# Recent questions tagged operator-grammar 1 vote
1
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?
2
Say I have a grammar, S→ AB A→ a B→ b This grammar is not operator grammar as 2 non terminals are lying side by side, but can be converted to an operator grammar. S→ ab , A→ a , B→ b here i have a doubt, operator grammar as the name suggests should have a ... right? how can we operate even two terminal symbols when placed side by side? Isn't it same as placing 2 non-terminal symbol side by side?
3
Given an operator grammar : $E -> E * F / F + E / F$ $F -> F - F / id$ Is this table correct for the above operator grammar ?
4
Can anyone explain why operator precedence parsing cannot handle unary minus,and what are the approach to handle it.
5
In Operator precedence parsing, precedence relations are defined, i. for all pair of non-terminals. ii. for all pair of terminals. iii. to delimit the handle. iv. only for certain pair of terminals.
6
Given an Operator Grammar as, E -> E*F / F+E / F F -> F-F / id How to determine associativity in this case? Since Operator grammar can be ambiguous also. Is the above question solved using associativity and precedence of operators we consider in C programming? or it is Different.
Which grammar rules violate the requirement of the operator grammar? A, B, C are variables and a, b, c are terminals $A \rightarrow BC$ $A \rightarrow CcBb$ $A \rightarrow BaC$ $A \rightarrow \epsilon$ (a) 1 only (b) 1 and 2 (c) 1 and 3 (d) 1 and 4