1. Lecture 2 SCOPE – Local and Global variables
Reference passed – passed by value or pointer
Recursion: calling itself
Stack pointer – temporary memory location
Subroutine
More about pointer
Data Representation
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2. SCOPE
Variables have scopes, that is, a variable name may refer to different values depending on where it is used. – variable a in subroutine c and subroutine d is different even the same name
If it is used inside a function that declares a local variable of that name, the name will refer to a data item that is private to that function
It is defined as global, it means it can be accessed by others.
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3. Example – abc program name
#include <stdio.h> int first; int second; void callee ( int first ) { int second; second = 1; first = 2; printf("callee: first = %d second = %d\n", first, second); } int main (int argc, char *argv[]) { first = 1; second = 2; callee(first); printf("caller: first = %d second = %d\n", first, second); return 0; }
Same variable “second”, but different memory location
DOS>abc 12 34
Here, argc = 3,
argv[0] = abc
argv[1]= 12
argv[2] =34
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4. Demonstration – memory dump
Address of first is 0x004235bc in main()
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5. Demonstration – memory dump
Address of first is 0x0065fda8 in callee()
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6. Formal Parameters: Value and Reference*
void callee ( int first ) { int second; second = 1; first = 2; printf("callee: first = %d second = %d\n", first, second); } first = 1
… callee(first) // pass the value of first to callee
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7. Example (passed by pointer) *
void callee ( int * first ) //not a variable, but an address { int second; second = 1; *first = 2; printf("callee: first = %d second = %d\n", *first, second); } int main (int argc, char *argv[]) { first = 1; second = 2; callee(&first); //passed by address printf("caller: first = %d second = %d\n", first, second); return 0; }
Content by address
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8. Why should we dump memory?
Look at the program the memory , variable a is 12ff7c, b[0] 12ff68…, it will loop more than 5, interesting
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9. Why should we dump memory?
The software will modify the content of variable a such that this program will loop more than 6.
b[0] ff68
b[1] ff6c
……….
b[6] ff7c
variable a
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10. Recursion* Calling itself, say
callee (n) {
if n = 0 return
printf(“,,,,,,,”);
callee(n-1); }
It will create local variable (n) for each call, n , n-1, n – 2…. Until 0
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11. Activation Records and Stacks
When a function is called, it will put the variables and return the address to the stack ( a special memory chunk in the operating system).
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12. The stack pointer
It holds the address where the stack ends. It is here that a new activation record will be allocated.
The frame pointer holds the address where the previous activation record ends. It is to this value that the stack holds.
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13. Procedure for stack operation
When a function is called, the compiler and hardware:
push the frame pointer into the stack
set the frame pointer equal to the stack pointer
decrement the stack pointer by as many memory addresses as are required to store the local state of the callee.
No need to memorise the sequence
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First-in last-out principle for stack

14. Diagram - stack push (create)
Before
After
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15. Diagram - stack pop (return)
Before
After
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16. How to find the return address in a routine?
For the prorgam,
the return address is next
address of the variable in the
stack pointer
Return Address
004010AF
Address of
variable first
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17. The code itself
It means the program code that occupies the address, how to find it?
You can dump the memory content or write a simple program.
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18. Example of this simple prorgam – here, assumes that the proggrams start from 0x0040b7d8
#include <stdio.h> int a; int main (int argc, char *argv[]) { int i; unsigned char * c; a = 5; c = ((unsigned char *) 0x0040b7d8); for (i = 0; i < 10; i++) { printf("%02X ", *c); c++; } printf("\n"); }
Note the address here
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19. Assembly Code The language that uses by computer directly.
You should understand the difficulty and its relationship with high level language
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20. Its equivalent assembly language
C and assembly
if (a == 0) { a = 5; }
High level C language
Its equivalent assembly language
Very complicate
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21. Pointers to an integer and a character
int *pointer
char *pointer
Can be a function as follows:
int (*newvar)(char)
newvar = &existing_function;
int existing_function (char c) { }
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22. The CPU also has Memory
The CPU also maintains its own banks of memory called registers.
They temporarily hold the data
As a result, the program is faster.
faster
register
Memory
hierarchy
Cache memory
Main memory
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Disk

23. Example of register, look at the following program, a = b + 3*c
int a; int b = 4; int c = 3; int main ( int argc, char * argv[]) { a = b + 3 * c; return 0; }
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24. Set a break Point at a = b + 3*c;
Mov eax, [c (00421b98)]
where eax is the register (CPU’s memory)
register
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25. Register Allocation and Compiler Optimization
CPU has a few registers, the compiler needs to decide when to assign the registers to speed up the operation.
Compiler can set the optimisation level.
Question, why don’t we set it faster?, The answer: the program code might be faster.
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26. Example of Optimisation level
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27. Representation of Data
Bits and Bit Manipulation
Integers – 32 bits
Character – 8 bits
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28. Bits and Bit Manipulation
The world in Zeros and Ones
Grouping Bits into Words
Bit Operations
Computers use binary 0 and 1 to represent data and manipulate data
Human finds it difficult to understand 1s or 0s
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29. The World in Zeros and Ones
A single unit of this form of data is called a binary digit, or bit for short
It can be 1 or 0
It is used to represent the voltage of “high” or “low”
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30. Grouping Bits into Words
It is to group bits together, For example,
Boolean a; //binary 1 or 0
int i; // 32 bits
char a; //8 bits
float f; // 32 bits
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31. Data representation – group bits
1 or 0 //binary
// 0, hex, group 4 bits
// 1
// 2
0011 ……
//f (15 decimal)
Group 4 bits together
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32. 0x9A0477F3 is represented as a 32-bit word:
In memory is different, say the address of this data is 0x0065fdf4
0x0065fdf f3
0x0065fdf
0x0065fdf
0x0065fdf A
Memory location in reverse order
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33. Example showing the data in reverse order
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34. In memory, it will be arranged in
Code
void main() { int i = 0x9a0477f3; }
Understand this, as this is related to your exam.
In memory, it will be arranged in
f a, interesting
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35. Exercise – to display the contents
#include <stdlib.h> #include <stdio.h> void main() { int i = 0x9a0477f3; char *ptr; ptr = (char *)&i; //type casting for (int j = 0; j <4; j++) printf("The address is %x , value of byte %d is %x\n", ptr + j, j, *(ptr +j) ); }
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36. Summary SCOPE – can use the same name, but different memory locations
Reference passed – when a routine has been called, values of variables will be destroyed.
Recursion: very elegant, but wastes CPU time
Stack pointer – first-in and last out
Subroutine – return address
Pointer – very powerful, but difficult to implement
Data representation in memory
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