GATE CSE 2017 | Set 1 | Question: 36

Question

Consider the C functions foo and bar given below:

int foo(int val) {
    int x=0;
    while(val > 0) {
        x = x + foo(val--);
    }
    return val;
}

int bar(int val) {
    int x = 0;
    while(val > 0) {
        x= x + bar(val-1);
    }
    return val;
}

Invocations of $foo(3)$ and $bar(3)$ will result in:

• A. Return of $6$ and $6$ respectively.
• B. Infinite loop and abnormal termination respectively.
• C. Abnormal termination and infinite loop respectively.
• D. Both terminating abnormally.

Comments

Let's analyze both functions carefully.

Function foo

int foo(int val) {
    int x = 0;
    while(val > 0) {
        x = x + foo(val--);
    }
    return val;
}

The key point is post-decrement:

foo(val--)

passes the current value of val to the recursive call, and only then decrements val.

Suppose we call:

foo(3)

Inside the loop:

foo(val--)

becomes

foo(3)

and only after the call returns would val become 2.

But the call is foo(3) again, which immediately calls foo(3) again, and so on.

Thus the recursion never progresses toward a base case.

Result: infinite recursion → stack overflow → abnormal termination.

---

Function bar

int bar(int val) {
    int x = 0;
    while(val > 0) {
        x = x + bar(val - 1);
    }
    return val;
}

For bar(3):

while(val > 0)

and notice that val is never modified inside the loop.

The recursive call:

bar(val - 1)

eventually reaches:

bar(0)

which returns normally.

However, after that return, we are still inside:

while(val > 0)

with val == 1 (or 2, or 3 depending on the level), and val never changes.

Therefore the loop runs forever.

Result: infinite loop.

---

Conclusion

• foo(3) → abnormal termination (infinite recursion / stack overflow)
• bar(3) → infinite loop

Answer: C. Abnormal termination and infinite loop respectively.

---

 

Taking  val=1 ;  makes this loop work infinite,  basically bar(0) will  be called again and again , we are not calling bar(1) multiple times,  it is while loop which is calling bar(0) again and again.

while(1> 0) { //val=1

        x= x + bar(val-1);

    }

---

$$\text{bar(3)}$$

 

Answer 1

while(val>0)
{
x = x + foo(val--);
}
In this case foo(val--) is same as foo(val) & val-- ;
Because the recursive function call is made without changing the passing argument and there is no Base condition which can stop it.
It goes on calling with the same value ‘val’ & the system will run out of memory and hits the segmentation fault or will be terminated abnormally.
The loop will not make any difference here.
while(val>0)
{
x = x + bar(val-1);
}
bar(3) calls bar(2)
bar(2) calls bar(1)
bar(1) calls bar(0) ⇾ Here bar(0) will return 0.
bar(1) calls bar(0)
bar(1) calls bar(0)……..
This will continue.
Here is a problem of infinite loop but not abrupt termination.
Some compilers will forcefully preempt the execution.

Comments

thanks got it

 

Answer 2

Hello use this link Video Solution: https://www.youtube.com/watch?v=fW_vmo8A3Lg&ab_channel=JatinGupta

Answer 3

https://youtu.be/uB_vN8tMORw?feature=shared                best answer found

Answer 4

YOUTUBE LINK

Gate 2017 pyq C Programming | Consider the C functions foo and bar given below:int foo(int val)