1.4k views

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 ?

1. $x = 1 + x;$
2. $x = 1 - x;$
3. $x = x - 1;$
4. $x = 1\% x;$
| 1.4k views

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

Let's consider options one by one:

1. $x= 1+x$
• For $x = 1$, it gives $2$ So, False

2. $x= 1- x$
• Here, B is correct, as
• For $x= 0$, it gives $1$.
• For $x= 1$, it gives $0$.

3. $x = x - 1$
• For $x=0$ , it gives $-1$. So, False

4. $x = 1 \% x$
• 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.

by Boss (15.5k points)
edited
+1 vote
option b.

if x=0 we get x=1

if x=1 we get x=0..

2 nd option satisfies this constraint
by Junior (647 points)

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 )

by Active (1.9k points)
0
it should be x%1 rt?
+1
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
0
ya..arjun sir i think so...it should be x%1...
0
ya.. mitali... that can also be true if x is allowed to take a value of 0 or 1 only....
0
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?
0

x=1%0

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

0
Yes i understand
0

why (D) is wrong?

?

0
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.

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