GATE IT 2004 | Question: 15

Question

Let $x$ be an integer which can take a value of $0$ or $1$. The statement

if (x == 0) x = 1; else x = 0;

is equivalent to which one of the following ?

• A. $x = 1 + x;$
• B. $x  = 1 - x;$
• C. $x = x - 1;$
• D. $x = 1\% x;$

Comments

basically finding 1's complement of x

and 1's complement is nothing but 1 - x , as if x = 1 ; 1-1 = 0 and if x = 0; 1-0 = 1 

Answer 1

Firstly, our requirement is for $x=1$ it makes '$0$' and for $x= 0$ it makes '$1$'

Let's consider options one by one:

• A. $x= 1+x$
  • For $x = 1$, it gives $2$ So, False
     
• B. $x= 1- x$
  • Here, B is correct, as 
    • For $x= 0$, it gives $1$.
    • For $x= 1$, it gives $0$.
       
• C. $x = x - 1$    
  • For $x=0$ , it gives $-1$. So, False
     
• D. $x = 1 \% x$
  • For  $x= 0$ , it gives $1 \% 0$ . I think it is undefined
  • Even if we consider $x = x\%1 $   
    • for $x= 0$ ,it gives $0\%1 = 0$ But we require $1$. 

So, Option (B) is correct.

Comments

Dont assume its boolean if its not given

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1 % 0 is undefined behavior 

• 1 / 0 → undefined

• 1 % 0 → also undefined 

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but it is given it will only 0 or 1

Answer 2

option b.

if x=0 we get x=1

if x=1 we get x=0..

2 nd option satisfies this constraint

Answer 3

B is Correct.

This code is nothing but flipping the value of X.

If Content of X is 0 or 1 then we can flip it using following logic

X = 1 - X

or

X = X ^ 1      (^ is xor operation)

or

X = !x         (! is logical not operation)

So option B is the answer.

Answer 4

option D,

as,

x=1%x

assigns remainder after dividing with 1 and hence x has to be 0 for getting remainder value anything other than 0.(as 1 divides every number>0 )

Comments

it should be x%1 rt?

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sir it should be option B right?

x= 1-x

if we take x=0 den we will get x =1 and for x=1, we will get x=0

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ya..arjun sir i think so...it should be x%1...

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ya.. mitali... that can also be true if x is allowed to take a value of 0 or 1 only....

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I have the doubt regarding option (D)

      x=1%x

if x=0,then we get x=1

x=1%0

x=1

and if x=1, then we get x=0

x=1%1

x=0

why $(D)$ is wrong?

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x=1%0

It will not give 1. "Divide by 0 exception"  

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Yes i understand

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why (D) is wrong?

?  

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if(x=0)
  x=1;
else 
  x=0; 

 

$(D)$ x = 1% x

       when x=0

             x = 1%0

                  i.e.$x=\frac{1}{0}=\infty$

                    it does not give any remainder.

 when x=1

             x = 1%1

                  i.e.$x=\frac{1}{1}$

                    it  gives remainder $= 0$.

Both cases are not satisfied. So, this is false.

Answer 5

option B is the correct answer 

x=1-x 

for x=0 :

x=1-0

=1

for x=1:

x=1-1

=0

no other options are matching the requirement