# shift operator

nachagoni rishi

Greenhorn

Posts: 26

posted 11 years ago

The following code will print

1: int i = 1;

2: i <<= 31;

3: i >>= 31;

4: i >>= 1;

5:

6: int j = 1;

7: j >>= 31;

8: j >>= 31;

9:

10: System.out.println("i = " +i );

11: System.out.println("j = " +j);

A) i = 1

j = 1

B) i = -1

j = 1

C) i = 1

j = -1

D) i = -1

j = 0

the answer is d. can u please give me an explanation for the answer.

1: int i = 1;

2: i <<= 31;

3: i >>= 31;

4: i >>= 1;

5:

6: int j = 1;

7: j >>= 31;

8: j >>= 31;

9:

10: System.out.println("i = " +i );

11: System.out.println("j = " +j);

A) i = 1

j = 1

B) i = -1

j = 1

C) i = 1

j = -1

D) i = -1

j = 0

the answer is d. can u please give me an explanation for the answer.

Vipin Das

Ranch Hand

Posts: 47

posted 11 years ago

Hi,

when you push 1 to 31 times left it overwrite the sign bit thus turning it to a neg number(1 << 31)

It will looks like 10000000...(put 31 zeroes) etc. To find out the real negative number take the two's complement which will give -2147483648. Now it is shifting to the right using the

>> signed shift operator, 111111111.... Taking the 2's complement we will get it is -1. Now if it is shifted

right it will continues to 1111111... making it -1. Thus the value of i is -1

In the second question the bit position is 00000000....1(put 31 zeroes), a single right shift itself push the 1 out and

it becomes 0. Thus further right shift makes no change. so j becomes 0.

For more info :

javaranch

when you push 1 to 31 times left it overwrite the sign bit thus turning it to a neg number(1 << 31)

It will looks like 10000000...(put 31 zeroes) etc. To find out the real negative number take the two's complement which will give -2147483648. Now it is shifting to the right using the

>> signed shift operator, 111111111.... Taking the 2's complement we will get it is -1. Now if it is shifted

right it will continues to 1111111... making it -1. Thus the value of i is -1

In the second question the bit position is 00000000....1(put 31 zeroes), a single right shift itself push the 1 out and

it becomes 0. Thus further right shift makes no change. so j becomes 0.

For more info :

javaranch

Arnab karmakar

Ranch Hand

Posts: 46

posted 11 years ago

hi rishi,

there is no complicacy.

U could guess the answer from the second option also.

In the second option,

j=1;

whose binary is

0000 0000 0000 0000 0000 0000 0000 0001

if we shift it by 31 bits, the out put is

0000 0000 0000 0000 0000 0000 0000 0000

So j =0;

and u can see that out of four options only one contains j=0;

U have to be tricky while answering any question

Arnab

there is no complicacy.

U could guess the answer from the second option also.

In the second option,

j=1;

whose binary is

0000 0000 0000 0000 0000 0000 0000 0001

if we shift it by 31 bits, the out put is

0000 0000 0000 0000 0000 0000 0000 0000

So j =0;

and u can see that out of four options only one contains j=0;

U have to be tricky while answering any question

Arnab

Netty poestel

Ranch Hand

Posts: 131

posted 11 years ago

Hi there

even I was struggling sometime with these bit juggling game till some kind souls explained me how it's done :-

int i = 1;

2: i <<= 31;

3: i >>= 31;

4: i >>= 1;

line 2 results :-

10000000000000000000000000000000

line 3 results:-

11111111111111111111111111111111 [all gaps preceeding the '1' on line 2 results are prefixed with a '1']

line 4 results in:

11111111111111111111111111111111 [ drop the last '1' on line 3 result, and fill the empty space created LHS with a '1']

this # is -ve , see the '1' on the extreme LHS , which decides this fact.

so flip all bytes:-

00000000000000000000000000000000

and add 1

resulting in

00000000000000000000000000000001

thus the final outcome is -ve 1 or -1.

also go for Arnab's theory of elimination, in this situation that seems to be the most smartest thing to do...

apart from giving the Exam and cracking it maybe in style,what 'real life value' this bit game has, confounds reasoning.

even I was struggling sometime with these bit juggling game till some kind souls explained me how it's done :-

int i = 1;

2: i <<= 31;

3: i >>= 31;

4: i >>= 1;

line 2 results :-

10000000000000000000000000000000

line 3 results:-

11111111111111111111111111111111 [all gaps preceeding the '1' on line 2 results are prefixed with a '1']

line 4 results in:

11111111111111111111111111111111 [ drop the last '1' on line 3 result, and fill the empty space created LHS with a '1']

this # is -ve , see the '1' on the extreme LHS , which decides this fact.

so flip all bytes:-

00000000000000000000000000000000

and add 1

resulting in

00000000000000000000000000000001

thus the final outcome is -ve 1 or -1.

also go for Arnab's theory of elimination, in this situation that seems to be the most smartest thing to do...

apart from giving the Exam and cracking it maybe in style,what 'real life value' this bit game has, confounds reasoning.