ECE 160 – Comp Programming I
Data Structures with C
Prof. P. H. Viall
Office: Research 217
Tel: (508) 999-8240
Web: http://portal.ece.umassd.edu/~pviall/ece160
Book:
Computer Science: A Structured Programming Approach Using C
B. A. Forouzan and Richard F. Gilberg
Second Edition (or latest); Brooks/Cole, 2001.
Tentative Grading
• Exams (75%)
Best 3 of 4 exams (no makeups)
Exams are 1 hour in class exams
Exams MAY be open book/note
• Homework/Programs/Labs (25%)
Note – scheme is tentative – grading may be
adjusted ±2%
Working Together (programs)
• is OK - TO A POINT
• Working together appropriately means…
– Talking about a problem in general terms
– One person helping another to fix a particular error
• Working together INAPPROPRIATLY means…
– One person working, and one copying.
– Person A does program 1 and person B copies, then
person B does program 2 and A copies.
– One person using the mouse and one using the
keyboard to input a program, and then submit two
copies of the same program (with just the persons name
altered).
Working Together (programs)
• If you do work together, please indicate
who you were working with
• Penalty for INAPPROPRIATLY working
together
– first offense: written warning; both on
homework and email to UMD account
– second offense: grade of homework halved
– third offense: grade of zero on homework
– further offenses will result in lowering of
course grade, possibly to 'F'
UMD Policy on Academic Dishonesty
Taken from UMD Catalog 2003-2004, page 51
•
•
•
•
All UMass Dartmouth students are expected to maintain high standards of academic integrity
and scholarly practice. The University does not tolerate academic dishonesty of any variety,
whether as a result of a failure to understand proper academic and scholarly procedure or as an
act of intentional dishonesty.
A student found guilty of academic dishonesty is subject to severe disciplinary action which
may include expulsion from the University. Refer to the Student Handbook and Student Judicial
Code for due process.
A high standard of academic integrity promotes the pursuit of truth and learning and respect for
the intellectual accomplishments of others. These are values that are fundamental to the mission
of this University. Such values are undermined by academic dishonesty.
Academic dishonesty is defined as attempting to obtain academic credit for work that is not
one’s own. Examples include:
copying another student’s answers on an examination;
obtaining, or attempting to obtain, the answers to an examination in advance;
submitting a paper that was written by someone else;
submitting a paper that includes phrases, sentences and paragraphs that were copied verbatim, or almost
verbatim, from a work written by someone else, without making this clear without indicating that these
words were someone else’s through the use of quotation marks or other appropriate citation conventions;
5.
collaborating on a homework assignment when this has been expressly forbidden by the professor;
6.
using unauthorized materials in completing assignments or examinations;
7.
submitting the same paper for more than one class without the express permission of the instructors
