CS 1400

Pick ups from chapters 4 and 5

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

Example Menu

char choice;

cin >> 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;

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

}

Example Menu (bis)

char choice;cin >> 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;

}

Compound boolean expressions• Two relational (or Boolean) expressions

can be combined using&& (and)

|| (or)

• examples:if ((age > 21) && (age < 35))

cout << “You could be drafted!”;

if ((age < 0) || (age > 120))cout << “Incorrect age!”;

(Don’t confuse with & and |)

Precedence

• && has precedence over ||

• relational operators have precedence over &&if (a<b && c>d || e==f) same as

if (((a<b) && (c>d)) || (e==f))

Caution!!

• Be sure &&, || are between relational expressions…if (a < b||c) ??

if (b&&a == c) ??

Comparing two strings…• The following will have unexpected results:

char name1[20], name2[20];

cin >> name1 >> name2;

if (name1 == name2)

cout << “these names are the same!”;

The reasons will become clear later on… This is correct:

if (strcmp (name1, name2) == 0)

cout << “these names are the same!”; *FYI

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 << “this file could not be opened!”;

• Alternate example:if (!fin)

cout << “this file could not be opened!”;

Other loop statements…

• The do-while loopdo

{ 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

Example;

• Input validation – a program must prompt for an age and only accept a value greater than 0 and less than 120.

do

{ cout << “Enter a valid age: “;

cin >> age;

} while ((age > 0) && (age < 120));

Other loop statements.

• The for loopfor ( 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

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.

Example;

• Add 20 numbers input by the user;float num, sum = 0.0;

int n;

for (n=0; n<20; n++)

{ cout << “Enter a number: “;

cin >> num;

sum += num;

}

Example variation…• The test variable can actually be declared

within the loopfloat num, sum = 0.0;

for (int n=0; n<20; n++)

{ cout << “Enter a number: “;

cin >> num;

sum += num;

}

A programmer has a choice…• Convert the following while loop to a for loop

int count = 0;

while (count < 50)

{ cout << “count is: “ << count << endl;

count ++;

}

• Convert the following for loop to a whilefor (int x = 50; x > 0; x- - )

{ cout << x << “ seconds to launch.\n”;

}

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

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

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)

Pseudocode…1. Prompt and input loan amount, annual interest rate,

and number of years

2. Calculate and display monthly payment

3. Print report header

4. For each month in the loan period:a) calculate the monthly interest

b) calculate the principle

c) display the month number, interest, principle, and balance

d) calculate the new balance

1

2

3

for ( int month = 1 ; month <= num_years * 12 ; month++) // 4.

{

}

d)

c)

b)

a)

First refinement…

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

…

Pseudocode – rough outline1. 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

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

1

for ( int student = 0 ; student < num_of_students ; student++) //2

{

}

c)

b)

a)

First refinement…

e)

f)

d)

c)

Second refinement…

e)

f)

…

for (int count=0; count<=num_of_grades; count++) // d)

{

}

…

ii--

i--