1. EECE.2160 ECE Application Programming
Instructor:
Dr. Michael Geiger
Summer 2018
Lecture 2:
Operators
Basic variable output with printf()
Basic variable input with scanf()

2. ECE Application Programming: Lecture 2
Lecture outline
Announcements/reminders
Program 1 due Friday, 5/25
10 points: register for access to the course textbook
10 points: introduce yourself to your instructor
30 points: complete simple C program
Sign up for the course discussion group!
Review
Basic program structure
Data types
Variables
Today’s lecture
Operators
Basic variable output with printf()
Basic variable input with scanf()
11/8/2019
ECE Application Programming: Lecture 2

3. Review: Basic C program structure
Preprocessor directives
#include: typically used to specify library files
Main program
Starts with: int main() or void main()
Enclosed in block: specified by { }
Ends with return 0;
Indicates successful completion
Optional if main() is void
Doesn’t return value: return;
Basic output
Call printf(<string>);
<string> can be replaced by characters enclosed in double quotes
May include escape sequence, e.g. \n (new line)
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ECE Application Programming: Lecture 2

4. ECE Application Programming: Lecture 2
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 2

5. ECE Application Programming: Lecture 1
Variables - assigning
varname = expression;
Declared variable
single variable on left side of =
expression
any legal expression
Expression can be constant, variable, function call, arithmetic operation, etc.
Variable type (int, float, etc) and expression result type should match
If not, funny things can happen ...
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ECE Application Programming: Lecture 1

6. ECE Application Programming: Lecture 1
Variables (cont.)
var name
memory loc
main() { float hours, payrate; float grosspay; int j;
hours = 40.0;
hours
40.0
4278
payrate ?
427C
4280
grosspay ? j ?
4284
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ECE Application Programming: Lecture 1

7. ECE Application Programming: Lecture 1
Variables (cont.)
var name
memory loc
main() { float hours, payrate; float grosspay; int j;
hours = 40.0;
payrate = 20.00;
hours
40.0
4278
payrate
20.0
427C
4280
grosspay ? j ?
4284
11/8/2019
ECE Application Programming: Lecture 1

8. ECE Application Programming: Lecture 1
Variables (cont.)
var name
memory loc
main() { float hours, payrate; float grosspay; int j;
hours = 40.0;
payrate = 20.00;
grosspay = hours * payrate;
hours
40.0
4278
payrate
20.0
427C
4280
grosspay
800.00 j ?
4284
note: referencing a variable only "reads" it (non-destructive). Assigning to a variable overwrites whatever was there (destructive).
11/8/2019
ECE Application Programming: Lecture 1

9. ECE Application Programming: Lecture 1
Variables (cont.)
var name
memory loc
main() { float hours, payrate; float grosspay; int j;
hours = 40.0;
payrate = 20.00;
grosspay = hours * payrate;
j = 5;
hours
40.0
4278
payrate
20.0
427C
4280
grosspay
800.00 j 5
4284
note: referencing a variable only "reads" it (non-destructive). Assigning to a variable overwrites whatever was there (destructive).
11/8/2019
ECE Application Programming: Lecture 1

10. ECE 160 - Introduction to Computer Engineering I
02/09/2005
Variables (cont.)
var name
memory loc
main() { float hours, payrate; float grosspay; int j;
hours = 40.0;
payrate = 20.00;
grosspay = hours * payrate
j = 5;
j = j + 1;
hours
40.0
4278
payrate
20.0
427C
4280
grosspay
800.00 j
5 6
4284
note: referencing a variable only "reads" it (non-destructive). Assigning to a variable overwrites whatever was there (destructive).
11/8/2019
ECE Application Programming: Lecture 1
(c) 2005, P. H. Viall

11. ECE Application Programming: Lecture 1
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 1

12. ECE Application Programming: Lecture 1
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
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ECE Application Programming: Lecture 1

13. Arithmetic Operations
Operator
Addition +
Subtraction -
Multiplication *
Division /
Modulus Division
(Remainder) %
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ECE Application Programming: Lecture 2

14. 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 2

15. ECE Application Programming: Lecture 2
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 2

16. 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 2

17. ECE Application Programming: Lecture 2
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 2

18. 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 2

19. ECE Application Programming: Lecture 2
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 2

20. 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 2

21. ECE Application Programming: Lecture 2
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 2

22. 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 2

23. ECE Application Programming: Lecture 2
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 2

24. ECE Application Programming: Lecture 2
Example: printf()
Show the output from each programs(assume #include <stdio.h> for all)
void main() {
int i=2, j=3, k, m;
k = j * i;
m = i + j;
printf("%d %d %d %d\n", i, j, k, m); }
double f, g;
f = 1.0 / 4.0;
g = f * 20;
printf("f = %lf,\ng = %.2lf\n", f, g);
int a = 5, b = 2;
printf("Output%doesn't%dmake%dsense", a, b, a + b);
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ECE Application Programming: Lecture 2

25. ECE Application Programming: Lecture 2
Example solution
void main() { int i=2, j=3, k, m; k = j * i; k = 2 * 3 = 6 m = i + j; m = = 5 printf("%d %d %d %d\n", i, j, k, m); } Output:
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ECE Application Programming: Lecture 2

26. Example solution (cont.)
void main() { double f, g; f = 1.0 / 4.0; f = 0.25 g = f * 20; g = 0.25 * 20 = 5 printf("f = %lf,\ng = %.2lf\n", f, g); } Output: f = , g = 5.00 (remember, 6 places after decimal point printed by default with floating-point data)
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ECE Application Programming: Lecture 2

