CS136 Winter 2015 - Tutorial 3
CS136 ISAs: [email protected]
27th, 28th of January, 2015
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Goals of this Tutorial
The goal of this tutorial is to reinforce the following material:
• different systems of numeration (binary, decimal, hexadecimal
and octal)
• memory capacity
• bits, bytes, and chars in memory
• overflow
• modules in C (testing and interface & implementation files)
• working with floats
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Systems of Numeration
Recall three systems of numeration: binary (base 2), decimal (base
10), and hexadecimal (base 16). We will also introduce one new
system of numeration, octal (base 8).
When we write the number 97 in decimal, we interpret it as
97 = 9 × 101 + 7 × 100 .
We can represent the number 97 in other numerical systems;
consider the following alternate represenations of 97:
11000012 ,
and 6116 .
Each base is denoted by a subscript. Conventionally, if a number
is in base 10 then it will not have a subscript.
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Binary notation
In a binary representation (also known as base 2) there are two
distinct digits (01).
When we write the number 1011010 in binary, we interpret it as
1 × 26 + 0 × 25 + 1 × 24 + 1 × 23 + 0 × 22 + 1 × 21 + 0 × 20
= 64 + 16 + 8 + 2 = 90 (in base 10).
4 binary digits, can represent 24 (16) different possible values
(0 . . . 1111) or (0 . . . 15) in base 10.
In CS 251 / CS 230 you will learn more about binary notation.
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Hexadecimal notation
In a hexadecimal (hex) representation (also known as base 16)
there are sixteen distinct digits (0123456789ABCDEF).
When we write the number 2A9F in hex, we interpret it as
2 × 163 + 10 × 162 + 9 × 161 + 15 × 160
= 8192 + 2560 + 144 + 15 = 10911 (in base 10).
The reason hex is so popular is because it is easy to switch between
binary and hex representation. A single hex digit corresponds to
exactly 4 bits.
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Conversion table
Dec
Bin
Hex
Dec
Bin
Hex
0
0000
0
8
1000
8
1
0001
1
9
1001
9
2
0010
2
10
1010
A
3
0011
3
11
1011
B
4
0100
4
12
1100
C
5
0101
5
13
1101
D
6
0110
6
14
1110
E
7
0111
7
15
1111
F
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Octal
In an Octal representation (also known as base 8) there are eight
distinct digits (01234567).
We write the number 0452 in octal, we interpret it as
4 × 82 + 5 × 81 + 2 × 80
= 256 + 40 + 2 = 298 (in base 10).
Consider the following equivalent representations of 97.
const int a = 0141; // base 8.
const int b = 97;
// base 10.
const int c = 0x61; // base 16.
Be careful, 0141, 0x141 and 141 are not the same!
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CQ 1:
You are Bart from the Simpsons (a tv show that had its debut in
1989). In the Simpsons each character has four fingers on each
hand (total of 8 fingers). If Bart created his own numerical system, it
would likely be in base 8. Represent 85 in base 8.
0105
0414
0232
D 0125
E 0101
A
B
C
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Memory Capacity
To have a better understanding of the C memory model, we provide
a brief introduction to working with bits and bytes.
You are probably aware that internally, computers work with bits.
A bit of storage (in the memory of a computer) is in one of two
states: either 0 or 1.
A traditional light switch can be thought of as a bit of storage.
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Early in computing it became obvious that working with individual
bits was tedious and inefficient. It was decided to work with 8 bits of
storage at a time, and a group of 8 bits became known as a byte.
Each byte in memory is in one of 256 possible states.
With today’s computers, we can have large memory capacities:
• 1 KB = 1 kilobyte = 1024 (210 ) bytes∗
• 1 MB = 1 megabyte = 1024 KB = 1,048,576 (220 ) bytes∗
• 1 GB = 1 gigabyte = 1024 MB = 1,073,741,824 (230 ) bytes∗
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∗
The size of a kilobyte can be 1000 bytes or 1024 bytes,
depending on the context. Similarly, a megabyte can be 106 or
220 bytes, etc..
Manufacturers often use the measurement that makes their
product appear better. For example, A terabyte (TB) drive is
almost always 1012 bytes instead of 240 .
To avoid confusion in scientific use, a standard was established
to use KB for 1000 bytes and KiB for 1024 bytes, etc..
In general use, KB is still commonly used to represent both.
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Primary Memory
Modern computers have primary memory in addition to secondary
storage (hard drives, solid state drives, flash drives, DVDs, etc.).
The characteristics of primary memory and secondary storage
devices vary, but in general:
Primary Memory:
Secondary Storage:
• very fast (nanoseconds)
• (≈20x-1000x) slower
• medium capacity (≈GB)
• large capacity (≈TB)
• high cost ($) per byte
• low cost ($) per byte
• harder to remove
• removable or portable
• erased on power down
• persistent after power down
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In practice, programs can only “run” in primary memory.
