2Bytesprog2 course_2014_c5_pointers

Prog_2 course- 2014
2 bytes team
Kinan keshkeh
IT Engineering-Damascus University
3rd year

Static Variables && Dynamic Variables
Static Variables :
var x:integer;
The memory for this variables x still until program End , and doesn’t free even we don’t use it
Ex:
X
Memory
We determine their places at memory first of all
value

Dynamic Variables :
Memory
The Heap
X
var x: ^integer;
Ex:
the address
the address
value
The memory for this variables x in the Heap, we can take it or dispose it, so it can be free!!
We determine their places at memory during the program

Why Dynamic??
1) better usage to memory
2) to create advance data structure like trees , lists….atc

What is the pointer?
The Definition
The pointer is an address to a memory place
Var ptr: ^Type;
Lets Have an explication!! but…..

P1
X
Value..
P1^
P1
Var p1,p2: ^Type; x :Type;
P1^
P1
P1^
X:=value
P2
xx
P1^
P2
New(p1);
P1^:=x
Value..
P2:=P1
Value..
P1:=nil;
Value..
+1

xx
P1^
P2
Dispose(p2);
Value..
xx
P1^
xx
P2:=nil;
Value..
+1

Program pointer on Soso;
There is wrong !
Type
pEmp=^Emp;
Emp=Record
eno:integer; ename:string[20];
esal:real;
end;
Var
E:Emp; E1:pEmp;
n,m,y:^integer;
EXAMPLE:

Begin i:=5; new(y); y^:=50; write(y); write(^y); write(y^); y:=i y^:=i ; y^:=20; i:=y^; y:=0; dispose(y); y:=nil; new(m); new(y); m^:=30; y^:=80; n:=y; write(y^); write(n^); E1:=nil; new(E1); E1^:=E ; Read(E1^.Eno); Dispose(E1); End.
false
false
TRUE
NOTE: Without “ New(n)” !!
80 80
false
TRUE

STACK {FILO}
First In Last Out
pringles

The Data
next
Nil
TOP
Stack=Record
the Data: Type;
next: Pstack;
end;
The Definition
Pstack=^stack;
Var
Top: Pstack;
D:Type;
……….

Stack procedures :
Push(s,ch)
Pop(s,ch)
S_Top(s,ch)
Is_Empty(s)
3)
2)
1)
4)

EX: You have a string “KiKi” How to put/take it in /from the STACK ?!
i
next
TOP
K
i
K
Nil

Stack=Record ch: char; next: Pstack; end;
Type Pstack=^stack;
Var Top: Pstack; st: string[4]; i:integer; x:char;
BEGIN
Readln(st); top:=nil;
For i:=1 to (length(st) ) do
push( top, st[i] ) ;
While(top<>Nil) do
begin
Pop( top, x) ; write(x,’ ’);
end;
END.
Program fifi;

PUSH
Procedure push (var Top:Pstack; c:char); Var temp:Pstack; Begin end;
new(temp);
temp^.ch:=c;
temp^.next:=top;
top:=temp;
i
next
TOP
K
i
K
Nil
temp
K
TOP
i
TOP
K
TOP
i

Stack=Record
ch: char;
next: Pstack;
end;
Type Pstack=^stack;
BEGIN
For i:=1 to (length(st) )
While(top<>Nil) do
begin
end;
END.
Program fifi;

POP
Procedure pop (var Top:Pstack; var c:char);
Var temp:Pstack;
Begin
end;
Dispose(temp);
Temp:=Top;
Top:=Top^.next ;
C:=top^.ch;
i
next
TOP
K
i
K
Nil
The output:
C
i
temp
K
TOP
temp
TOP
i
temp
TOP
K
temp
i
K
i
K

Stack=Record ch: char; next: Pstack; end;
Type Pstack=^stack;
BEGIN
For i:=1 to (length(st)-1)
begin
end;
END.
Program fifi;
While(top<>Nil) do

Is_Empty
Function Is_Empty (top:Pstack):boolean begin if top =Nil then Is_Empty:=True; else Is_Empty:=false; end;

S_Top
Function s_Top (top:Pstack): integer/…Type begin if not is_Empty(top) then s_Top :=top^.key; end;
NOTE: Don’t write then : top:=top^.next
Cause that function just return the element without change in stack!!

NOTE:
You can reach to the data , at a determinate place by pointers without do pop() cause with pop() you cant restoring your ‘poped’ data .. You do: S:=top; {bring the third element 64} i:=s^.next^.next^.key; OR: S:=top; {bring the 11 element} For i:=1 to 10 do s:=s^.next; i:=s^.key
52
next
TOP
35
64
76
Nil
S
+1

Homework:
لديك تعبير حسابي مثلا :
(( (1 * (3+2)) * 1) +3 )
المطلىب :
حساب قيمة التعبير الحسابي وطباعتها على
الشاشة باستخدام المكدس pop/push
ملاحظة :مكدس واحد فقط !
+20 points

Group : group link Mobile phone- Kinan : 0994385748 Facebook account : kinan’s account
2 bytes team

2Bytesprog2 course_2014_c5_pointers

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