involved.
This list of examples should not be considered exhaustive.
1.
2.
3.
4.
UMD Policy on Academic Dishonesty
•
This definition of academic dishonesty applies to information submitted in other forms besides
paper. Submitting a project of a musical or artistic nature where all or part of the project is
someone else’s work, without acknowledging this fact, constitutes academic dishonesty.
Submitting computer files that do not represent one’s own work is also considered
to be academic dishonesty; examples of computer-based academic dishonesty
would include submitting a computer program or text file created by someone else
as one’s own, or submitting the output of a computer program written by someone
else, and claiming to have written the program that generated the output.
•
•
•
•
•
For all forms of academic dishonesty, students who knowingly allow other students to use their
work are themselves considered to be academically dishonest. Examples would include students
who knowingly allow other students to copy their exam answers, and students who give papers
that they have written to other students so that the other students can submit them for credit.
A faculty member is appointed by the Faculty Senate to act as an Academic Ethical Matters
Facilitator. This individual will offer advice to both students and faculty about the issues involved
in penalizing academic dishonesty, and the process of appealing such penalties.
Penalties assessed by faculty members for academic dishonesty generally consist of a reprimand, a
requirement to resubmit the work in a more acceptable form, a lowering of a grade, failure in the
course in which the alleged infraction took place, or a combination of these.
Instead of (or in addition to) assessing such penalties, a faculty member may refer the matter to the
UMass Dartmouth Student Judiciary. Decisions made by the Student Judiciary may be appealed to
the University Appellate Board.
A form acknowledging that each student has read, understands, and will abide by these standards
will be signed by the student and placed into the student’s permanent file.
Working Together (exams)
• Exams are definitely an individual endeavor.
• You may not work together during exams.
• You may not share any materials during an
exam.
• In general the penalty for cheating on an
exam or otherwise covertly attempting to
raise your grade on an exam shall be an ‘F’
for the course.
Canceled Classes and Tests
• If class is cancelled the day an exam is
scheduled, we will have the exam the next
time the class meets
• If class is cancelled the session prior to the
exam (the day you have to ask questions),
then the next class meeting will be the
“question time”, and the class meeting after
that will be the exam
Course Objectives
1.
To learn how a computer stores data, and conversion between
various number bases
2. To learn the fundamentals of using the c compiler and c
preprocessor
3. To learn how to use a modern environment to create, compile,
execute and debug C programs
4. To learn how to use procedures to modularize a program, and
how to pass parameters by value and by address
