1. EECE.2160 ECE Application Programming
Instructors:
Dr. Michael Geiger
Fall 2018
Lecture 4:
Operators; printf() basics

2. Announcements/reminders
Program 1 due today
10 points: register for access to the course textbook
10 points: introduce yourself to your instructor
30 points: complete simple C program
DON’T FORGET BLACKBOARD “ASSIGNMENT”
Program 2 to be posted; due Friday, 9/21
Won’t cover scanf() until Friday, 9/14
Suggestions
Start working on program design now
Test equations/output by assigning values to variables
Once you understand scanf(), replace assignments with scanf() calls
5/26/2019
ECE Application Programming: Lecture 4

3. ECE Application Programming: Lecture 4
Lecture outline
Review
Variables
Today’s lecture
Operators
Basic variable output with printf()
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ECE Application Programming: Lecture 4

4. ECE Application Programming: Lecture 4
Review: Variables
Four basic data types
int, float, double, char
Variables
Have name, type, value, memory location
Variable declarations: examples
int x;
float a, b;
double m = 2.35;
Assignments: examples with variables above
a = 7.5;
x = a + 2; x = 9, not 9.5
m = m – 1; m = 1.35
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ECE Application Programming: Lecture 4

5. ECE Application Programming: Lecture 4
Example: Variables
What values do w, x, y, and z have at the end of this program?
int main() {
int w = 5;
float x;
double y;
char z = ‘a’;
x = 8.579;
y = -0.2;
w = x;
y = y + 3;
z = w – 5;
return 0; }
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ECE Application Programming: Lecture 4

6. ECE Application Programming: Lecture 4
Example solution
int main() {
int w = 5;
float x;
double y;
char z = ‘a’;
x = 8.579;
y = -0.2;
w = x;
y = y + 3;
z = w – 5;
return 0; }
w = 5
z = ‘a’ (ASCII value 97)
x = 8.579
y = -0.2
w = 8 (value is truncated)
y = (-0.2) + 3 = 2.8
z = 8 – 5 = 3
(ASCII value 3 = "end of text" character)
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ECE Application Programming: Lecture 4

7. Arithmetic Operations
Operator
Addition +
Subtraction -
Multiplication *
Division /
Modulus Division
(Remainder) %
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ECE Application Programming: Lecture 4

8. Results of arithmetic operations
(using non-integer makes result double precision)
.08*
12.0/
10/5 2
10/3 3 (not 3.333…)
10 % 3 1
12 % 5 2
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ECE Application Programming: Lecture 4

9. ECE Application Programming: Lecture 4
Operators (cont.)
Previous operators are binary
Deal with two values
C also supports some unary operators
For now, we’ll simply deal with unary negation
e.g., if x = 3, the statement
-x;
produces the value -3
Important note: The statement above does not change the value of x
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ECE Application Programming: Lecture 4

10. Operators and variables
Operators can be used either with constants or variables
Examples:
int main() {
int w, x, y, z;
w = 3 + 2; // w = 5
x = -w; // x = -5
y = x – 7; // y = -12
z = w * y; // z = -60
return 0; }
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ECE Application Programming: Lecture 4

11. ECE Application Programming: Lecture 4
Operators (cont.)
More complex statements are allowed
e.g. x = ;
Parentheses help you prioritize parts of statement
Makes difference with order of operations
x = * 3;
is different than
x = (1 + 2) * 3;
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ECE Application Programming: Lecture 4

12. Example: Arithmetic operations
Evaluate each of the following expressions, including the type (int or double) in your answer
19/3
3/19
19%3
3%19
5 + 7/2 /2 /2
5 * 3 % 3 / / 2
5 * (3 % 3) / /3
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ECE Application Programming: Lecture 4

13. ECE Application Programming: Lecture 4
Example solution
19/3 = 6 (integer division)
3/19 = 0 (integer division)
19%3 = 1
3%19 = 3
5 + 7/2 = = 8
/2 = = 8.0
/2 = = 8.5
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ECE Application Programming: Lecture 4

