Operator Precedence Parsing.

Question

In operator precedence parsing we have the rule that production cannot have two adjacent non-terminals or an epsilon production, so this production, S--> ab is allowed but not S--> AB, A->a and B->b, though they are giving us the same output. Why so?

Comments

because operator precedence parser can work on only operator grammar. And that rule is one of the conditions for being an operator grammar.

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Okay, Thank u.. :)

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See page 3 from this

http://www.math.uni.lodz.pl/~robpleb/Wyklad_9_ang.pdf

Answer 1

Rule of Operator precedence Parser is Any two variable should not adjacent to each other.If they are adjacent to each other then convert it!

eg- In C compiler,no two variables are adjacent 

            ab!=a*b this is not allowed in our c compiler

Answer 2

because for first grammar: S→ ab( adjacent terminals) is an operator grammar

but the second one is S->AB,  A->a, B->b (adjacent non-terminals (AB)) is not operator grammar (Yes we all know that we can make it operator grammar by replacing A and B’s Value in S but first hand  it is an non-operator grammar, So we can’t make its Relation Table).