RAM Allocation

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Data Structures in C for Non-Computer Science Majors

Kirs and Pflughoeft

RAM Allocation

RAM AllocationChapter 3

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What you What you MUSTMUST know before we start: know before we start:

• What basic data types there are:

• Signed/Unsigned Characters• Signed/Unsigned Shorts/Integers/Longs

• The difference between characters & numbers

• How the sign-bit is used• How negative values are stored• What precision and magnitude are and how they

affect real numbers

(Remember: The topics in this course build on each other)

• float/double

• How many bits each data type requires

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Storing Storing CharactersCharacters in RAM in RAMWe know that characters are really numbers:• signed char (or char, by default) on 8-bits:

8-bits = 1-byte = X X X X X X X X

Sign bitValue

• unsigned charValue

Regardless of whether signed or unsigned, the data type char requires 1-byte of storage per variable

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Consider the following C DeclarationConsider the following C Declaration

charchar a, b == ‘f’, ‘f’, c c = 87= 87;;

What is the effect of the command ???What is the effect of the command ???

1. We are reserving 3-bytes of storage

2. We are associating each location with a variable name (LOCATIONS a, b, and c)

3. We are initializing location b with the Character ‘f’= ASCII 102= 11001102

4. We are placing the value 87 (= 10101112) into location c

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Where in RAM will we find locations Where in RAM will we find locations aa, , bb, and , and cc ??? ???

We Don’t Know:We Don’t Know:• Address allocations are made at RUN-TIME and are based on available locations.

Assume that at run-time, we find that Addresses 5010, 5014, and 5015 are Available:• Variable a will be assigned to address 5010• Variable b will be assigned to address 5014• Variable c will be assigned to address 5015

How Will this appear in RAM ???How Will this appear in RAM ???

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5005

001010015006

110001105007

101110015008

001001115009

000001005010 (a)

111000015011

000000005012

011100115013

000000015014 (b)

011001105015 (c)

010101115016

110100115017

001000105018

100100105019

000000005020

00110110

Given: charchar a, b == ‘f’, ‘f’, c c = 87= 87;;

Where: Variable a => 5010: UnassignedVariable b => 5014: ‘f’ = 10210 = 011001102

Variable c => 5015: 8710 = 010101112

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Why Does Location Why Does Location aa (address 5010) contain (address 5010) contain something?? We did NOT initialize the variable.something?? We did NOT initialize the variable.

Whatever was previouslypreviously stored at that location is still there.

If we were now (after variable allocation) to issue the command:

printf (“%c %d”, a, a);

We would produce the output:

-31

That Makes NO Sense at ALL !!That Makes NO Sense at ALL !!

Let’s Examine why this would occur

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Location 5010 (variable a) contains: 11100001

Because location 5010 contains a (signed) char, the numeric value is:

1 1 1 0 0 0 0 1Left-most bit = ‘1’:Value is negative:COMPLIMENT

0 0 1 1 1 1 0Since we are using 2’s Compliment:ADD 1: + 1

1111100

= -(24 + 23 + 22 + 21 + 20) = -(16 + 8 + 4 + 2 + 1)

= -31-31

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That still doesn't explain why the ASCII character That still doesn't explain why the ASCII character is printed !!! is printed !!!

Even though the numeric value is -31, The value of location a is checked against the ASCII table as if it were an unsigned char:

1 1 1 0 0 0 0 1Evaluates to

27 + 26 + 25 + 20 = 128 + 64 + 32 + 1 = 225225

Which corresponds to the ASCII Character

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I don’t know if I believe – I know about your I don’t know if I believe – I know about your propensity to lie to us !!!propensity to lie to us !!!

OK – here’s a OK – here’s a program I wrote:program I wrote:

And here’s the output And here’s the output I received:I received:

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What if we try to store an illegal number, say What if we try to store an illegal number, say 837837, , into a character location in RAM ???into a character location in RAM ???

charchar illegal == 837;837;printf (“%c %d”, illegal, illegal);

If we were to issue the commands:

We would produce the output:

E 69

83710 = 11010001012

WHY ???WHY ???

BUTBUT requires 10-bits of storage

WE RESERVED ONLY 8-BITS !!!WE RESERVED ONLY 8-BITS !!!

