Dynamic Scope

Question

What is the output of the following program if dynamic scoping is used?
 

int a,b,c;
void func1()
{
    int a,b;
    a=6;
    b=8;
    func2();
    a=a+b+c;
    print(a);
}

void func2(){
    int b,c;
    b=4;
    c=a+b;
    a+=11;
// here in case of dynamic scoping will 
//variable a of func1() be updated to a=6+11 ?
    print(c);
}

void main(){
    a=3;
    b=5;
    c=7;
    func1();
}

1. 7 19
2. 10 1 
3. 10 23 
4. 10 32

Comments

10 ,32 ??

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But answer to this question is given as - 7 19 is this answer possible

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in dynamic scoping the when calling func2 a=6 is referred so it will update func1's local variable a?won't it keep on seraching for the referred func till end i.e main here to get a?

Answer 1

Initial values:
$a = 3,\quad b = 5,\quad c= 7$

In dynamic scoping when a function is called the called function gets the variable of the calling function even if not explicitly passed (and not the global one)

Here, in main, all $a,b,c$ are the global ones as there is no declaration here.

In func1:
New $a,b$ are declared. $a = 6, b = 8.$ func2 is called with these new values and the original global $c$.

In func2:
New $b,c$ declared. $b=4.$ $c = a + b$ assigns $6+4 = 10$ to $c$. $10$ will be printed and $a$ becomes $6+11 = 17$

In func1:
$a+b+c$ is printed which will be $17 + 8 + 7 = 32.$

Answer 2

But iam getting 7 ,21 as answer because in func1()  a,b are declared as local and local variables have higher precedence over global variables s0 6+8+7=21

Comments

Your answer is wrong because in func2 when it c=a+b is done it has b=4 but no a is declared so it will first check the value of a in calling function i.e.. func1 sunce it is dynamic scoping not static scoping.

So at func2 c=a+b=6+4=10.. Now a+=11will increment the a of func1 by 11 such that a becomes 17 so when we return back from func2 to func1 a=a+b+c its a=17 b=8 and c=7 such that a=a+b+c=17+8+7=32

So ans is 10,32

Note-dont run this code in c compiler since c has static scoping not dynamic scoping

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When we return back from func2 a=17 , then why b is not 4 (as b has value 4 in func2) why we are taking a from func2 and b from func1?

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b and c are local to func2 (as they are initiated as int b,c), so it will not update value of b and c of func1 but a is not local to func2, so value of a will be returned back to func1.