27 March, 2000

CS1001 Lecture 9

• Program Debugging• Previews of Arrays, Functions, and Subroutines• More Project Design and code segments

27 March, 2000

Errors in program development• Common errors:

– syntax or compiler errors -- errors discovered during compilation; e.g., punctuation, unbalanced ( )

– Run-time errors -- errors detected during execution of the program; e.g., division by zero, overflow

– Logic errors -- errors in design of algorithms and in coding the algorithms

• System error messages help solve syntax and run-time errors

• Programmers must debug and solve logic error by using their knowledge of the problem, the algorithm and the coding.

27 March, 2000

Range of noise level example

Text pages 223-228LargestLevel = 0SmallestLevel = 999DO

Print * “Noise level?”Read *, NoiseLevelIF (NoiseLevel > LargestLevel) THEN

LargestLevel = NoiseLevelElse IF (NoiseLevel < SmallestLevel) THEN

SmallestLevel = NoiseLevelEND IFIF (NoiseLevel < 0) EXIT

END DOPRINT *, LargestLevel-SmallestLevel

Input data9410288-1

27 March, 2000

Desk checking

• Trace Table - recording in a table, step by step, the important variables

Loop NoiseLevel LargestLevel SmallestLevelFirst pass 94 94 999Second pass 102 102 999Third pass 88 102 88Fourth pass -1 102 -1

Therefore LargestLevel-SmallestLevel = 102- (-1) = 103

• Use temporary print statement in code • Use special system debugging software

27 March, 2000

Modify and retest

LargestLevel = 0SmallestLevel = 999DO

Print * “Noise level?”Read *, NoiseLevelIF (NoiseLevel < 0) EXIT IF (NoiseLevel > LargestLevel) THEN

LargestLevel = NoiseLevelElse IF (NoiseLevel < SmallestLevel) THEN

SmallestLevel = NoiseLevelEND IF

END DOPRINT *, LargestLevel-SmallestLevel

Input data9410288-1output is 14

What if?Input data8894102-1

27 March, 2000

Desk Check of revision

Loop NoiseLevel LargestLevel SmallestLevelFirst pass 88 88 999Second pass 94 94 999Third pass 102 102 999Fourth pass -1 102 999Therefore LargestLevel-SmallestLevel = 102 - (999) = -897

Loop NoiseLevel LargestLevel SmallestLevelFirst pass 94 94 999Second pass 102 102 999Third pass 88 102 88Fourth pass -1 102 88Therefore LargestLevel-SmallestLevel = 102- (88) = 14

27 March, 2000

Modify & Retest

LargestLevel = 0SmallestLevel = 999DO Print * “Noise level?” Read *, NoiseLevel IF (NoiseLevel < 0) EXIT IF (NoiseLevel > LargestLevel) LargestLevel = NoiseLevel IF (NoiseLevel < SmallestLevel) SmallestLevel = NoiseLevelEND DO

PRINT *, LargestLevel-SmallestLevel

27 March, 2000

Desk Check of revision 2

Loop NoiseLevel LargestLevel SmallestLevelFirst pass 88 88 88Second pass 94 94 88Third pass 102 102 88Fourth pass -1 102 -1Therefore LargestLevel-SmallestLevel = 102 - (88) = 14

Loop NoiseLevel LargestLevel SmallestLevelFirst pass 94 94 94Second pass 102 102 94Third pass 88 102 88Fourth pass -1 102 -1Therefore LargestLevel-SmallestLevel = 102- (88) = 14

27 March, 2000

More Efficient Implementation

Print * “Noise level?”Read *, NoiseLevelLargestLevel = NoiseLevelSmallestLevel = NoiseLevelDO IF (NoiseLevel < 0) EXIT IF (NoiseLevel > LargestLevel)

THEN LargestLevel = NoiseLevel ELSE IF (NoiseLevel < SmallestLevel)

SmallestLevel = NoiseLevel END IF Print * “Noise level?” Read *, NoiseLevelEND DO

27 March, 2000

Preview of Arrays

• Primitive data types vs structured data types• An array is a data structure

– elements are of the same type– access to elements by index -- elements are subscript variables– Specification and declaration:

