GATE IT 2006 | Question: 50

Question

Which one of the choices given below would be printed when the following program is executed?

#include <stdio.h>
void swap (int *x, int *y)
{
    static int *temp;
    temp = x;
    x = y;
    y = temp;
}
void printab ()
{
    static int i, a = -3, b = -6;
    i = 0;
    while (i <= 4)
    {
        if ((i++)%2 == 1) continue;
        a = a + i;
        b = b + i;
    }
    swap (&a, &b);
    printf("a =  %d, b = %d\n", a, b);
}
main()
{
    printab();
    printab();
}

• A. $a = 0, b = 3$
  $a = 0, b = 3$
• B. $a = 3, b = 0$
  $a = 12, b = 9$
• C. $a = 3, b = 6$
  $a = 3, b = 6$
• D. $a = 6, b = 3$
  $a = 15, b = 12$

Comments

option D is the correct answer

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Solution

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Never saw more clearer and better solution than this
I dont know why it doesnt have more upvotes

Answer 1

First of all, the swap function just swaps the pointers inside the function and has no effect on the variables being passed. 

Inside printab, a and b are added odd integers from $1$-$5$, i.e., $1+3+5 = 9$. So, in first call to printab, $a = -3 + 9 = 6$ and $b = -6 + 9 = 3$. 

Static variables have one memory throughout program run (initialized during program start) and they keep their values across function calls. So, during second call to printab, $a = 6 + 9 = 15$, $b = 3 + 9 = 12$.

Hence, (D) is choice. 

Comments

@ankitgupta.1729 @Abhrajyoti00

How does the static variable behave if used in two different function.

e.g.

printab()

{ 

static int a=-3;

}

main()

{

static int a = 5;

printab();

}

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@GateOverflow04 

Static has scope within the body (where it is initialized) but it’s lifetime is throughout the program. Thus for the main function variable $a$ will be initialized in data segment with value $5$ which is local only to the main function.

And another variable $a$ will be initialized in data segment with value $-3$ which is local only to the printab() function.

Since it is in data segment it's value can be altered any time, but can't be initialized more than once. Also the value of 'a' in main() and the value of 'a' in the printab() function are independent of each other  and can't be used anywhere outside their own scope.

This will be useful for static variables : https://gateoverflow.in/2176/gate-cse-2012-question-48?show=386635#c386635

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If anyone has doubt regarding the parenthesis part here is the answer, during the evaluation yes, (i++) will be evaluated first but evaluating it doesn't mean I value be increased at this point, the main concept of i++ operator is that, current value of I is used in the current expression and once the usage of I in expression ends it is incremented.

Answer 2

while (i <= 4)
{
    if ((i++)%2 == 1) continue; // key point of program here if comdition true then go to while loop directly 
    a = a + i; // will process when i= even 
    b = b + i; // will process when i= even 
 }

Inside printab, a and b are added odd integers from

for i= 1 a= -3 + 1 = -2  and b = -6+1= -5 

for i= 3 a= -2 + 3 = 1  and b = -5+3= --2 

for i= 5 a= 1 + 5 = 6  and b = -2 +5= 3

here printf(a,b) = 6,3 

if static int i= 0; given in code then only old value of i is used. 
since static int i and i= 0 is given so for next printab () i= 0 is taken.
 By looking 1st  printf(a,b) = 6,3 

Ans is D

reffer for continue:https://www.codingunit.com/c-tutorial-for-loop-while-loop-break-and-continue

Comments

@ prashant in the above link it says  " It is also possible to use ++i or --i. The difference is is that with ++i (prefix incrementing) the one is added before the “for loop” tests if i < 10. With i++ (postfix incrementing) the one is added after the test i < 10."

is it right ....??? plz clear my doubt.

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Why i is reintianlize to 0?

, as i is static variable and intialize only once, in second printable() i should star with 5,

Correct me if wrong somewhere!

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when i=5 then loop terminates then  how i=5 is calculated?

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In for loop prefix increment and postfix increment both are same

But in while loop make difference 

i++ means first test condition then increment 

++i means first increment then test condition

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@Ram Swaroop

since the brackets have more precedence then why (i++) is not incrementing first and then checking the condition ?

In your example given you have used ++i and not (++i).

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@ Satbir same doubt ...did u got it?