5. To learn the syntax of the c language, including arrays,
structures, dynamic allocation, and file access
6. To learn how to use and manipulate strings using the C language
7. To learn to use system libraries within a program
8. To learn how to program, and how to design well written,
maintainable programs
Base Conversions
Convert 123|10 to Base 10 (gotta start someplace)
1 2 3 |10
| | |
| | `--| `----`-------
= ? |10
3 x 100 = 3 x
1 =
3
2 x 101 = 2 x 10 = 20
1 x 102 = 1 x 100 = 100
123
Therefore, 123|10 = 123|10
Base Conversions
Convert 123|8 to Base 10
1 2 3 |8 = ? |10
| | |
| | `--- 3 x 80 = 3 x
| `----- 2 x 81 = 2 x
`------- 1 x 82 = 1 x
1 =
8 =
64 =
Therefore, 123|8 = 83|10
3
16
64
83
Base Conversions
Convert 2BAD|16 to Base 10
2 B A D |16
| | | |
| | | `--| | `----| `------`---------
= ? |10
D
A
B
2
x
x
x
x
160
161
162
163
= 13 x
1
= 10 x
16
= 11 x 256
= 2 x 4096
Therefore, 2BAD|16 = 11181|10
=
=
=
=
13
160
2816
8192
11181
Base Conversions
Convert 20DD|16 to Base 10
??
Base Conversions
Convert 20DD|16 to Base 10
2 0 D D |16
| | | |
| | | `--| | `----| `------`---------
= ? |10
D
D
0
2
x
x
x
x
160
161
162
163
= 13 x
1
= 13 x
16
= 0 x 256
= 2 x 4096
Therefore, 20DD|16 = 8413|10
=
=
=
=
13
208
0
8192
8413
Base Conversions
Convert FACE|16 to Base 10
??
Base Conversions
Convert FACE|16 to Base 10
F A C E |16
| | | |
| | | `--| | `----| `------`---------
= ? |10
E
C
A
F
x
x
x
x
160
161
162
163
=
=
=
=
14
12
10
15
x
1
x
16
x 256
x 4096
Therefore, FACE|16 = 64206|10
=
14
=
192
= 2560
= 61440
64206
Base Conversions
Convert 123|10 to Base 8
Base you are converting to
8 )123
8 )15 R 3
8 )1 R 7
0 R 1
Read Up!
Therefore, 123|10 = 173|8
Base Conversions
Convert 64206|10 to Base 16
Base you are converting to
16 )64206
16 )4012
16 )250
16 )15
16 )0
R
R
R
R
14
12
10
15
(E)
(C)
(A)
(F)
Therefore, 64206|10 = FACE|16
Read Up!
Base Conversions
123|7 = ?|9 (Can’t do directly)
Step 1: 123|7=?|10
1 2 3
| | |
| | `-3x70= 3
| `--2x71=14
`-----1x72=49
66
\ 123|7=66|10
Step 2: 66|10=?|9
9 )66
9 ) 7 R 3
0 R 7
\ 66|10=73|9
\ 123|7=73|9
Base Conversions
• Converting between
bases which are
powers of 2 (base 2,
4, 8, and 16)
• Each Digit in base
4, 8 or 16 represents
2, 3 or 4 bits
Dec
Bin
Base 4
Octal
Hex
0
0000
00
00
0
1
0001
01
01
1
2
0010
02
02
2
3
0011
03
03
3
4
0100
10
04
4
5
0101
11
05
5
6
0110
12
06
6
7
0111
13
07
7
8
1000
20
10
8
9
1001
21
11
9
10
1010
22
12
A
11
1011
23
13
B
12
1100
30
14
C
13
1101
31
15
D
14
1110
32
16
E
15
1111
33
17
F
Base Conversions
Dec
Bin
Bs 4
Oct
Hex
0
0000
00
00
0
1
0001
01
01
1
2
0010
02
02
2
3
0011
03
03
3
4
0100
10
04
4
5
0101
11
05
5
6
0110
12
06
6
7
0111
13
07
7
8
1000
20
10
8
9
1001
21
11
9
10
1010
22
12
A
11
1011
23
13
B
\ 10101110001101|2=25615|8
12
1100
30
14
C
13
1101
31
15
D
NOTE: # is a place holder for zero
14
1110
32
16
E
15
1111
33
17
F
10101110001101|2=?|8
Work from right to left
Divide into 3 bit groups
#10 101 110 001 101
2
5
6
1
5
Base Conversions
Dec
Bin
Bs 4
Oct
Hex
0
0000
00
00
0
1
0001
01
01
1
2
0010
02
02
2
3
0011
03
03
3
4
0100
10
04
4
5
0101
11
05
5
6
0110
12
06
6
7
0111
13
07
7
8
1000
20
10
8
9
1001
21
11
9
10
1010
22
12
A
11
1011
23
13
B
\ 10101110001101|2=2B8D|16
