GATE CSE 2019 | Question: 27

Question

Consider the following C program:

#include <stdio.h>
int r() {
       static int num=7;
       return num--;
}
int main() {
       for (r();r();r())
              printf(“%d”,r());
       return 0;
}

Which one of the following values will be displayed on execution of the programs?

• A. $41$
• B. $52$
• C. $63$
• D. $630$

Comments

thanks @merohan17` this helped me better then the answers below.

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Question checks whether you know the control flow of for loop or not..If one of the option was 40 then many(including me) would have marked it in hurry.

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Initialization in done only once in for loop,
1. then we check condition ,
2. then we execute the for loop body
3. then we update loop variable
 go to step 1 & repeat till condition fails

Answer 1

Basic Points:-

1. After every expression statement  in the for loop, there is a sequence point
2. After the return value is copied, there is a sequence point.

for loop execution :-  for(e1;e2;e3)

On the first iteration, expression$1$ executed ( generally termed as initialization expression,)

next expression$2$ executed ( generally termed as condition check expression, if it evaluates to non-zero then only, for loop body executed, otherwise for loop terminates.)

After the first iteration, expression$3$ is executed  ( generally termed as increment expression. ), after that $e2$ evaluates and the process continues!

$\text{for}(\color{green}{r()};\color{red}{r()};\color{blue}{r()})$

{

    printf("%d",r());

}

before the main starts, the execution num is initialized with $7$ ( note that it is stored under static memory due to it is a static number.)

on first iteration :- $\color{green}{r()} \implies$ return $7$ but num changed to $6$.

$\color{red}{r()} \implies$ return $6$ but num changed to $5. \implies$ condition evaluate to true $\implies$ for loop body executes !

  printf("%d",r()); $\implies$ return $5$ but num changed to $4. \implies$ print $5$ on the screen.

$\color{blue}{r()} \implies$ return $4$ but num changed to $3$.

$\color{red}{r()} \implies$ return $3$ but num changed to $2. \implies$ condition evaluate to true $\implies$ for loop body executes !

  printf("%d",r()); $\implies$ return $2$ but num changed to $1. \implies$ print $2$ on the screen.

$\color{blue}{r()} \implies$ return $1$ but num changed to $0$.

$\color{red}{r()} \implies$ return $0$ but num changed to $-1. \implies$ condition evaluate to false $\implies$ for loop over !

Hence option $B$ : $52$

Comments

Amazing explanation!

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One key thing is that $num--$ here is post decrement  so only after returning value it will be decremented

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first it prints 5 and then 2 ,thats why @Jain07

Answer 2

May be this works : 

Must know the property of loop that :

(i) for first run initialization part run for first and last time 

(2) condition part runs each time.

(3) updation part runs after one entry in code

(4) In last run where condition fails still for statement part run i.e. condition part fails surely no entry in code part but updation part runs.

Answer 3

It will print 52

See the function starts from main which has a for loop .

for (r();r();r())

So first r() will return 7 and then 6 will go to second r() which is a condition (why 6 then notice its num-- so first it will return value then decrement , Now second r() which is a condition will return 6 and decrement to 5 ,further as this condition satisfies 5 will go for print :

printf("%d", r());

Now again here there is r() so it will print 5 and decrement again to 4

Which will go to increment/decrement condition of for that is third r() and as we see its a decrement

num--

so it will return 4 and decrement to 3 which will again go to second r() and will return 3 with decrementing to 2 and that 2 will go to print r() which will print 2  .

so final print 52

Comments

why is the ans not 41 i feel it should be 41 because printf will print whatever r() is returning

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184640 ktr.cse.14 updated my response :)

Answer 4

52. It will be decremented only after value is sent

Comments

No it will print 41.

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What is the answer for convert function.

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I checked it on ideone. It is 52

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initialization, comparison

print

increment/decrement

and so on

Answer 5

52

https://ideone.com/z3LEj9

Answer 6

As it is given return num-- ; so the value retuned will be that of num before decreasing it. 

So, in first call, num = 7, returns 7 and becomes 6. Next call finds num = 6, returns 6 and makes num 5.

 

The maechanism of for loop:

     Initialize expression -> Test -> Body -> Update -> Test -> Body -> Update .................continues until test expression is false.

So the values will be :

    Initialize expression (returned 7, num = 6)-> Test (returned 6, num = 5) ->

    Body (returned 5 printed, num = 4  ) ->

    Update (returned 4, num = 3) -> Test (returned 3, num = 2) ->

    Body (returned 2 printed, num = 1) ->

    Update (returned 1, num = 0) -> Test (returned 0 You are terminated, num = -1) .

Values printed 52