1. 16.216 ECE Application Programming
Instructor: Dr. Michael Geiger
Fall 2011
Lecture 3: Number system basics
Representing data in C

2. ECE Application Programming: Lecture 3
Lecture outline
Announcements/reminders
Course home page:
Disc. gp.:
Should have been invited
All course announcements here
Be careful when replying to !
Assignment 1 due 11:59 PM today
File names matter!
Submit only .c file
Assignment 2 coming Monday, due 9/19
Review: C program basics
Today: representing data in C
Number systems
Variables and constants
Will start operators (time permitting)
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ECE Application Programming: Lecture 3

3. Review: Basic C program structure
Preprocessor directives
#include: typically used to specify library files
#define: generally used to define macros or constants
Main program
Starts with: int main() (or void main())
Enclosed in block: specified by { }
Ends with return 0;
Basic output
Call printf(<string>);
<string> can be replaced by characters enclosed in double quotes
May include escape sequence, e.g. \n (new line)
Comments
Improve readability
Multi-line: enclosed by /* */
Single line: starts with // , runs to end of line
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ECE Application Programming: Lecture 3

4. Number representation: bases
Humans operate in decimal (base 10)
Why don’t computers?
Computers operate in binary (base 2)
Each digit is a bit (binary digit)
Can also use octal (base 8) or hexadecimal (base 16)
Hex digits: 0-15, with A = 10, B = 11, C = 12, D = 13, E = 14, F = 15
Base conversion
Binary  hex: start with LSB and make 4-bit groups
e.g = 1B716 = 0x1B7
Note that an extra 0 is implied for the first group: 0010001
Binary  decimal: multiply bit 0 by 20, bit 1 by 21, etc.
e.g = (0 x 23) + (1 x 22) + (1 x 21) + (1 x 20) =
= = 710
Decimal  binary / hex: repeated integer division, where remainder gives you each digit, starting with LSB
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ECE Application Programming: Lecture 3

5. Decimal  binary / hex example
3510 = ?2
35 / 2 = 17 R1 (LSB)
17 / 2 = 8 R1
8 / 2 = 4 R0
4 / 2 = 2 R0
2 / 2 = 1 R0
1 / 2 = 0 R1 (MSB)
 3510 =
3510 = ?16
35 / 16 = 2 R3 (LS digit)
2 / 16 = 0 R2 (MS digit)
 3510 = 2316
Double-check: does this answer match the binary value we found?
 2316
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ECE Application Programming: Lecture 3

6. ECE Application Programming: Lecture 3
Base Conversions
Convert 2BAD|16 to Base 10 2 B A D |16 = ? |10 | | | | | | | `--- D x 160 = 13 x 1 = 13 | | `----- A x 161 = 10 x 16 = 160 | ` B x 162 = 11 x 256 = 2816 ` x 163 = 2 x 4096 = Therefore, 2BAD|16 = 11181|10
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ECE Application Programming: Lecture 3

7. ECE Application Programming: Lecture 3
Base Conversions
Convert 20DD|16 to Base 10 ??
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ECE Application Programming: Lecture 3

8. ECE Application Programming: Lecture 3
Base Conversions
Convert 20DD|16 to Base D D |16 = ? |10 | | | | | | | `--- D x 160 = 13 x 1 = 13 | | `----- D x 161 = 13 x 16 = 208 | ` x 162 = 0 x 256 = 0 ` x 163 = 2 x 4096 = Therefore, 20DD|16 = 8413|10
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ECE Application Programming: Lecture 3

9. ECE Application Programming: Lecture 3
Base Conversions
Convert FACE|16 to Base 10 ??
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ECE Application Programming: Lecture 3

10. ECE Application Programming: Lecture 3
Base Conversions
Convert FACE|16 to Base 10 F A C E |16 = ? |10 | | | | | | | `--- E x 160 = 14 x 1 = 14 | | `----- C x 161 = 12 x 16 = 192 | ` A x 162 = 10 x 256 = 2560 ` F x 163 = 15 x 4096 = Therefore, FACE|16 = 64206|10
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ECE Application Programming: Lecture 3

