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Lecture 4: Calculating by
Iterating
The while Repetition Statement
Repetition structure
Programmer specifies an action to be repeated while
some condition remains true.
Pseudocode (go shopping)
While there are more items on my shopping list
Purchase next item and cross it off my list
condition
The body of the while
while loop repeated until condition becomes false
Can loop a known or unknown number of times for
repeating calculations or function calls (printf).
Flowchart
condition
while body
Counter-Controlled Repetition
Problem
Write a power algorithm (ex. 2x, where x is a user
defined number)
0
1
Product
2
Formulating algorithm
2*product product
21
Pseudocode
2*product product
22
2*product product
Set product to 1
2*product product
Type in exponent
2*product product
Set counter to 1
While counter is less than or equal to exponent
product = 2 * product
add one to the counter
Print out product
23
24
25
x times
Counter-Controlled Repetition
Formulating algorithm
Flowchart
begin
product = 1;
counter = 1;
Input exponent
counter< exponent
false
Print product
end
true
product = 2*product;
counter = counter+1;
Counter-Controlled Repetition
C code
/*
Computing 2^x, where x is a user defined number
*/
#include <stdio.h>
/* function main begins program execution */
int main( void )
{
int product; /* product of 2^x */
int counter; /* number of 2's multipled */
int exponent; /* user defined the power of 2 */
product = 1;
counter = 1;
printf( "Enter the exponent:\n" );
/* prompt */
scanf( "%d", &exponent );
/* read an integer */
while ( counter <= exponent ) {
product = 2 * product;
counter = counter + 1;
}
Pseudocode
Set product to 1
Type in exponent
Set counter to 1
While counter is less than or
equal to exponent
Counter to control while loop
product = 2 * product
add one to the counter
Print out product
Initialize counter to 1
while loop iterates as long as counter <=
exponent
Increment the counter
printf("2^%d equals to %d\n", exponent, product);
return 0; /* indicate that program ended successfully */
} /* end function main */
Do we really need the variable power? Can we save one variable?
Counter-Controlled Repetition
C code - using one less variable
/*
Computing 2^x, where x is a user defined number
*/
#include <stdio.h>
/* function main begins program execution */
int main( void )
{
int product; /* product of 2^x */
int exponent; /* user defined the power of 2 */
product = 1;
printf( "Enter the exponent:\n" );
/* prompt */
scanf( "%d", &exponent );
/* read an integer */
printf("2^%d equals to ", exponent );
while ( exponent > 0 ) {
product = 2 * product;
exponent = exponent - 1;
}
printf("%d\n", product);
return 0; /* indicate that program ended successfully */
} /* end function main */
Counter-Controlled Repetition
Loop repeated until counter reaches a certain value
Definite repetition: number of repetitions is known
In-Class Programming Exercise
Write a program that completes the following:
1. Read in an integer from the user.
2. Print out that number of numbers from the Fibonacci
sequence. This would require you to calculate the values of the
sequence, not hard code them in your program.
(Hint: You should calculate the Fibonacci sequence using two
variables.)
3. Use a while loop counter to print the values (you should need
only 1 print statement in your loop).
(Hint: The Fibonacci number looks like this: 1 1 2 3 5 8 13 21
34... )
Submit your program in the dropbox called Fibonacci
Sequence and complete Program Quiz 3 using your
program.
Fibonacci Sequence
1 1 2 3 5 8 13 21 34 ...
F1 = 1
F2 = 1
Fn = Fn-1 + Fn-2
Fibonacci Sequence
Program
assumes that
input of 2 or
higher is used.
begin
Initializing first two Fibonacci numbers;
Enter # of Fibonacci numbers, num
Print first two Fibonacci numbers
counter = 3;
counter<=num
false
end
true
Compute next Fibonacci number;
Print the Fibonacci number;
counter = counter +1;
Fibonacci Sequence
Program
assumes that
input of 2 or
higher is used.
begin
last = 1;
current = 1;
Enter # of Fibonacci numbers, num
Print first two Fibonacci numbers
counter = 3;
next = last + current;
counter<=num
false
end
true
Print the Fibonacci number next;
last = current;
current = next;
counter = counter +1;