Draw dependency graphs corresponding to each grammar rule (given in the below table) and for expression 7/4/3.0 Describe the two passes needed to compute the attributes on the syntax tree of 7/4/3.0, including a possible order in which the nodes could be visited and the attribute values computed at each point. Grammar RuleSemantic RulesS- expеxp.etype Dif exp.isFloat then float else intS.val - exp.valexp, .isFloat =exp2 .isFloat or exp, .isFloatexp2 .etype = exp .etypeexp3 .etype - exp, .etypeexp,.val =if exp, .etype - intthen exp, .val div exps .valelse expz .val / exp3 .valexp.isFloat = falseexp.val =if exp.etype = int then num.valelse Float(num.val)exp.isFloat = trueexp.val = num.num.valexp, - expz / exp3exp + numexp num.num

Draw dependency graphs corresponding to each grammar rule (given in the below table) and for…

Answered: Describe the two passes needed to comp

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Draw dependency graphs corresponding to each grammar rule (given in the below table) and for expression 7/4/3.0
Describe the two passes needed to compute the attributes on the syntax tree of 7/4/3.0, including a possible order in which the nodes could be visited and the attribute values computed at each point.

Grammar Rule
Semantic Rules
S- exp
еxp.etype D
if exp.isFloat then float else int
S.val - exp.val
exp, .isFloat =
exp2 .isFloat or exp, .isFloat
exp2 .etype = exp .etype
exp3 .etype - exp, .etype
exp,.val =
if exp, .etype - int
then exp, .val div exps .val
else expz .val / exp3 .val
exp.isFloat = false
exp.val =
if exp.etype = int then num.val
else Float(num.val)
exp.isFloat = true
exp.val = num.num.val
exp, - expz / exp3
exp + num
exp num.num

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