ENGR/CS 101 CS Session
Lecture 5
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Log into Windows/ACENET (reboot if in Linux)
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Start Microsoft Visual Studio 2010
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Has everyone finished the exercise from last
time so that their program will encipher one
uppercase letter?
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Outline
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Problem: input more than one letter at a time
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C# programming language
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Strings and indexing
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For-loops
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Problem Specification
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Today we will modify
our GUI program to
accept and encipher
whole words in
uppercase letters,
rather than just a
single letter.
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Strings
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Enciphering one letter at a time is tedious.
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Want to handle an entire word (or sentence).
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Design: Encipher each letter in the plaintext
box. Need to use a string, rather than a
character.
A string is a sequence of characters. The Text
property of a textbox is a string. We can
access it using
plaintextBox.Text;
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Indexing
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Individual characters are accessed by giving
the index of the character in the string.
'A' 'C' 'E' 'S'
[0] [1] [2] [3]
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The syntax for accessing an individual
character is: <string var>[<index>]
Example: plaintextBox.Text[2] is the
character 'E'
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Indexing
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As shown on the previous slide, the indexes of
a string start at 0.
We can use indexing to get the key letter, too.
It is the first character in the keyBox.Text string.
shiftKey = keyBox.Text[0];
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For-Loops
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An index used to access string characters may
be any integer expression. In particular, it may
be a variable. So we can make a variable
count from 0 to the last index of the string to
access each character one at a time.
We do this with a for-loop (also called a
counting loop). A for-loop consists of 4 parts:
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Lecture 5
A loop control variable and its initial value
A loop condition test of the loop control variable
A loop control variable update
A body of steps that are repeated
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For-Loops
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For our Encipher handler design, we would use
the for-loop in the following way (changes in
bold).
1. Clear ciphertext box
2. For index i from 0 to length of plaintextBox.Text by 1
2.1. Get the shift key from keyBox.Text[0]
2.2. Get the plaintext letter from plaintextBox.Text[i]
2.3. Compute the corresponding ciphertext letter
2.4. Append the ciphertext letter to the ciphertext box.
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The loop control variable is i, with initial value of
0. It is updated by adding 1 to i until it reaches
the length of the plaintextBox.Text string.
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For-Loops
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The syntax of a for-loop in C# is:
for ( <lcv declaration/initialization>; <loop condition>; <lcv update statement>)
{
<steps to be repeated>
}
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For our program, we would write:
for (int i = 0; i < plaintextBox.Text.Length; i++)
{
// steps to be repeated
}
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For-Loops
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Some notes:
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Lecture 5
Clearing the ciphertext box should only be done
once, so it happens before the loop starts.
The length of a string is obtained using the Length
property, accessed with the dot notation (e.g.,
plaintextBox.Text.Length). It is one more
than the last index, so the loop condition uses <
rather than <=. More on conditions next class.
The loop body is computing the ciphertext letter and
appending it to the ciphertext box in the same way
as the last program.
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In-class Exercise
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Modify today's program to encipher a line of
plaintext and display the ciphertext using a forloop (code shown on the last two slides):
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Lecture 5
Add a for-loop around the statements that
correspond to steps 2.1 - 2.4 on Slide 8. Note that
the body of the for-loop is exactly the same code as
before, except for computing shiftKey and plainLetter using indexing instead of
char.Parse( )
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In-class Exercise
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Test your program with the string "ACES" =>
"IKMA"
See what happens if you type in "Go Aces!"
We'll fix this "problem" next class.
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Putting the Code Together
// The modified parts from the previous program are bold
// Everything is the same on this page.
// Variable declarations
char shiftKey, // key letter plainLetter, // user input
cipherLetter;// result
int shiftNumber, // # of shift places
index; // of cipher letter
// Clear the result box
ciphertextBox.Text = "";
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Putting the Code Together
// For each letter in plaintext box, indexed by i
for (int i = 0; i < plaintextBox.Text.Length; i++)
{
// Use indexing to get shift key and plaintext letter
shiftKey = keyBox.Text[0]; plainLetter = plaintextBox.Text[i];
// The rest is exactly the same as before // Compute the corresponding ciphertext letter
shiftNumber = shiftKey ­ 'A';
index = (plainLetter ­ 'A' + shiftNumber) % 26;
cipherLetter = (char)((int)'A' + index);
// Append the enciphered letter to ciphertext box
ciphertextBox.AppendText(cipherLetter.ToString());
} // don't forget the closing curly brace
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