12
1100
30
14
C
13
1101
31
15
D
NOTE: # is a place holder for zero
14
1110
32
16
E
15
1111
33
17
F
10101110001101|2=?|16
Work from right to left
Divide into 4 bit groups
##10 1011 1000 1101
2
B
8
D
Base Conversions
1357|8=?|2
1
3
5
7
001 011 101 111
\ 1357|8=001011101111|2
Note: one need not
write the leading zeros
Dec
Bin
Bs 4
Oct
Hex
0
0000
00
00
0
1
0001
01
01
1
2
0010
02
02
2
3
0011
03
03
3
4
0100
10
04
4
5
0101
11
05
5
6
0110
12
06
6
7
0111
13
07
7
8
1000
20
10
8
9
1001
21
11
9
10
1010
22
12
A
11
1011
23
13
B
12
1100
30
14
C
13
1101
31
15
D
14
1110
32
16
E
15
1111
33
17
F
Base Conversions
FACE|16=?|2
F
A
C
E
1111 1010 1100 1110
\ FACE|16=1111101011001110|2
Dec
Bin
Bs 4
Oct
Hex
0
0000
00
00
0
1
0001
01
01
1
2
0010
02
02
2
3
0011
03
03
3
4
0100
10
04
4
5
0101
11
05
5
6
0110
12
06
6
7
0111
13
07
7
8
1000
20
10
8
9
1001
21
11
9
10
1010
22
12
A
11
1011
23
13
B
12
1100
30
14
C
13
1101
31
15
D
14
1110
32
16
E
15
1111
33
17
F
Base Conversions
A1CE|16=?|8
2
1
C
E
1010000111001110
1 2 0 7 1 6
\A1CE|16=120716|8
Dec
Bin
Bs 4
Oct
Hex
0
0000
00
00
0
1
0001
01
01
1
2
0010
02
02
2
3
0011
03
03
3
4
0100
10
04
4
5
0101
11
05
5
6
0110
12
06
6
7
0111
13
07
7
8
1000
20
10
8
9
1001
21
11
9
10
1010
22
12
A
11
1011
23
13
B
12
1100
30
14
C
13
1101
31
15
D
14
1110
32
16
E
15
1111
33
17
F
Terminology
•
•
•
•
Bit – a 0 or a 1
Nibble – 4 bits
Byte – 8 bits
Word – Machine dependent
(8086/88 = 16 bits; 386/486/Pent = 32 bits)
• Doubleword – Machine dependent
(8086/88 = 32 bits; 386/486/Pent = 64 bits)
Terminology
•
•
•
•
•
•
K – Kilo;
M – Mega;
G – Giga;
T – Tera;
P – Penta;
E – Exa;
210 =
220 =
230 =
240 =
250 =
260 =
1,024
1,048,576
1,073,741,824
1,099,511,627,776
1,125,899,906,842,624
1,152,921,504,606,846,976
Rough Cost for RAM/Disk
R
A
M
1956 1978 1985 2000 2004 2005
K
$19.00
$ 0.16
M
$19000.
$160.00
G
$ .50
$ .20
$500.00
$200.00
80.00
$10.00
$ .83
.10
$9999.99
$830.00
100.00
T
D
i
s
$250
$7.00
k
K
M
G
T
$9999.99
$7000.00
See http://www.littletechshoppe.com/ns1625/winchest.html for more info
ASCII-American Standard Code
for Information Interchange
•
•
•
•
Used to represent characters
Seven bit code used to represent character
‘0’ through ‘9’ are characters
IBM used 8 bits in PC and added 128
additional characters (line drawing, Greek)
• Problem with ASCII – the A
• Unicode addresses this problem
ASCII-points to note
•Values 30-39=Chars ‘0’-’9’
•‘a’ = ‘A’ + 32 (Bit 5)
•‘A’ = Ctrl/A+64 (bit 6)
•‘A’ < ‘Z’ < ‘a’ < ‘z’
•If character is represented in a
byte, bit 7 is usually 0
•Chars 0-31d are not printable,
but may influence how other
chars are printed.
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
00
01
02
03
04
05
06
07
08
09
0A
0B
0C
0D
0E
0F
10
11
12
13
14
15
16
17
18
19
1A
1B
1C
1D
1E
1F
C-@
C-A
C-B
C-C
C-D
C-E
C-F
C-G
C-H
C-I
C-J
C-K
C-L
C-M
C-N
C-O
C-P
C-Q
C-R
C-S
C-T
C-U
C-V
C-W
C-X
C-Y
C-Z
ESC
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
20
21
22
23
24
25
26
27
28
29
2A
2B
2C
2D
2E
2F
30
31
32
33
34
35
36
37
38
39
3A
3B
3C
3D
3E
3F
!
"
#
$
%
&
'
(
)
*
+
,
.
/
0
1
2
3
4
5
6
7
8
9
:
;
<
=
>
?
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
40 @
41 A
42 B
43 C
44 D
45 E
46 F
47 G
48 H
49 I
4A J
4B K
