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+13 votes

Consider the following C-program:

void foo (int n, int sum) {
    int k = 0, j = 0;
    if (n == 0) return;
    k = n % 10; j = n/10;
    sum = sum + k;
    foo (j, sum);
    printf ("%d,",k);

int main() {
    int a = 2048, sum = 0;
    foo(a, sum);
    printf("%d\n", sum);

What does the above program print?

  1. 8, 4, 0, 2, 14

  2. 8, 4, 0, 2, 0

  3. 2, 0, 4, 8, 14

  4. 2, 0, 4, 8, 0

asked in Algorithms by Veteran (69k points) | 1.5k views

3 Answers

+16 votes
Best answer

Option D.

foo is printing the lowest digit. But the printf inside it is after the recursive call. This forces the output to be in reverse order

2, 0, 4, 8

The final value "sum" printed will be 0 as C uses pass by value and hence the modified value inside foo won't be visible inside main.

answered by Loyal (3.3k points)
edited by
@Arjun sir plzz elaborate this solution.
@Neha Singh

just execute a program in pen and paper u will get D as an answer...
+6 votes

Quick soln :-Option Elimination

We will try to analyse o/p from last.

Last line of program is to print sum which is passed by value so it will retain its value 0. So option A & C eliminated.

Now call foo(2048,0) which push 8 into stack first so it will pop at last so 8 will print as 2nd last o/p.

Hence B is eliminated and Option D is Ans.

answered by Veteran (23.4k points)
0 votes
the last print statement to be executed is in the main(). Since every time foo() is called, we are doing pass by value, the value stored in variable sum is local to foo() function calls.

So when control returns to main(), the value of sum will be 0. (as initialized in main()'s body). That eliminates (a) & (c).

Recursive calls(values stored in stack -> LIFO) on foo(), when returned is printing k values in the reverse order : 2->0->4->>8

Hence, answer : (d)
answered by Veteran (10.1k points)

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