When you “launch” a program, it is copied from secondary storage
to primary memory before it is “run”.
In this course, we are always referring to primary memory, which is
also known as Random Access Memory (RAM). With RAM you
can access any individual byte directly and you can access the
memory in any order you desire (randomly).
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Traditional secondary storage devices (hard drives) are faster if
you access data sequentially (not randomly).
Primary memory became known as RAM to distinguish it from
sequential access devices.
The term “RAM” is becoming outdated, as solid state drives and
flash drives use random access. Also, modern RAM can be
faster when accessed sequentially (in “bursts”).
Regardless, when you encounter the term “RAM”, you should
interpret it as “primary memory” .
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Bits, Bytes, and C Characters in Memory
Recall that in C, single quotes (') are used to indicate an ASCII
character. Moreover, in C there is no difference between the
following two variables:
const char letter_a = 'a';
const char ninety_seven = 97;
In memory, letter_a is represented in 0s and 1s. Recall that there
are 8 bits in a byte and a char reserves 1 byte in our RunC
environment.
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CQ 2:
Consider the following definition for i.
const int i = 'b'; // equivalent to 98 in decimal.
How many bytes does i reserve on the stack and how many 1s
(bits) are in the binary representation of i?
A
B
C
D
E
4 bytes and 1 bit
4 bytes and 2 bits
4 bytes and 3 bits
4 bytes and 4 bits
1 byte and 3 bits
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Overflow and Underflow
Recall that integer overflow (or underflow) occurs when we try to
represent values outside of the integer limits.
By carefully specifying the order of operations you can sometimes
avoid overflow.
If you #include <limits.h>, the constants INT_MIN and
INT_MAX are defined with those limit values.
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example: overflow
k!
.
Consider the equation m!
(1)(2)(3)...(k−2)(k−1)(k)
We can express the former equation as (1)(2)(3)...(m−2)(m−1)(m) .
If we let k
= 13 and m = 10 then
(1)(2)...(9)(10)
1
13!
=
(11)(12)(13) = (1716) = 1716
10!
(1)(2)...(9)(10)
1
k! may overflow for sufficiently large values of k , however if we
simplify the expression before we evaluate the factorial then we
can mitigate overflow.
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Modules in C
Recall that a C module consists of an interface (.h file) and an
implementation (.c file) file.
The interface consists of: function declarations, variable
declarations, and structure definitions; whereas the implementation
consists of variable and function definitions.
If a function is declared in the interface then the design recipe goes
in the interface with the function declaration, otherwise it goes with
the function definition in the implementation file.
From assignment 03 onward, you are expected to assert your
required restrictions.
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example: safe overflow module
The CS136 C master has promoted you to project lead for project
Safe Overflow. This responsibility entails writing a C module which
provides a function that safely determines if the sum of two integers
will overflow.
Fortunately, the previous project lead wrote some of the interface
and implementation; it is your task to finish writing the interface and
implementation.
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example: test module for the safe overflow module
Great work, you have completed the C module; the CS136 C master
is pleased. Oops! The CS136 C master noticed that you did not
submit your test module.
The CS136 C master is asking you to write a test module that
verifies the correctness of your module. Write the test module and
send it to the CS136 C master.
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Working with Floats
Consider a bank account with 1 billion dollars and 0 cents; observe
what happens when we withdraw $100.
const float balance = 1000000000.0;
const float withdraw = 100.0;
const float new_balance = balance - withdraw;
printf("balance = %f\n", balance);
printf("withdraw = %f\n", withdraw);
printf("new balance = %f\n", new_balance);
balance = 1000000000.0
withdraw = 100.0
new balance = 999999872.000000
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CQ 3:
From the previous example, the new balance is not accurate
because:
A floats are stored in memory
B one billion is outside the range of a float
C floats have a mantissa and an exponent
D floats are not precise
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It is dangerous to compare floats using the equality operator (==).
Typically, we only want to know if a float is within some small
distance (i.e. an epsilon) of another float.
Bad
Good
const float a = 10.0/3.0;
const float b = 3.333333;
const float a = 10.0/3.0;
const float b = 3.333333;
const float epsilon = 0.000001;
printf("a = %f\n", a);
printf("b = %f\n", b);
printf("a = %f\n", a);
printf("b = %f\n", b);
assert(a == b); // failed.
a = 3.333333
b = 3.333333
Program finished with errors
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assert((a - b) <= epsilon);
assert((a - b) >= -epsilon);
a = 3.333333
b = 3.333333
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