Write an expression that will cause the following code to print "Equal" if the value of sensorReading is "close enough" to targetValue. Otherwise, print "Not equal". Hint: Use epsilon value 0.0001. Ex: If targetValue is 0.3333 and sensorReading is (1.0/3.0), output is:
Write an expression that will cause the following code to print "Equal" if the value of sensorReading is "close enough" to targetValue. Otherwise, print "Not equal". Hint: Use epsilon value 0.0001. Ex: If targetValue is 0.3333 and sensorReading is (1.0/3.0), output is:
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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Question
Write an expression that will cause the following code to print "Equal" if the value of sensorReading is "close enough" to targetValue. Otherwise, print "Not equal". Hint: Use epsilon value 0.0001. Ex: If targetValue is 0.3333 and sensorReading is (1.0/3.0), output is:
Expert Solution
Step 1
Introduction:
The programme set up the two values, tragedy value and sensor reading.
Another near difference value is defined and initialised at 1E-4.
The application will check whether the difference between sensor reading and targer value is more than the close difference set.
If not, it prints equal; otherwise, it prints NOT Equal.
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