CS 1400

1
CS 1400

• Pick ups from chapters 4 and 5

2
if-else-if

• Chained if statements can be useful for menus
• Enter savings plan choice (A) economy
  rate
• (B) saver rate
• (C) preferred rate
• (D) special rate

3
Example Menu

• char choice
• cin gtgt choice
• if (choice A)
• rate 2.15
• else if (choice B)
• rate 3.25
• else if (choice C)
• rate 4.86
• else rate 5.22

4
The switch statement

• General form
• switch (ordinal variable)
• case constant value1 statements
• break
• case constant value2 statements
• break
• case constant value3 statements
• break
• default statements

5
Example Menu (bis)

• char choice
• cin gtgt choice
• switch (choice)
• case A rate 2.15
• break
• case B rate 3.25
• break
• case C rate 4.86
• break
• default rate 5.22

6
Compound boolean expressions

• Two relational (or Boolean) expressions can be
  combined using
• (and)
• (or)
• examples
• if ((age gt 21) (age lt 35))
• cout ltlt You could be drafted!
• if ((age lt 0) (age gt 120))
• cout ltlt Incorrect age!

(Dont confuse with and )
7
Precedence

• has precedence over
• relational operators have precedence over
• if (altb cgtd ef) same as
• if (((altb) (cgtd)) (ef))

8
Caution!!

• Be sure , are between relational
  expressions
• if (a lt bc) ??
• if (ba c) ??

9
Comparing two strings

• The following will have unexpected results
• char name120, name220
• cin gtgt name1 gtgt name2
• if (name1 name2)
• cout ltlt these names are the same!
• The reasons will become clear later on This is
  correct
• if (strcmp (name1, name2) 0)
• cout ltlt these names are the same!

FYI
10
Testing for file errors

• A file variable may be tested for the failure of
  the most recent file operation using the member
  function .fail()
• Example
• ifstream fin
• fin.open(myfile.txt)
• if (fin.fail())
• cout ltlt this file could not be opened!
• Alternate example
• if (!fin)
• cout ltlt this file could not be opened!

11
Other loop statements

• The do-while loop
• do
• statements
• while ( expr )
• This is a post-test loop The expression is
  tested at the bottom of the loop and always
  performs at least one iteration

12
Example

• Input validation a program must prompt for an
  age and only accept a value greater than 0 and
  less than 120.
• do
• cout ltlt Enter a valid age
• cin gtgt age
• while ((age gt 0) (age lt 120))

13
Other loop statements.

• The for loop
• for ( initialization test expr update )
• statements
• This is a pre-test loop The test expression is
  tested at the top of the loop. The
  initialization, test expression and update
  segments can be any valid C expression

14
Explanation of segments

• initialization
• This segment is executed once prior to beginning
  the loop
• test expr
• This segment is executed prior to each iteration.
  If this segment is false, the loop terminates
• update
• This segment is executed at the bottom of each
  iteration.

15
Example

• Add 20 numbers input by the user
• float num, sum 0.0
• int n
• for (n0 nlt20 n)
• cout ltlt Enter a number
• cin gtgt num
• sum num

16
Example variation

• The test variable can actually be declared within
  the loop
• float num, sum 0.0
• for (int n0 nlt20 n)
• cout ltlt Enter a number
• cin gtgt num
• sum num

17
A programmer has a choice

• Convert the following while loop to a for loop
• int count 0
• while (count lt 50)
• cout ltlt count is ltlt count ltlt endl
• count
• Convert the following for loop to a while
• for (int x 50 x gt 0 x- - )
• cout ltlt x ltlt seconds to launch.\n

18
Example loan amoritization

• A case study (5.14, pg 295)
• Write a loan amoritization program
• inputs principle, interest_rate, num_years
• outputs
• monthly payment
• for each month month number, interest,
  principle, remaining_balance

19
Formulas

• payment
• (principle interest_rate/12 term) / (term-1)
• where term
• (1 interest_rate/12) num_years 12
• monthly_interest
• (annual_rate / 12) balance
• principle
• payment monthly_interest

20
Example execution

• Enter loan amount 2500
• Enter annual Interest rate 0.08
• Enter number of years 2
• Monthly payment 113.07
• month interest principle balance
• 1 16.67 96.40 2403.60
• 2 16.02 97.04 2306.55
• 3 15.38 97.69 2208.86
• (a total of 24 month lines)

21
Pseudocode

• Prompt and input loan amount, annual interest
  rate, and number of years
• Calculate and display monthly payment
• Print report header
• For each month in the loan period
• calculate the monthly interest
• calculate the principle
• display the month number, interest, principle,
  and balance
• calculate the new balance

22
First refinement
for ( int month 1 month lt num_years 12
month) // 4.

23
Example loops within loopsinput table of grades

• Student count 64
• John 9 scores 9 7 8 9 6 10 10 8 10
• Fred 13 scores 7 8 5 6 7 6 8 9 6 10 9 10 3
• Emily 8 scores 8 4 5 3 9 7 5 4

24
Pseudocode rough outline

• 1. Input number_of_students
• 2. Repeat for number_of_students iterations
• a-b) Input student_name and count_of_grades
• c-d) sum grades for this student
• e) Calculate average
• f) Output student_name and average

25
Pseudocode (more detailed)

• 1. Input number_of_students
• 2. Repeat for number_of_students iterations
• a) Input student name
• b) Input count_of_grades
• c) initialize sum to 0.
• d) Repeat for count_of_grades iterations
• i Input grade
• ii sum grade
• e) Calculate average
• f) Output student_name and average

26
First refinement
for ( int student 0 student lt num_of_students
student) //2

27
Second refinement

for (int count0 countltnum_of_grades count)
// d)