short to bit ?

[nimo frey]

Hello,

Is there a method to convert a short-value to bit-value?

I want to do something like:

short a = 2;
short b = 8;

short c = shortToBit(a) + shortToBit(b);

So you see, I want to add the bit value of a and b and convert the summary to a short value.

How can I do that?

[Henry Wong]

What do you mean by "bit value"? Do you mean you want a specific bit?

If you do, the easiest way is to shift the bit, and AND out the rest of the bits.

For example, if you want the fourth bit of the short...

boolean bit = (((s >> 4) & 0x01) == 1);

or if you don't want to shift ...

boolean bit = ((s & 0x10) == 0x10);

Henry

[nimo frey]

for example:

Short a = 2; // in byte: 0000'0010
Short b = 8; // in byte 0000'1000

the sum is: 0000'0010+0000'1000 = 0000'1010, which is in (decimal): 10

So you see, I want to sum the byte value of two short values and convert the byte value to decimal value (in my case, a short).

How can I do that in Java?


By the way:
Byte byte = a.byteValue()*8; // byte is "2" instead of "0000'0010", why is that so?

// for what is 0x10 ??
boolean bit = ((2 & 0x10) == 0x10); // this returns false, so bit=0

[Henry Wong]

nimo frey wrote:for example:

Short a = 2; // in byte: 0000'0010
Short b = 8; // in byte 0000'1000

the sum is: 0000'0010+0000'1000 = 0000'1010, which is in (decimal): 10

So you see, I want to sum the byte value of two short values and convert the byte value to decimal value (in my case, a short).

How can I do that in Java?

I am definitely missing something here.... 2 + 8 = 10

That is how numbers are added in Java. How is what you are describing any different? What's wrong with...

sum = a + b;

Henry

[Henry Wong]

By the way:
Byte byte = a.byteValue()*8; // byte is "2" instead of "0000'0010", why is that so?

I think you are confusing a byte with how it is displayed. A byte is a 8 bit value -- that is all. Whether that value is displayed (when you print it) as "2", or "0x2", or "0000'0010" depends on how you display it. In all three cases, the byte has the same value -- there is no "instead".

Henry

[Dave Brown]

Maybe the loss of precission compiler error is what's deterring the OP.. An addition will result in an int... so if you really needed a short as the result you must cast the addition i.e. short sum = (short) (a + b);

Dave.

Henry Wong wrote:

nimo frey wrote:for example:

Short a = 2; // in byte: 0000'0010
Short b = 8; // in byte 0000'1000

the sum is: 0000'0010+0000'1000 = 0000'1010, which is in (decimal): 10

So you see, I want to sum the byte value of two short values and convert the byte value to decimal value (in my case, a short).

How can I do that in Java?

I am definitely missing something here.... 2 + 8 = 10

That is how numbers are added in Java. How is what you are describing any different? What's wrong with...

sum = a + b;

Henry

[Henry Wong]

// for what is 0x10 ??
boolean bit = ((2 & 0x10) == 0x10); // this returns false, so bit=0

0x10 is the hexidecimal representation for 16, which is the fourth bit, ie... 0001'0000 in binary representation. And yeah, with 2, the fourth bit is zero.

Henry

[Henry Wong]

Dave Brown wrote:Maybe the loss of precission compiler error is what's deterring the OP.. An addition will result in an int... so if you really needed a short as the result you must cast the addition i.e. short sum = (short) a + b;

Dave.

Yeah, we will definitely need some clarification from the OP.

Henry

[Campbell Ritchie]

Dave Brown wrote: . . . short sum = (short) a + b;

If you want a short, surely you mean short sum = (short) (a + b); ?

[Dave Brown]

Campbell Ritchie wrote:

Dave Brown wrote: . . . short sum = (short) a + b;

If you want a short, surely you mean short sum = (short) (a + b); ?

yes thats what I mean, and stop calling me Shirly ;-)

[Campbell Ritchie]

Dave Brown wrote: . . . stop calling me Shirly ;-)

I lost a pair of glasses today, so I can't understand that joke any more

[nimo frey]

I have a couple of questions:

Imagine, I want the SECOND Bit beginning from the RIGHT Side of a Short-Value:

Short: 2
In Bits(Dualsystem): 0000'0010 (1 is the second bit beginning from the right side)

Is this shift-operation right, to get the second bit beginning searching from the right side ?

int bit= 2 >> 1;
System.out.println("bit: "+bit); // this returns 1, which should be right

int bit= 2 >> 0;
System.out.println("bit: "+bit); // this returns 2, why not 0 ?

[Campbell Ritchie]

Because shift operations do not produce a single bit. Look at this old thread.

If you want a particular bit try

/**
 * Gets the <i>n</i>th bit from the right, where the least-significant bit is no 0 and the most-significant bit is no 31.
 *@param input The number to investigate
 *@param n How many bits from the right to go to find the bit, eg 3 = bit no 3 which is the bit marked with an x in
 * 0000_0000_0000_0000_0000_0000_0000_x000
 * If the value of n is outwith the range 0 - 31, n % 32 is used.
 *@return 1 if the bit is set, 0 otherwise
 */
public static int getNthBitFromRight(int input, int n)
{
    int result = (input & (1 << n)) == 0 ? 0 : 1;
    return result;
}//end getNthBitFromRight

[nimo frey]

thank you, the thread is good and helped me to solve my problem!!

[Campbell Ritchie]

You're welcome