4C L
4D M
4E N
4F O
50 P
51 Q
52 R
53 S
54 T
55 U
56 V
57 W
58 X
59 Y
5A Z
5B [
5C \
5D ]
5E ^
5F _
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
60 `
61 a
62 b
63 c
64 d
65 e
66 f
67 g
68 h
69 i
6A j
6B k
6C l
6D m
6E n
6F o
70 p
71 q
72 r
73 s
74 t
75 u
76 v
77 w
78 x
79 y
7A z
7B {
7C |
7D }
7E ~
7F DEL
upper 3 bits
00000
00001
00010
00011
00100
00101
00110
00111
01000
01001
01010
01011
01100
01101
01110
01111
10000
10001
10010
10011
10100
10101
10110
10111
11000
11001
11010
11011
11100
11101
11110
11111
lower 5 bits
000
001
010
011
100
101
110
111
Binary Addition
0+0=0
0+1=1
1+0=1
1 + 1 = 0 (with carry of 1)
1011011 + 10110 = ?
1 0 1 1 0 1 1
+
1 0 1 1 0
---------------1 1 1 0 0 0 1
100111001 + 11011 = ?
100111001 + 11011 = ?
1 0 0 1 1 1 0 0 1
+
1 1 0 1 1
-----------------
100111001 + 11011 = ?
1 0 0 1 1 1 0 0 1
+
1 1 0 1 1
----------------1 0 1 0 1 0 1 0 0
Hex Addition
4BA978
+ D2B84
58D4FC
•
8+4=C
•
7+8=F
•
9+B=4 w/ carry
• 1+A+2=D
•
B+D=8 w/ carry
• 1+4+0=5
Hex Addition Table
00
01
02
03
04
05
06
07
08
09
0A
0B
0C
0D
0E
0F
10
00
00
01
02
03
04
05
06
07
08
09
0A
0B
0C
0D
0E
0F
10
01
01
02
03
04
05
06
07
08
09
0A
0B
0C
0D
0E
0F
10
11
02
02
03
04
05
06
07
08
09
0A
0B
0C
0D
0E
0F
10
11
12
03
03
04
05
06
07
08
09
0A
0B
0C
0D
0E
0F
10
11
12
13
04
04
05
06
07
08
09
0A
0B
0C
0D
0E
0F
10
11
12
13
14
05
05
06
07
08
09
0A
0B
0C
0D
0E
0F
10
11
12
13
14
15
06
06
07
08
09
0A
0B
0C
0D
0E
0F
10
11
12
13
14
15
16
07
07
08
09
0A
0B
0C
0D
0E
0F
10
11
12
13
14
15
16
17
08
08
09
0A
0B
0C
0D
0E
0F
10
11
12
13
14
15
16
17
18
09
09
0A
0B
0C
0D
0E
0F
10
11
12
13
14
15
16
17
18
19
0A
0A
0B
0C
0D
0E
0F
10
11
12
13
14
15
16
17
18
19
1A
0B
0B
0C
0D
0E
0F
10
11
12
13
14
15
16
17
18
19
1A
1B
0C
0C
0D
0E
0F
10
11
12
13
14
15
16
17
18
19
1A
1B
1C
0D
0D
0E
0F
10
11
12
13
14
15
16
17
18
19
1A
1B
1C
1D
0E
0E
0F
10
11
12
13
14
15
16
17
18
19
1A
1B
1C
1D
1E
0F
0F
10
11
12
13
14
15
16
17
18
19
1A
1B
1C
1D
1E
1F
10
10
11
12
13
14
15
16
17
18
19
1A
1B
1C
1D
1E
1F
20
Positive/Negative Numbers
1100
1011
-5
1110
1101
-4
-3
0000
1111
-2
-1
0010
0001
0
1
0100
0011
2
3
0101
4
5
Notes
• We are in “4-bit” mode
• Think odometer math (-1=1111, -2=1110,
-3=1101, -4=1100, -5=1011, -6=1010, etc.)
• At some point a big positive number+1
becomes a big negative number
• Really a number “circle”
4-bit Number Circle
0000
1111
1110
-1
0
0001
1
-2
1101
1100
2
0010
-3
-4
4
-5
1011
0100
5
0101
-6
1010
0011
3
6
-7
1001
-8
1000
7
0110
0111
Observations (Part 1)
0110
6
+ 0011 + 3
1001
9

0100
4
0010 + 2
0110
6

0110
6
1101 +-3
10011
3

1010 -6
1101 +-3
10001
9

1100 -4
1110 +-2
11010 -6

1010 -6
0011 + 3
1101 -3

Observations (Part 2)
0110
6
+ 0011 + 3
1001
9

0100
4
0010 + 2
0110
6

0110
6
1101 +-3
10011
3

1010 -6
1101 +-3
10001
9

1100 -4
1110 +-2
11010 -6

1010 -6
0011 + 3
1101 -3

Observations (Part 3) & Conclusions
• Note when a carry into the sign bit occurs
• Note when a carry out of the sign bit occurs
• If one occurs and not the other – WRONG
• If both occur – RIGHT
• If neither occur – RIGHT
• Therefore:
OF = carry into sign bit carry out of sign bit
Organization of a computer
Address Bus
CPU
RAM
ROM
Printer
Disk
Data Bus
Control Bus
CRT
Kbd