2Bytesprog2 course_2014_c8_units

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

Introduction
How is the compilation process ??
the principle of compilation is based on fragmentation the program into several files then the compiler compile
each file separately then Connect all parts of the compiled to configure executable file (conversion program form machine language to “code” language )

Unit stop here
compilation
source file in Pascal language
Text files for some procedure or function
Unit programing ready to use
Libraries ready to use
linked
Execution

unit
Unit unitname
Interface
Implementation
Definition :
Uses
Const
Type
Var
The header of procedure or function
Uses
Const
Type
Var
The object of procedure or function founded in interface part
End.
After last end we have to put (.) not (;)
Unit divided into two sections :
1) Interface :
2) Implementation :

Interface : or what we call it <<public part>> , in this part we can define the visible things to any program or another unit for example :
Uses wincrt;
Const pi=3.14;
Type arr=array[1..100] of integer;
var x:integer;
Function sum(x,y:integer):integer;
Function max(a,b:integer):integer;
Note: include interface part we only define the header of procedure or function

Implementation : or what we call it <<private part>> in this part we can define invisible things, we can‟t deal with them when we go out from the definition of the unit
Uses wincrt;
Const max=100;
Type person=record
fname,lname:string;
end;
var c:char ;
Function sum(x,y:integer): integer;
Begin
Sum:=x+y;
end;
Function max(a,b:integer):integer;
Begin
if (a > b) then
Max:=a;
Else
Max:=b;
end;

After writing unit what do we have to do ?
1.Correct the errors, if any
2.Compilation
3.Save this file under unit‟s name with (TPU) <<unitname.TPU>>
For example :
Unit my_unit;
Save as:<<my_unit.TPU>>
4.Recall this unit in main program using
reserved word (uses)
uses my_unit ;
Note : unit is subprogram unexecutable we cannot run it ……

Unit my_unit1;
Interface
Const val=200;
Var w : integer ;
procedure swap(var x,y:integer);
Implementation
var temp :integer ;
begin
temp:=x;
x:=y;
y:=temp;
end;
End.
Compilation
Attention !!!

Unit my_unit1;
Interface
Const val=200;
Var w : integer ;
Implementation
var temp :integer ;
readln(x);
begin
temp:=x;
x:=y;
y:=temp;
end;
End.
{false}
There is no Instructions in unit

Program prog1;
Uses my_unit1;
Var a,b :integer;
Begin
w:=5; temp:=10; a:=2; b:=3;
End.
{true}
{false}
a= 3
b= 2
swap(a,b);
Writeln(„a= „,a);
Writeln(„b= „,b);
w,b);
w= „,w);
w= 5

Program prog1;
Uses my_unit1;
Var a,b,w :integer;
Begin
w:=5; temp:=10; a:=2; b:=3;
End.
swap(w,b);
Writeln(„w= „,w);
Writeln(„b= „,b);
My_unit1.w
If you want to deal with w in my_unit
You should write
Which w ????
{this}
my_unit1.w:=10

Exercise :
أنشئ وحذة بشهج تٍ وعشف ف هٍا تابع قٌىم بحساب ) a^n ( وإجشائ تٍ
تقىم بحساب هعكىس عذد طب عٍ وإجشائ تٍ تقىم بطباعت عناصش سلسلت
أحاد تٌ بشكل هعكىس )هن النها تٌ حتى البذا تٌ ( ) على أن تتن كتابت
الإجشائ اٍث السابقت بشكل عىدي ( ثن استخذم هزه الىحذة ف بشناهج
سئ سٍ ورلك لاستذعاء التىابع السابقت

Recursion
It‟s process that procedure or function recall itself within the body of that procedure or function
 Recursion always need to stop condition

Unit recursion_unit;
Interface
type list=^node;
node=record
value:integer;
next : list;
end;
Function y(a,n:integer): longint ;
Procedure reverse(z:integer ; var z1:integer);
Procedure printlist (p: list );
Implementation
Begin
If n=0 then
y:=1
else
y:=y(a,n-1)*a;
end;

Procedure reverse(z:integer ;var z1:integer);
Begin
If (z<>0) then
begin
Z1:=z1*10 + z mod 10;
Reverse (z div 10 , z1);
end;
end;
Procedure printlist(p : list );
Begin
If (p^.next <> nil ) then
Printlist(p^.next);
Write(p^.value:5);
end;
End.

Program pro2bytes_team;
uses recursion_unit;
var a,n,z,z1,m,x,i:integer;
p,temp:list;
Begin
Readln(a,n);
If (n < 0) then
Writeln(„a^n = „,1/y(a,n))
Else
Writeln(„a^n = „,y(a,n));
Writeln(„ input positive number „);
Readln(z);
z1:=0;
Reverse (z,z1);
Writeln(„z1 = „,z1);

Writeln(„input the number of nodes „);
Readln(m);
Writeln(„input the value „);
Readln(x);
new(p);
p^.value:=x;
temp:=p;
for i:=2 to m do
begin
new(temp^.next) ;
temp:=temp^.next;
Writeln(„input the value „);
Readln(x);
temp^.value:=x;
end;
temp^.next:=nil;
Printlist(p);
End.

Execution :
Begin
If n=0 then
y:=1
else
y:=y(a,n-1)*a;
end;
n=3
a=2
a =
n
y= * 2
n= 3
n= 2
n= 1
n= 0
y= * 2
y= * 2
y = 1
waiting
waiting
waiting
stop condition
8
4
2
1
End.

Procedure printlist(p : list );
Begin
If (p^.next <> nil ) then
Printlist(p^.next);
Write(p^.value:5);
end;
End.
Execution :
next
20
15
10
5
p
nil
p
p
p
stop condition !!!
output
20
15
10
5

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

2Bytesprog2 course_2014_c8_units

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