27. Example solution (cont.)
void main() {
int a = 5, b = 2;
printf("Output%doesn't%dmake%dsense",
a, b, a + b); }
Output: Output5oesn't2make7sense
(Every %d gets replaced with a number, which is underlined above to show what happens—in practice, the console isn’t going to underline your output!)
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ECE Application Programming: Lecture 2

28. ECE Application Programming: Lecture 2
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 2

29. ECE Application Programming: Lecture 2
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
11/8/2019
ECE Application Programming: Lecture 2
(c) 2005, P. H. Viall

30. ECE Application Programming: Lecture 2
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 2

31. ECE Application Programming: Lecture 2
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 2

32. ECE Application Programming: Lecture 2
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 2

33. ECE Application Programming: Lecture 2
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 2

34. ECE 160 - Introduction to Computer Engineering I
02/09/2005
scanf() function
Used to get input from user
Returns number of items successfully assigned
First argument is format specifiers
Essentially same as printf() format string
Every format specifier (%d, %lf, etc.) corresponds to an input value to be read
Format string can contain other characters, which will be ignored if they are present
If they’re not, you have a problem …
Remaining arguments are variable addresses
Use “address of” operator: &
For example, given: int a;
 The address of a is: &a
11/8/2019
ECE Application Programming: Lecture 2
(c) 2005, P. H. Viall

35. ECE 160 - Introduction to Computer Engineering I
02/09/2005
scanf() function
Documentation info:
int scanf(const char *format [,argument] ...)
format - is format specifiers similar to printf() specifiers
arguments - are ADDRESSES of where to store what the user enters
11/8/2019
ECE Application Programming: Lecture 2
(c) 2005, P. H. Viall

36. ECE 160 - Introduction to Computer Engineering I
02/09/2005
scanf() function
int hours; float rate;
scanf("%d %f",&hours,&rate);
If user types:
34 5.7
hours ?
1284
rate ?
1288
hours 34
1284
rate
5.7
1288
11/8/2019
ECE Application Programming: Lecture 2
(c) 2005, P. H. Viall

37. scanf() format strings
scanf() will skip space characters for all types but %c
Read input until it finds something that’s not a space, then see if it matches the desired type
If type matches, value will be stored in specified variable
If type doesn’t match, nothing stored; function stops
Space in string only matters if using %c
%c will read any character
Includes spaces, newlines, etc.
Example: given scanf("%d%c", &i, &c);
Input: 3a  i = 3, c = 'a'
Input: 3 a  i = 3, c = ' '
Input: 3
a  i = 3, c = '\n' (assuming newline directly after 3)
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ECE Application Programming: Lecture 2

38. ECE Application Programming: Lecture 2
scanf() return value
scanf() returns # of successfully read items
Ex.: given scanf("%d%d", &x, &y);
Input: 3 7  x = 3, y = 7, return value = 2
Input:  x = 3, y = 7, return value = 2
Input:  x = 3, y = ?, return value = 1
y is unchanged
Input: x1 7  x = ?, y = ?, return value = 0
x, y both unchanged
Can assign return value to variable
Example:
int numRead; // # input values read
numRead = scanf("%d%d", &x, &y);
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ECE Application Programming: Lecture 2

39. ECE Application Programming: Lecture 2
Example
Variables: int i; double d; char c;
What values are read for each of the following inputs and scanf() calls? Assume the input is as follows:
scanf("%d%lf", &i, &d)
scanf("%d %lf", &i, &d)
scanf("%lf%d", &d, &i)
scanf("%d%c", &i, &c)
scanf("%d %c", &i, &c)
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ECE Application Programming: Lecture 2

40. ECE Application Programming: Lecture 2
Example solution
What values are read for each of the following inputs and scanf() calls?
scanf("%d%lf", &i, &d)
 i = 34, d = 5.7
scanf("%d %lf", &i, &d)
scanf("%lf%d", &d, &i)
 d = 34, i = 5
scanf("%d%c", &i, &c)
 i = 34, c = ' ' (space)
scanf("%d %c", &i, &c)
 i = 34, c = '5'
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ECE Application Programming: Lecture 2

41. Using scanf() and printf() together
ECE Introduction to Computer Engineering I
02/09/2005
Using scanf() and printf() together
#include <stdio.h>
int main() { int hours; float rate; float grosspay; printf("Enter hours: "); scanf("%d",&hours); printf("Enter pay rate: "); scanf("%f",&rate); grosspay = hours * rate; printf("You earned $%f\n",grosspay); }
11/8/2019
ECE Application Programming: Lecture 2
(c) 2005, P. H. Viall

42. scanf() function - Payroll Ver 2
ECE Introduction to Computer Engineering I
02/09/2005
scanf() function - Payroll Ver 2
#include <stdio.h>
int main() { double hours; double rate; double grosspay; printf("Enter hours: "); scanf("%lf",&hours); printf("Enter pay rate: "); scanf("%lf",&rate); grosspay = hours * rate; printf("You earned $%lf\n",grosspay); }
11/8/2019
ECE Application Programming: Lecture 2
(c) 2005, P. H. Viall

43. ECE Application Programming: Lecture 2
Final notes
Next time
PE1 (Flowcharts)
Conditional statements
Reminders:
Program 1 due Friday, 5/25
10 points: register for access to the course textbook
10 points: introduce yourself to your instructor
30 points: complete simple C program
Sign up for the course discussion group!
11/8/2019
ECE Application Programming: Lecture 2