GATE CSE 2015 | Set 2 | Question: 15

Question

Consider the following function written in the C programming langauge :

void foo(char *a)
 {
    if (*a && *a != ' ')
      {
       foo(a+1);
       putchar(*a);
      }
}    

The output of the above function on input "$ABCD \ EFGH$" is

• A. $ABCD \ EFGH$
• B. $ABCD$
• C. $HGFE \ DCBA$
• D. $DCBA$

Comments

Since this is a recursion so characters are stored one by one on the stack and then printed right.?But if putchar(char *a) is placed before calling of the foo() then will the output be ABCD?

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plzz explain someone

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YES

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https://www.youtube.com/watch?v=6A1fVJN7dQo

Answer 1

Answer D as priority of $!=$ is greater than that of $\&\&$ in C. The execution happens as:

if ((*a) && (*a != ' '))

So, the if breaks either when $*a = 0$ (not '$0$' but ASCII $0$ or null character '\0'), or when $*a =$' '.

So, the recursive call goes like

$\text{'A' - 'B' - 'C' - 'D' -' '}$ (breaks) and then starts outputting

$DCBA$

Comments

@ASNR1010 Why should there be Runtime error for illegal access?

The program doesn’t even call $foo(105)$ {taking base address of *a as 100 and char as 1B}.

It will successfully call till $foo(104)$ and end after popping out the remaining segments in the stack containing previous calls of $putchar(*a)$.

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@Arjun sir

if ((*a && *a) != ' ')

wont this always be true?

 

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agreed, condition should be always true

Answer 2

THIS MIGHT HELP

 

Answer 3

as given in condition :  if *a = ' ' then it will not proceed further..hence we have to take only till ABCD ,

forget about EFGH.
aslo, it is tail recursion , means , your last called function's full body will execute first. Hence it will print DCBA.

Answer 4

Change the answer to D.

Answer 5

Output:-  DCBA