14. Example solution (cont.)
For each of the following, underlined part(s) evaluated first at each step
5 * 3 % 3 / / 2
= 15 % 3 /
= 0 /
= = 19
5 * (3 % 3) / /3
= 5 * 0 / /3
= 0 /
= = 17.0
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ECE Application Programming: Lecture 4

15. ECE Application Programming: Lecture 4
I/O basics
Need ability to
Print variables (or results calculated using them)
Read values from input
Output: printf()
Already seen basics
Input: scanf()
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ECE Application Programming: Lecture 4

16. Basic printf() formatting
To print variables/constants, insert %<type> (format specifier) in your format string
%c: single character
%d or %i: signed decimal integer
%f: float; %lf: double
Prints 6 digits after decimal point by default
To control # digits, use precision
"%.4lf" prints with 4 digits (4th digit rounds)
"%.0lf" prints with 0 digits (round to nearest integer)
When printed, format specifier is replaced by value of corresponding expression
If x is 3, printf("x + x = %d", x + x) prints:
x + x = 6
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ECE Application Programming: Lecture 4

17. ECE Application Programming: Lecture 4
printf() example
float a=67.49,b= ; printf("hello %f there %f\n",a,b); printf("%f%f%f%f\n",a,a,b,b); printf("a=%.2f, b=%.1f",a,b); printf("Cool huh?\n");
Printed:
hello there a=67.49, b=10.0Cool huh?
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ECE Application Programming: Lecture 4

18. ECE Application Programming: Lecture 4
printf() details
Detailed slides on printf() follow
Skip these if you don’t want to go overboard with the full details of how the function works
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ECE Application Programming: Lecture 4

19. ECE Application Programming: Lecture 4
02/07/2005
printf()
Documentation info:
int printf(const char *format [,argument] ...)
Type of value returned
Name of function
( ) indicate printf is a function
First arg type and formal name (required, since no brackets)
next argument type and name (in this case it may be any simple type)
[ ] indicate optional arguments
… indicates previous argument repeated zero or more times
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ECE Application Programming: Lecture 4
(c) 2005, P. H. Viall

20. ECE Application Programming: Lecture 4
printf()
int printf(const char *format [,argument] ...)
Type of value returned (int in this case)
All functions return at most one value.
The type void is used to indicate a function returns no value
There is no requirement to use the value returned.
The printf() function returns the number of characters printed (including spaces); returns negative value if error occurs.
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ECE Application Programming: Lecture 4

21. ECE Application Programming: Lecture 4
printf()
int printf(const char *format [,argument] ...)
Name of function; printf( ) in this case
A function name is ALWAYS followed by a set of (), even if the function takes no arguments
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ECE Application Programming: Lecture 4

22. ECE Application Programming: Lecture 4
printf()
int printf(const char *format [,argument] ...)
Type (const char *) and name (format) of first argument
For the moment, const char * can be thought of as a series of characters enclosed in double quotes
The name format may be thought of as a code indicating how the arguments are to be interpreted, and how the output should look.
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ECE Application Programming: Lecture 4

23. ECE Application Programming: Lecture 4
printf()
int printf(const char *format [,argument] ...)
zero of more optional arguments, each preceded by a comma
zero because of the …
optional because of the [ ]
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ECE Application Programming: Lecture 4

24. ECE Application Programming: Lecture 4
Final notes
Next time
scanf()
Reminders:
Program 1 due today
10 points: register for access to the course textbook
10 points: introduce yourself to your instructor
30 points: complete simple C program
DON’T FORGET BLACKBOARD “ASSIGNMENT”
Program 2 to be posted; due Friday, 9/21
Won’t cover scanf() until Friday, 9/14
Suggestions
Start working on program design now
Test equations/output by assigning values to variables
Once you understand scanf(), replace assignments with scanf() calls
5/26/2019
ECE Application Programming: Lecture 4