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Because we reserved only 1-byte, Because we reserved only 1-byte, ONLYONLY the the right-right-mostmost 8-bits will be stored: 8-bits will be stored:

1 1 0 1 0 0 0 1 0 1

= 26 + 22 + 20 = 64 + 4 + 1 = 6969

Which Corresponds to the ASCII Character ‘E’

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Integers (c data type Integers (c data type intint))Integers require 2 CONTIGUOUS bytes (16-bits) of storage

Consider the c declaration:

intint x = ‘W’, y, z = 5487;

In Fact we are (once again):

(LOCATIONS x, y, and z)3. Initializing location x with ‘W’

(= ASCII 87 = 10101112 = 00000000010101112 on 16-bits)

4. Initializing location z with 5487(548710 = 10101011011112 = 00010101011011112 on 16-bits)

(of course we all know it is really 4 Contiguous bytes)

1. Requesting 6-bytes (2 per variable) of RAM be reserved2. Associating each variable name with a reserved location

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Let’s Assume locations 7212, 7213, 7214, 7216, and 7219 through 7250 Are available:

• Variable x is assigned address 7212 (and 7213)• Variable y is assigned address 7219 (and 7220)• Variable z is assigned address 7221 (and 7222)

Why aren’t addresses Why aren’t addresses 72147214 and and 72167216 used??? used???

Because we need 2 contiguouscontiguous bytes of storage for integers:

• We can’t use location 7214 because location 7215 is NOT available

• We can’t use location 7216 because location 7217 is NOT available

Looking at RAM, We might see:

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intint x = ‘W’, y, z = 5487; x = 0000000001010111 y is unassigned z = 0001010101101111

• x stored at address 7212 (and 7213)• y stored at address 7219 (and 7220)• z stored at address 7221 (and 7222)

7211

001010017212 (x)

000000007213 (x)

010101117214

001001117215

000001007216

111000017217

000000007218

011100117219 (y)

000110017220 (y)

011001107221 (z)

0001010107222 (z)

011011117223

001000107224

100100107225

000000007226

00110110

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Once again, Notice that there is ‘garbage’ in location y (Addresses 7219 & 7220):

00011001 01100110

7219 7220

= 212 + 211 + 28 + 26 + 25 + 22 + 21 = 4096 + 2048 + 256 + 64 + 32 + 4 + 2= 6,5026,502

Which is the output we would obtain if we issued the command:

printf (“%d”, y);

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Lest we forget, since we know that integers are really stored on 32-bits:

intint x = ‘W’, y, z = 5487; x = 0000000001010111 y is unassigned z = 0001010101101111

• x stored at address 7212 (to 7215)• y stored at address 7219 (to 7222)• z stored at address 7223 (to 7226)

7211

001010017212 (x)

000000007213 (x)

000000007214 (x)

00000000 010101117215 (x) 7216

011000017217

000100107218

000000007219 (y)

000000007220 (y)

00000000 000110017221 (y) 7222 (y)

011001107223 (x)

000000007224 (x)

000000007225 (x)

000101017226 (x)

01101111 000100007227 7228

01111000

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What would happen if we tried to print an What would happen if we tried to print an integerinteger as as as an ASCII as an ASCII charactercharacter ??? ???

Depends:Depends: intint anumber == 104;104;printf (“%c”, anumber);• If we issue the commands:

The output will be: hh

Which is the ASCII character for decimal value 104

intint anumber == 6502;6502;printf (“%c”, anumber);• If we issue the commands:

The output will be: ff

WHAT ???WHAT ???

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Don’t forget how the integer 6502 was stored (on 16-bits):

0 0 0 1 1 0 0 1 0 1 1 0 0 1 1 0

When we try to print out a character according to the ASCII tables, we consider ONLY 8-bits (the right-mostright-most 8-bits)

= 26 + 25 + 22 + 21 = 64 + 32 + 4 + 2 = 102102

Which corresponds to the ASCII character ‘f’

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What if an illegal value is entered??What if an illegal value is entered??The same situation occurs as did when we entered an illegal value for the data type char:

If we make the declaration: intint badnumber == -52434;-52434;

Since: 52,43410 = 11001100110100102

Then: - 52,43410 = 00110011001011012 1’s Comp.+ 10011001100101110

= 213 + 212 + 29 + 28 + 25 + 23 + 22 + 21

= 8192 + 4096 + 512 + 256 + 32 + 8 + 4 + 2= 13,10213,102

Which is the output produced by the Statement: printf (“%d”, printf (“%d”, badnumberbadnumber));;

2’s Comp.

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I don’t know if I believe anymore -- About anything!!!I don’t know if I believe anymore -- About anything!!!

OK – let’s write another program – but instead of using the data type int, let’s use the data type short

WHY ????WHY ????