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If we use bracket then definitely  bracket have higher precedence than any increment or decrement operator

But in example we are using only i++ and ++Iso it work normal as post increment and pre increment operator work.

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@googlegoku what you're not getting

 

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@Ram Swaroop
int i=0;
" if ((i++)%2 == 1) "

why i is incremented after %2 operation is done even though i++ is in bracket 

 

 

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You are talking about the actual gate questionsorry I think wrong  you are asking about my comment

@googlegoku
We know unary operators (++,--) have higher precedence than arithmetic operators but here () have useless because of post increment operator
you can try this code
int main(void) {

int i =0;
while (i<=4)
{
if ((i++)%2==1) continue;

printf ("%d", i);
}
   
    return 0;
}

Answer 135

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i think () operator has higher presendence than % in case of  use it as function call

otherwise it has not affect expression.

Answer 3

$main()$

$\downarrow$

$printab()$

$\fbox{static i=0}$ $\fbox{static a=-3}$$\fbox{static b=-6}$

$i=0$  

$1.while(0<=4)\{$

$if(0\%2==1)// condition\ false$

$\fbox{i=1}$

$a=-3+1=-2$

$b=-6+1=-5$

$2.while(1<=4)\{$

$if(1\%2==1)\ continue;$

$\fbox{i=2}$

$3.while(2<=4)\{$

$if(2\%2==1)// condition\ false$

$\fbox{i=3}$

$a=-2+3=1$

$b=-5+3=-2$

$4.while(3<=4)\{$

$if(3\%2==1)\ continue;$

$\fbox{i=4}$

$5.while(4<=4)\{$

$if(4\%2==1)// condition\ false$

$\fbox{i=5}$

$a=1+5=6$

$b=-2+5=3$

$6.while(5<=4)// condition\ false$

$\downarrow$

$swap(\&a,\&b)$

Important to notice: It will swap the pointers not the values.

Print the values of $a,b=6,3$

---

Again call $printab();$

.

.

Print the values of $a,b=15,12$

$Ans: D$

Comments

Point to point explantion

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Thanks a lot sir

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but post increment is inside the brackets so its evaluated first before comparison anyway due to higher precedence

Answer 4

        if ((i++)%2 == 1) continue;

This is the heart of the code. continue will make the compiler skip rest of the lines in the loop, and go straight back to checking the loop condition again.
Post increment will increment i after checking the condition.

When i = 0, we won't hit continue. So, add 1 (and not 0, because post-increment)

When i = 1, we hit continue. So skip rest of the lines.

When i = 2, add 3.

When i = 3, we hit continue.

When i = 4, add 5.

Break out of while loop now. Because i = 5.

 

So finally; $a = -3 + 1 + 3 + 5 = 6$. And $b = -6 + 1 + 3 + 5 = 3$

Swap doesn't do anything to the values inside a and b, so a and b stay intact.

a = 6, b = 3

Option D.

 

Now, when printab() is called the second time, i = 5. But in the immediate next line, i = 0. So, while loop will run the same.
$a = 6 + 1 + 3 + 5 = 15$
$b = 3 + 1 + 3 + 5 = 12$

Comments

but post increment is inside the brackets so its evaluated first before comparison anyway due to higher precedence

Answer 5

answer should be:

a =  6, b = 3
a =  15, b = 12

i.e option D

Comments

@arjun In the 2nd printab() call, i value will be 5 since it is static variable and we cannot re-initialize it. So why a=15,b=12?

Answer 6

the given code can be simplified as the code below without changing its meaning and output.

#include <stdio.h>

/*void swap(int *x, int *y)
{
    static int *temp;
    temp = x;
    x = y;
    y = temp;
}*/

void printab()
{
    static int i, a = -3, b = -6;
    i = 0;

    while (i <= 4)
    {
        //we have to increment i irrespective of the number being odd or even
        if (i % 2 == 1) //if i is odd
        {

            i++; // i is incremented here when it is odd
            continue;
        }

        
        i++; //i is incremented here when it is even
        a = a + i;
        b = b + i;
    }



    //we can remove the swap function since it only swaps the pointers to a,b and has no effect on a,b values themselves.
    //swap(&a, &b);
    printf("a =  %d, b = %d\n", a, b);
}
int main()
{
    printab();
    printab();
    return 0;
}