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

there it should be num --,then it will be correct

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If it was --num, program would go into infinite loop.

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this should have been question number $52$

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

can you tell how we have written 52

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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

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