• Because, as we know, the data type short still requires 16-bits (2-bytes) of storage (As integers used to)

• We will NOT have to change our illustration

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Once again, here is the Once again, here is the program I wrote: program I wrote:

And here is the And here is the output I received:output I received:

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One more quick example:One more quick example:Suppose we make the declaration: intint badnumber == 72925;72925;

Where: 7292510 = 100011100110111012

Requiring 17-bits

Since an integer is stored on 16-bits:

= 212 + 211 + 210 + 27 + 26 + 24 + 23 + 22 + 20

= 4096 + 2048 + 1024 + 128 + 64 + 16 + 8 + 4 + 1= 7,3897,389

Which is the output produced by the Statement: printf (“%d”, printf (“%d”, badnumberbadnumber));;

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Other Integers:Other Integers:

Data Type

unsigned int

Bits Required

16 (2-bytes)

Legal Values

0 through 65,535(No sign bit)

longOr signed long

32 (4-bytes) -2,147,483,648 through 2,147,483,647

unsigned long 32 (4-bytes) 0 through 4,294,967,295

How would the data type How would the data type longlong appear in RAM ??? appear in RAM ???

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Long Integers (c data type Long Integers (c data type longlong))Integers require 4 CONTIGUOUS bytes (32-bits) of storage

Consider the c declaration:

longlong l1 = ‘3’, l2;

In Fact we are (once again):

1. Requesting 8-bytes (4 per variable) of RAM be reserved2. Associating each variable name with a reserved location

(LOCATIONS l1, and l2)3. Initializing location l1 with ‘3’

( = ASCII 51 = 1100112 = 0000000000000000000000001100112 on 32-bits)

(REMEMBER: The data types int and long are now the same)

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Let’s again assume locations 9011 through 9017, and 9021 through 9076 Are available:

• Variable l1 is assigned address 9011 (through 9014)• Variable l2 is assigned address 9021 (through 9024)

Why aren’t addresses Why aren’t addresses 90159015 through through 90179017 used??? used???

Because we need 4 contiguouscontiguous bytes of storage for longs:

• If we were to try and store variable l2 at location 9015, we would need addresses 9016 through 9018 to be available also

Looking at RAM, We might see:

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longlong l1 = ‘3’, l2; l1 = 0000000000000000 0000000000110011 l2 is unassigned

• l1 stored at address 9011 (through 9014)• l2 stored at address 9021 (through 9024)

9011

000000009012

000000009013

000000009014

001100119015

000001009016

111000019017

000000009018

011100119019

000110019020

011001109021

1111111119022

011011119023

001000109024

100100109025

000000009026

00110110

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What ‘Garbage’ will we find at location What ‘Garbage’ will we find at location l1l1 ??? ???At addresses 9021 through 9024, we find:

11111111011011110010001010010010 (On 32-bits)

Since the left-most bit = ‘1’ ( ==> the number is negative) we must compliment:

0000000100100001101110101101101+ 1

0000000100100001101110101101110

= -(223 + 220 + 215 + 214 + 212 + 211 + 210 + 28 + 26 + 25 + 23 + 22 + 21)= -(8,388,608 + 1,048,576 + 32,768 + 16,384 + 4,096 + 2,048 + 1,024 + 256 + 64 + 32 + 8 + 4 + 2)= -9,493,8709,493,870

(1’s Comp.)

(2’s Comp.)

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Floating-point numbers (real numbers)Floating-point numbers (real numbers)

data type float:• 4 CONTIGUOUSCONTIGUOUS bytes (32-bits) per variable• 7 decimals of precision• Sometimes also referred to as single precision reals

data type double:• 8 CONTIGUOUSCONTIGUOUS bytes (64-bits) per variable• ANSI: 10 decimals of precision

data type long double:• 16 CONTIGUOUSCONTIGUOUS bytes (128-bits) per variable• ANSI: 10 decimals of precision

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floatfloat f1 = -92.337, f2;

• f1 stored at address 8910 (through 8913)• f2 stored at address 8914 (through 8917)

8909

000000008910

101000108911

010011108912

001100118913

000001008914

101000108915

010111018916

001000108917

001000108918

011001108919

1011100118920

011011118921

001000108922

001000108924

000000008925

00110110

RAM Storage ???RAM Storage ???

Available: Addresses 8910 through 8993

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So what do we need to do ??So what do we need to do ??• Make sure you THOUROGHLY understand ALL of the

concepts covered in these slides

• Answer ALL of the relevant questions on the Review Page

• Submit your References

• Submit your Question(s)

• Look at the Bits/Bytes/ASCII C/C++ Programming Assignment (it’s not due yet, but it can’t hurt to look at it)

??? Any Questions ??? (Please!!)??? Any Questions ??? (Please!!)

• Get ready for Quiz 1 --- It’s coming

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RAM Allocation