GATE CSE 2004 | Question: 31, ISRO2008-40

Question

Consider the following C function:

int f(int n)
{
    static int i = 1;
    if(n >= 5) return n;
    n = n+i;
    i++;
    return f(n);
}

The value returned by $f(1)$ is:

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Comments

might this will help to better understand

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how many times the function will return 7?

one time or four times? And how it will return.Plz explain.

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only single time 7 will be returned. bcoz f(7) will return 7 to f(4), f(4) now will return 7 to f(2), and f(2) will return 7 to f(1) and f(1) will return 7 to main function.

Answer 1

Answer is $7$. As,

$f(1):n=2,i=2$

$f(2):n=4,i=3$

$f(4):n=7,i=4$

$f(7):print(n) \Rightarrow 7$ ans

Comments

I think it is wrong because ,

the function calls will go like this

f(1):n=2,i=2

f(2):n=4,i=3

f(4):n=7,i=4

f(7):in this case it will matches the base then return 7

f(7) function call  comes out of the stack and then it will comes to the f(4) function call in the stack, at this it not matching with base case .Now n:8(4+4) and i becomes 5.now it will call the function call f(8) .f(8) matches with the base case then it returns 8.

f(8) comes out the stack.In the f(4) function call last statement has returned then it will comes out of the stack.

now stack pointer comes to the f(2),at this it is not matching with base .So,n=7(n=2,i=5) ,now i becomes 6.now it will f(7) function call.in the f(7) function call ,it is matching with  base case it will return 7

f(7) comes out the stack.In the f(2) function call last statement has returned then it will comes out of the stack.

now stack pointer comes to the f(1),at this it is not matching with base .So,n=7(n=1,i=7) ,now i becomes 6.now it will f(7) function call.in the f(7) function call ,it is matching with  base case it will return 7.

f(7) returns 7 and comes out of the function call .and f(1) also comes out of the stack by returning 7.In f(1) it is executing the last statement and that statement return statement .So it will comes outb of the function call.

Note:In any recursion call ,function call comes out af the stack when it is executing last statement and that statement must be return statement.

Function calls:

 

 

 

Answer 2

Static storage class allows to initialize a variable for one time and re-initialization can't be done till end of program .

Here Iteration1: n(1)<=5; and  i=1;

                        n(2)=n(1)+i(1);

                        i(2)=i(1)+1;

Iteration2: n(2)<=5; and  i=2;

                        n(4)=n(2)+i(2);

                        i(3)=i(2)+1;

Iteration3: n(4)<=5; and  i=3;

                        n(7)=n(4)+i(3);

                        i(4)=i(3)+1;

Iteration2: n(7)<=5; and program execution terminates . So 7 is returned.

Answer 3

so how it is 7

Comments

Initialize k to 1 as you have to calculate f(1). Then you will get answer as 7.

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@Sheetal you are not initializing k as 1

Answer 4

C​​​​​​

f(1) is called .

static int i= 1 gets executed only after first func call after the subsequent func calls this line gets is not executed . 

if( n>=5) : this condition is false for n= 1 

n = n + i // n becomes 2 

i++; after this line i becomes 2 

return f(2) // f(2) is called 

again this condition ( n>= 5 ) becomes false 

Now n becomes 2+ 2 = 4

i becomes 3

f(4) is  called 

again this condition ( n>=5) becomes false 

Now n becomes 4+3= 7 

i becomes 4 

f(7) is called 

Now this condition (n>=5) becomes true 

return n // 7 gets returned