1. EECE.2160 ECE Application Programming
Instructors:
Dr. Michael Geiger
Fall2018
Lecture 28
P6 overview

2. ECE Application Programming: Lecture 28
Lecture outline
Announcements/reminders
Program 6 due Monday, 11/19
Program 5 regrades due Monday, 11/26
Today’s lecture
Program 6 overview
6/12/2019
ECE Application Programming: Lecture 28

3. ECE Application Programming: Lecture 28
Program 6 overview
Read lines of text from input
Use array to track # times each letter occurs in input text
Use array contents to generate bar graph showing relative frequencies of each letter
Gives you practice using arrays and functions
Does not require the use of strings
6/12/2019
ECE Application Programming: Lecture 28

4. Overall program structure
main() should contain
Array to track letter frequency: int myHist[26]
Maximum value in array: int myMax
Used to determine height of histogram output
Program uses four commands
‘R’, ‘r’: Read a single line of input
Call ReadText(myHist, &myMax); to read line
‘P’, ‘p’: Print histogram
Call DrawHist(myHist, myMax); to print histogram
‘C’, ‘c’: Clear histogram (and max value)
‘Q’, ‘q’: Quit program
Only error checking: invalid command
All other input: reading characters, so no formatting errors
You may ignore some characters, but they’re not errors
6/12/2019
ECE Application Programming: Lecture 28

5. ECE Application Programming: Lecture 28
ReadText() algorithm
Read a single character
Do not use a string for the input
If that character is a letter, update the appropriate entry in the histogram
If that character is not a newline, return to step 1 and read another character
Function should also update max value, either as it reads characters or after reading all input characters
6/12/2019
ECE Application Programming: Lecture 28

6. ECE Application Programming: Lecture 28
ReadText() hints
<ctype.h> functions will help in ReadText()
isalpha(ch): returns “true” if ch is letter, “false” otherwise
Directly applies to one step on previous slide
Converting each letter to same case makes it easier to find appropriate entry in histogram
toupper(ch): returns uppercase letter if ch is lowercase letter; returns original ch otherwise
toupper('x') = 'X'; toupper('A') = 'A'
tolower(ch): returns lowercase letter if ch is uppercase letter; returns original ch otherwise
tolower('x') = 'x'; tolower('A') = 'a'
6/12/2019
ECE Application Programming: Lecture 28

7. ReadText() hints (continued)
Finding appropriate entry in histogram does not require conditional statement
You shouldn’t need to compare your input character to anything to find correct array index
Very basic “transformation” between ASCII value of letter and histogram index
Can treat a char variable as either
Character to be printed, or
Integer value corresponding to printed character
ASCII values
Uppercase and lowercase letters separate
Each set of letters is consecutive
‘A’ = 65, ‘B’ = 66, … ‘Z’ = 90
‘a’ = 97, ‘b’ = 98, … ‘z’ = 122
6/12/2019
ECE Application Programming: Lecture 28

8. What not to do in ReadText()
Using brute force method to find appropriate histogram index will incur -10 deduction
Brute force methods basically compare input letter to all possible letters
Prohibited brute force method #1: giant conditional statement
switch(ch) { // ch = input char
case ‘A’: case ‘a’:
// modify histo[0]
break;
case ‘B’: case ‘b’:
// modify histo[1]
etc … }
6/12/2019
ECE Application Programming: Lecture 28

9. What not to do in ReadText() (cont)
Using brute force method to find appropriate histogram index will incur -10 deduction
Brute force methods basically compare input letter to all possible letters
Prohibited brute force method #2: loop
char test = ‘A’;
for (i = 0; i < 26; i++) {
if (ch == test)
// modify histo[i]
test++; // Change test
} // to next letter
6/12/2019
ECE Application Programming: Lecture 28

10. ECE Application Programming: Lecture 28
DrawText() algorithm
For each row of output
For each entry in histogram
If current entry is at least row #, print “| “ (bar & space)
Otherwise, print “ “ (two spaces)
Must print bar graph from top to bottom
# rows based on max value in histogram
Printing spaces necessary to get everything to line up
6/12/2019
ECE Application Programming: Lecture 28

11. ECE Application Programming: Lecture 28
Final notes
Next time
Review structures
PE3
Reminders:
Program 5 regrades due Monday, 4/9
Program 6 due today
Program 7 due Friday, 4/13
6/12/2019
ECE Application Programming: Lecture 28