Question

Consider the following pseudocode. int a = 9; //global variable void go {a = a +2; print a; } main { int a = 4; 80: a = a +1; print a; } a. What is printed out if we are using static scope rules? Show runtime stack. b. What is printed out if we are using dynamic scope rules? Show calling tree.

Answer 1

Answer

Here is your answer if you have any doubt please comment, i am here to help you.

Here ,The scope of a variable 'a' is the region of the program in which uses of a refers to its declaration.

Scoping is generally divided into two classes:

• Static scoping
• Dynamic scoping.

1) Mosy of the programming language such as c++,c and etc are supporting static scoping. Here above the code will give the following output.

11
5

Here is the explanation of each line.

#include<stdio.h>
int a=9; // global variable initilized with 9.
void g()
{
        a=a+2;
        printf("%d\n",a);
}
int main()
{
        int a=4; //local variable inside the main initilized with 4.
        g(); //calling function g(). This will print 11, because that function uses the global variable.
        a=a+1; //this will increment the variable inside the main(local varibale), because local variable hides the global variable
        printf("%d\n",a); //this will print the value 5
        //so the output of the above program will 11 and 5
}

This is done in C. I put comment in each line.

output

(base) root@kali:-# ./a.out 11 5

Explanation:

Here the programming language support the static scoping, So, when we created a global varibale(a).It can be accessible anywhere in the program. We can update the value of that variable. But we defined the a new variable inside main with same name of the global variable(a). Inside that main block it will treated as local variable. Local variable have great priority rather than global inside the block. So any updation will take place it will not affect the global part , it only affect the localpart. (local variable hides global varaible inside the block.).

go) Uses the global section variable.Update a=9+2 main() Localvariable a-4 main() Local varible a=5 main() empty stack Global variable a=9 Global variable a-9 Global variable a=11

arrow indicates calling the function and returning the values.

2) It is very uncommon in modern language, but perl supports dynamic scope.It does not care about how the code is written, but care about how it executes. every time of execution of new function, a new scope is pushed onto the stack.

Here the output is similar to static scoping . 11 and 5. In dynamic scoping the compiler first searches the current block and then successively all the calling functions.

Here is the perl script for above probelm

$a = 10; 
sub g  
{  

   $a =$a+2;  
   print $a."\n";  
} 
g(); 
# Since local is used, a uses dynamic scoping.
local $a=4;
$a=$a+1;
print $a;

This program doesn't show the dynamic scoping, I will another example for this.

$a = 10; 
sub r
{
        return $a;
}
sub g  
{  

   # dynamic scoping.  
   $a =$a+2;  
   print r()."\n";  #if it is static scope it will print value 10 but, here it will print 12
} 
g(); 
# Since local is used, a uses dynamic scoping.
local $a=4;
$a=$a+1;
print $a;

In the above , the function checks will take only its own variable.

output

12 5(base) root@kali:-#

But in c/c+ it doesn't support.

any doubt please comment, i am here to help you. I used perl because , i don't know other languages which support dynamic scoping

any doubt please comment.

Thanks in advance