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Question

What will be the output of following program if we use a) static scoping b) dynamic scoping?

int a= 1,b=2;

main( ) {

int a=20,b=30;

Print(a,b); C( );

Print(a,b); D( ); }

C( ) {

int a= 50;

print (a,b);

D( );

Print(a,b);

}

D( ) {

Print (a,b);

a=3; b=4;

Print(a,b);

F( );

}

F( ) {

int b =6;

Print( a,b);

a=7; b=8;

}

Comments

Dynamic Scoping:
20 30 50 30 50 30 3 4 3 6 7 4 20 4 20 4 3 4 3 6

Static is same as @kumar.dilip

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@ kumar.dilip Your answer for Static is correct. Shaijal's answer for dynamic is correct.

20 30        50 2    1 2      3 4     3 6      50 4          20 30        7 4      3 4          3 6 

How we get this 7 4(in bold)?

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Please explain the method to find answer for both parts. I am really not getting how to solve this

Answer 1

When in a function, a variable which hasn't been declared inside the function is used, the scope of the variable is ambiguous. This problem is resolved in two ways :

1.Static Scoping (when Compiler resolves the problem) It uses the memory of those variables (with same name) that have been allocated memory at compile time in the static area i.e global variables

2. Dynamic Scoping (when Processor resolves the problem) It uses the memory of those variables (with same name) that have been most recently used in the runtime stack. i.e local variables of the other functions.

Comments

Thanks

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In Static Scoping,

When D() states print a, b.... There are no declared variables 'a' and 'b'  inside D that will be printed. So, the compiler then takes global variables 'a' and 'b'  and prints their values, and further updates it to 3 and 4.

When function F() prints a, b... It already had a local variable 'b' declared inside as b=6. So, it has no conflict with respect to 'b'.  But for 'a' same problem arises, the compiler again uses the global variable a. So when F() executes a=7 and b=8, 'b' changes are made to the local variable it already had and 'a' value is updated in the global variable.