11. Base Conversions 10101110001101|2=?|16 ##10 1011 1000 1101 2 B 8 D
Dec
Bin
Oct
Hex
0000 00 1
0001 01 2
0010 02 3
0011 03 4
0100 04 5
0101 05 6
0110 06 7
0111 07 8
1000 10 9
1001 11
1010 12 A
1011 13 B
1100 14 C
1101 15 D
1110 16 E
1111 17 F
|2=?|16 ##
2 B D
|2=2B8D|16
NOTE: # is a place holder for zero
Work from right to left Divide into 4 bit groups
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ECE Application Programming: Lecture 3

12. ECE Application Programming: Lecture 3
Base Conversions
Dec
Bin
Oct
Hex
0000 00 1
0001 01 2
0010 02 3
0011 03 4
0100 04 5
0101 05 6
0110 06 7
0111 07 8
1000 10 9
1001 11
1010 12 A
1011 13 B
1100 14 C
1101 15 D
1110 16 E
1111 17 F
FACE|16=?|2
F A C E
\ FACE|16= |2
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ECE Application Programming: Lecture 3

13. Example 1: Base conversions
Perform the following base conversions
1110 = ?2 = ?16
3710 = ?2 = ?16
1116 = ?10
0x2F = ?2 = ?10
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ECE Application Programming: Lecture 3

14. ECE Application Programming: Lecture 3
Example 1 solution
Perform the following base conversions
1110 = = B16
3710 = = 2516
1116 = 1710
0x2F = = 4710
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ECE Application Programming: Lecture 3

15. ECE Application Programming: Lecture 3
Representing data in C
Two major questions (for now)
What kind of data are we trying to represent?
Data types
Can the program change the data?
Constants vs. variables
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ECE Application Programming: Lecture 3

16. Four Types of Basic Data
Integer int
Floating point (single precision) float
Double Precision double
Character char
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ECE Application Programming: Lecture 3

17. ECE Application Programming: Lecture 3
Integer Constants
Any positive or negative number without a decimal point (or other illegal symbol).
Legal values:
Illegal values: 2,523 (comma) 6.5 (decimal point) $59 (dollar sign) 5. (decimal point)
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ECE Application Programming: Lecture 3

18. Range of Integers (Machine Dependent)
unsigned signed
char 0   +127
(8 bits)
short int 0   short
(16 bits)
int 0 to  long long int
(32 bits)
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ECE Application Programming: Lecture 3

19. float/double Constants
Any signed or unsigned number with a decimal point
Legal values:
Legal (exponential notation): 1.624e e e23 1.0e e e e e+7
Illegal:
$ , E5
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ECE Application Programming: Lecture 3

20. float/double Constants
Range of float (32 bits) ± E – 38 ± E + 38
Range of double (64 bits) ± E – 308 ± E + 308
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ECE Application Programming: Lecture 3

21. ECE Application Programming: Lecture 3
Character Constants
Stored in ASCII or UNICODE
Signified by single quotes (’ ’)
Valid character constants ’A’ ’B’ ’d’ ’z’ ’1’ ’2’ ’!’ ’+’ ’>’ ’?’ ’ ’ ’#’
Invalid character constants ’GEIGER’ ’\’ ’CR’ ’LF’ ’’’ ’’’’ ’”’ ”Q”
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ECE Application Programming: Lecture 3

22. Character Escape Sequences
Meaning
’\b’
Backspace
’\’’
Single quote
’\n’
Newline
’\”’
Double quote
’\t’
Tab
’\nnn’
Char with octal value nnn
’\\’
Backslash
’\xnn’
Char with hex value nn
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ECE Application Programming: Lecture 3

23. Using #define with constants
Often makes sense to give constant value a symbolic name for readability
Don’t want constant wasting memory space
Use #define to define a macro
Macro: named code fragment; when name is used, the preprocessor replaces name with code
Syntax: #define <name> <code>
Common use: defining constants
By convention, start constant values with capital letters
e.g. #define NumberOne 1
At compile time, all uses of “NumberOne” in your program are replaced with “1” and then compiled
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ECE Application Programming: Lecture 3

24. ECE Application Programming: Lecture 3
Next time
Variables
Basic numeric output
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ECE Application Programming: Lecture 3