GO Classes Test Series 2024 | Programming | Test 2 | Question: 21

Answer 1

int rec(int x, int y) → this function returns some value only when x = 0;

Now, when we pass non-zero x, will x parameter for any future call ever become 0?

We’re recursively calling rec, in only one case we’re modifying parameter x, only when x > y, and that to x – y.

Now, this x – y can never become 0.

Therefore, rec(x, y) for all non-zero x, will result in infinite recursive calls.

Answer :- D.

Comments

Suppose x > y, we will repetitively decrease x by y. Three cases are possible.

Case 1: We reach the stage where x < y.

In this case, “if” condition will never be true and x gets no chance to become zero and infinite loop/recursion happens.

Case 2: x = y. In this case, “else” block runs and y = y -x =  0.

In this case, x will remain whatever it is currently and would get no chance to become zero and infinite loop/recursion happens.

Case 3: 2*y > x > y.

In this case, x = x – y and now, x < y and we go back to case 1.

Hence, we can conclude that infinite loop/recursion is inevitable unless x = 0 is passed originally.

 

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if (x > = y)

this equal is needed to stop the infinite recursion.

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Or we can add a condition if(y==0)

Answer 2

Answer:D

The given program calculates the greatest common divisor (GCD) of two integers x and y using recursion. However, it lacks a termination condition for when x and y are equal, leading to an infinite loop. Consequently, the program will not produce any output and will continue executing indefinitely.

Answer 3

The function terminates only when x = 0 

but here when it is rec(2,0) => x > y and again rec(x-y, y) = rec(2-0,0) = rec(2,0) executes and this function keeps on repeating and it gets stackoverflow.