This document discusses scoping in programming languages. It defines scoping as the visibility of variables within a program. There are two main types of scoping: static scoping and dynamic scoping. Static scoping determines visibility at compile time based on the block structure, while dynamic scoping determines it at runtime based on the call stack. Examples are provided to demonstrate the differences between static and dynamic scoping by showing the output for different variable references under each scoping rule.

1. - presentation by
Nusrat.Mohammed.Hanif

2. What is Scoping ?
1. Scoping refers to the visibility of variables . In simple
words , we can say a textual region of program in which
binding is active is its scope.
2. A scope is the body of module class , subroutine or
structured control flow statement sometimes called as
block .

3. Types of Scopes:
There are 2 types of scopes :
1.Static Scoping
2. Dynamic Scoping
• Static Scoping : it is also called as lexical scoping. It is binding between
names and object can be determined at compile time. In static scoping
the compiler checks for say variable “x” if it is present in local block if it
is not there then it search in the main or parent block.
• Dynamic Scoping : It is binding between names and object can be
determined at run time. In dynamic scoping the compiler checks for say
variable “x” if it is present in local block if it is not there then it search
in the sibling functions block or from where it is called.

4.  Example : 1
i n t x ;
i n t m a i n ( )
{
x = 2 ;
f ( ) ;
g ( ) ;
}
V o i d f ( )
{
i n t x = 3 ;
h ( ) ;
}
v o i d g ( )
{
i n t x = 4 ;
h ( )
v o i d h ( )
{
p r i n t f ( “ % d ” , x ) ;
}
Output :
• By static rule :
x=2  main’s x
x=2  main’s x
• By dynamic rule :
x=3  f()’s x
x=4  g()’s x

5.  Example : 2
B e g i n
i n t e g e r m , n ;
P r o c e d u r e h a r d y :
b e g i n
p r i n t ( “ i n h a r d y n = “ , n ) ;
e n d ;
P r o c e d u r e l a u r e l ( n : i n t ) :
b e g i n
p r i n t ( “ i n l a u r e l m = “ ,
m ) ;
p r i n t ( “ i n l a u r e l n = “ , n ) ;
h a r d y ;
e n d ;
/ / m = 5 0 ; n = 1 0 0 ;
p r i n t ( i n m a i n p r o g n = “ , n ) ;
l a u r e l ( 1 ) ;
h a r d y ;
e n d ;
Out put :
• By st atic r ule :
n=100  main’s n
m=50  main’s m
n=1  l aurel’s n
n=100  main’s n
n=100  main’s n
• By dynamic r ule ( sibling)
n=100  main’s n
m=50  main’s m
n=1  l aurel’s n
n=1  ( hardy)laurel ’s n
n=100  main’s n

6.  Example : 3
i n t b = 5 ;
i n t f o o ( )
{
i n t a = b + 5 ;
r e t u r n a ;
}
i n t b a r ( )
{
i n t b = 2 ;
r e t u r n f o o ( ) ;
}
i n t m a i n ( )
{
f o o ( ) ;
b a r ( ) ;
r e t u r n 0 ;
}
Output :
• By static rule :
foo=10 (foo‘s return a)
bar =10 (foo’s return a)
• By dynamic rule :
foo=10 (foo‘s return a)
bar =7 (bar‘s return foo)