REAL, DIMENSION(l:u) :: array-name

– Example:INTEGER, DIMENSION(1:10) :: Input

INTEGER :: Range, Choice

DO Range=1, 10, 1

READ *, Input(Range)

END DO

27 March, 2000

Array Preview Continued

• Why arrays? Because you can operate on the subscript.• Example, continued:

PRINT *, “Pick a box from 1 to 10”

READ *, Choice

PRINT *, “Box ”, Choice, “ contains number ”, Input(Choice)

• You can perform much more complex operations on the variable used for the subscript

27 March, 2000

Preview of Functions and Subroutines• Fortran provides library functions for many common

operations and functions. (Chap 2)• User defined functions and subroutines

– For large complex problems, divide the problem into a number of simpler problems.

– Each subproblem can be considered separately, designed and coded separately.

– These codes of subproblems are then combined to form the whole solution.

– Each of these separate set of code -- functions and subroutines– Functions and subroutines are controlled by the main program

27 March, 2000

Project Introduction

Checkout stations

Servers check customers out at an average of average-service-time

Customers arrival rate

27 March, 2000

Project Introduction (cont’d)

Grocery Simulation

Output StatisticsGet parameters Run Simulation

SimulationClock

GenerateCustomers

CheckoutCustomers

GatherStatistics

27 March, 2000

Run Simulation

• At time zero, set parameters and initialize • Simulation Clock

Given Sim_Time: The simulation clock may be

DO Time = 1, Sim_Time

Generate Customers

Checkout Customers

Gather Statistics for time step Time

END DO

27 March, 2000

Generate Customers

• Define a function to generate new customers: NArrivals = CustomerGeneration ( )

• Put new customers into queueIF (NArrivals .NE. 0) then

DO I = 1, NArrivals

! Find Shortest or appropriate queue

! Add to queue

END DO

END IF

27 March, 2000

Queue

• Represent each check_out station as a FIFO queue• Should have

1. ways to have customers come in and go out

2. A idle_time counter to keep track of number of minutes when there is no customer in the queue.

3. At a time step (a minute), if no customer in queue, then we can increase idle_time counter by 1 minute.

4. Should we have a counter to see how many customers are in the queue? Probably more efficient.

27 March, 2000

Queue

• Same data type structure, at most 15 of them• Represent some of its properties by arrays• e.g., number of customers in queue

INTEGER, DIMENSION(15) : Ncustomers

Ncustomers(1) is the number of customers in queue 1,

Ncustomers(2) is the number of customers in queue 2,

etc.

• Given nstations as user input for the number of check out stations ==> number of queues needed

27 March, 2000

Queue (Ncustomers)• initialization:

DO I=1,nstations

Ncustomers(I) = 0

END DO

• Finding the line with least number of customers:shortestline=1

DO I=2, nstations

IF (Ncustomers(I) .LT. Ncustomers(shortestline)) & shortestline=I

END DO

27 March, 2000

Add_to_Queue

• Use Subroutine to implement this subproblemSUBROUTINE Add_to_queue

! *** coding details later

! You will have to tell this subroutine which

! queue to add the customer (i.e., pass in a parameter

! shortestline)

! *** you will be provided with a subroutine to add

! customer to a queue

Call AddTo (shortestline)

! Other book keeping activities

Ncustomers(shortestline) = Ncustomers(shortestline)+1

END SUBROUTINE Add_to_queue

27 March, 2000

Checkout Customers

• Use Subroutine to implement this subproblemSUBROUTINE Check_out

! *** coding details later

! *** you will be provided with a subroutine to remove

! customer from a queue

DO I = 1, nstations

IF (Ncustomers(I) .NE. 0 .AND. Finished)

THEN Call POP_queue (I)

Ncustomers(I) = Ncustomers(I)-1

END IF

END SUBROUTINE Check_out

27 March, 2000

Gather statistics• Idle_time can be handled the same way as Ncustomers

INTEGER, DIMENSION(15) : Ncustomers , Idle_time

Idle_time(1) is the number of idle minutes in queue 1,

Idle_time(2) is the number of idle minutes in queue 2, etc.

• InitializationDO I=1,nstation

Idle_time(I) = 0

END DO

• Track Idle_timeDO I=1,nstation

IF (Ncustomers(I) .EQ. 0) Idle_time(I) = Idle_time(I)+1

END DO