Post on 06-Aug-2015
transcript
Prog_2 course- 2014
2 bytes team
Kinan keshkeh
IT Engineering-Damascus University
3rd year
POINTERS
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;
Var Top: Pstack; st: string[4]; i:integer; x:char;
BEGIN
Readln(st); top:=nil;
For i:=1 to (length(st) )
push( top, st[i] ) ; While(top<>Nil) do
begin
Pop( top, x) ; write(x,’ ’);
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;
Var Top: Pstack; st: string[4]; i:integer; x:char;
BEGIN
Readln(st); top:=nil;
For i:=1 to (length(st)-1)
push( top, st[i] ) ; begin
Pop( top, x) ; write(x,’ ’);
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 مثال
(( (1 * (3+2)) * 1) +3 )
:المطلىب التعبير الحسابي وطباعتها على قيمةحساب
pop/pushالشاشة باستخدام المكدس
!مكدس واحد فقط :مالحظة
+20 points
Group : group link
Mobile phone- Kinan : 0994385748
Facebook account : kinan’s account
2 bytes team