the above grammars in this file tokens that are all uppercase are terminals that match the same string in lower case. For instance in the above grammar: if ( exp ) xx else xx ; is a legal statement. Unfortunately the grammar has a problem when compiled in bison:
+ bison -v -t -d check.y check.y contains 1 shift/reduce conflict.
Checking the .output file shows:
State 11 contains 1 shift/reduce conflict. Grammar rule 1 stmts -> stmts stmt ';' rule 2 stmts -> stmt ';' rule 3 stmt -> IF '(' EXP ')' stmt ELSE stmt rule 4 stmt -> IF '(' EXP ')' stmt rule 5 stmt -> XX state 11 stmt -> IF '(' EXP ')' stmt . ELSE stmt (rule 3) stmt -> IF '(' EXP ')' stmt . (rule 4) ELSE shift, and go to state 12 ELSE [reduce using rule 4 (stmt)] $default reduce using rule 4 (stmt)It is clear that a shift would move the ELSE using rule 3 but where did the problem with the reduce come from? The problem is that rule 3 also shows that a stmt stmt can be followed by ELSE because the then-part is a stmt and can be followed by ELSE. But rule 4 is a stmt therefore rule 4 can be followed by an ELSE! So if an ELSE token is seen then it should reduce AND it should shift it. Hence an shift/reduce error.
The fix is this we need to make sure there is no ambiguity in our grammar that would suggest that both parses that shift and reduce are viable. The classic ambiguous case is:
if ( exp ) if ( exp ) xx else xx ;does that mean (case 1) as indicated by the brackets:
if ( exp ) [ if ( exp ) xx else xx ];or does that mean (case 2):
if ( exp ) [ if ( exp ) xx ] else xx ;? The fix is to be sure that the else is either always with the nearest if (case 1) or farthest if (case 2). Classically the fix is to do case 1.
We enforce this by using a grammar like this:
%token IF EXP ELSE XX %% stmts : stmts stmt ';' | stmt ';' ; stmt : matched | unmatched ; matched : IF '(' EXP ')' matched ELSE matched | XX ; unmatched : IF '(' EXP ')' matched | IF '(' EXP ')' unmatched | IF '(' EXP ')' matched ELSE unmatched ; %%Statements that match matched are like:
if ( exp ) xx else xx if ( exp ) xx else if ( exp ) xx else xx if ( exp ) if ( exp ) xx else if ( exp ) xx else xx else if ( exp ) xx else xxStatements that match unmatched are like:
if ( exp ) xx if ( exp ) xx else if ( exp ) xx if ( exp ) if ( exp ) xx else xx else if ( exp ) xxMost importantly is what is left out of the grammar above! There are 4 posssible combinations of if-else:
IF '(' EXP ')' matched ELSE matched IF '(' EXP ')' matched ELSE unmatched IF '(' EXP ')' unmatched ELSE matched IF '(' EXP ')' unmatched ELSE unmatchedHowever, the last two do not occur in the grammar and are not allowed. It is this that prohibits the occurrence of an else that is NOT matched to the nearest if! That is the grammar simply does not allow an unmatched if to come between an else and its if. This removes the ambiguity of two parses and leaves only the one we want.
Now that we understand what is going on, how can we add a while statement? While statements are of the form:
WHILE '(' EXP ')' stmtbut stmt can be either matched or unmatched. Consider the expression:
if ( exp ) while ( exp ) if ( exp ) xx else xxAs before we have two cases (parsings) indicated by the brackets:
if ( exp ) while ( exp ) [ if ( exp ) xx else xx ] if ( exp ) [ while ( exp ) if ( exp ) xx ] else xxWe need to write our grammar to preclude the second of these two parses. That is a while with an unmatched if must not occur between an if and its else. This removes the ambiguity and allows the grammar to parse only the first meaning (the one we want).
This suggests that there are two kinds of while. A while that has an unmatched after it and a while that has a matched after it. The while with the unmatched must not be allowed to come before an else in the same way that the if without the else must not.
I have all but told you the answer but most importantly the reasoning behind it. I must leave something for you to figure out so I will stop here except to point out that the grammar I gave above for solving the dangling if could be slightly shorter by replacing
unmatched : IF '(' EXP ')' matched | IF '(' EXP ')' unmatchedwith
unmatched : IF '(' EXP ')' stmt
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|Robert Heckendorn||Up One Level||Last updated: