Java Programming - Operators and Assignments - Discussion
Discussion Forum : Operators and Assignments - Finding the output (Q.No. 4)
4.
What will be the output of the program?
class BitShift
{
public static void main(String [] args)
{
int x = 0x80000000;
System.out.print(x + " and ");
x = x >>> 31;
System.out.println(x);
}
}
Answer: Option
Explanation:
Option A is correct. The >>> operator moves all bits to the right, zero filling the left bits. The bit transformation looks like this:
Before: 1000 0000 0000 0000 0000 0000 0000 0000
After: 0000 0000 0000 0000 0000 0000 0000 0001
Option C is incorrect because the >>> operator zero fills the left bits, which in this case changes the sign of x, as shown.
Option B is incorrect because the output method print() always displays integers in base 10.
Option D is incorrect because this is the reverse order of the two output numbers.
Discussion:
17 comments Page 1 of 2.
Fakru said:
8 years ago
If you convert 0x80000000 (which is in hexadecimal form) to decimal form, it gives 2147483648. But, the range of integer in java is -2147483648 to 2147483647.
Since, the converted decimal value exceeds the range(positive value) of integer, it takes the negative value and the values go in a cycle from positive to negative. As the converted decimal value exceeds the integer range by 1, the 1st value from -2147483648 is taken, which is -2147483648.
Since, the converted decimal value exceeds the range(positive value) of integer, it takes the negative value and the values go in a cycle from positive to negative. As the converted decimal value exceeds the integer range by 1, the 1st value from -2147483648 is taken, which is -2147483648.
(1)
Suresh Bandam said:
1 decade ago
Int x = 0x80000000;
Here, x value is in Hexadecimal and if we convert into in decimal -2147483648.
If we convert into in binary, the value is 1000 0000 0000 0000 0000 0000 0000 0000.
x = x >>> 31;
X is stored using 32 bit 2's complement form.
By this Unsigned right shift operation,
Before: 1000 0000 0000 0000 0000 0000 0000 0000.
After: 0000 0000 0000 0000 0000 0000 0000 0001.
Here, x value is in Hexadecimal and if we convert into in decimal -2147483648.
If we convert into in binary, the value is 1000 0000 0000 0000 0000 0000 0000 0000.
x = x >>> 31;
X is stored using 32 bit 2's complement form.
By this Unsigned right shift operation,
Before: 1000 0000 0000 0000 0000 0000 0000 0000.
After: 0000 0000 0000 0000 0000 0000 0000 0001.
Keshav Kumar said:
9 years ago
Binary equivalent of 0X80000000 is 10000000000000000000000000000000. In int data type if 32nd bit is off, then decimal equivalent is positive and if 32nd bit is on then the equivalent is negative.
Santhosh said:
1 decade ago
How do you perform shift operation? Can you explain?
How did you get -2147483648 and 1
what is mean by it ?
How did you get -2147483648 and 1
what is mean by it ?
Manisha said:
7 years ago
How to convert 0x80000000 in hexadecimal to decimal?
Can you explain the process?
What is the value of x?
Can you explain the process?
What is the value of x?
Nagalakshmi said:
9 years ago
How to convert hexadecimal to decimal such huge number very quickly?
Please give me the short method.
Please give me the short method.
Hima said:
10 years ago
int x = 0x80000000;
'x' in the value 0x80000000; indicates it as a hexadecimal value.
'x' in the value 0x80000000; indicates it as a hexadecimal value.
Sriram said:
10 years ago
How the 0000 0000 0000 0000 0000 0000 0000 0001 converts to -2147483648 and 1?
Ricky said:
9 years ago
0X80000000 to 2147483648(2^31) is clear but how this '-' sign comes in it?
Amar said:
9 years ago
But hexadecimal representation of Number allow only 4 Digits after 0X?
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