With Regards Of Data Representation In The Memory

[Ranji Sura]

byte a,b;
a = 10;
b = a;

Hi,
Please help me here.. I want to create my note today, but I am bit confuse here..

a = 10; a is byte variable.. If I am right .. here 10 is as a int.. that means it has stored here in the 32 bit variable temporarily and then copy to the bits to the byte array..


0 000 0000 0000 0000 0000 0000 0000 1010 ==>>10
^ sign bit
when assigning to the 8 bit (a byte variable) last 8 bits copied to a byte variable
0000 1010 ==>>a now a contains 10;


what I want to know is... when assigning a to b, is it copied directly (8-bits of a) to (8-bits of b) ?
or
is it copied (8-bits of a) to (32-bit int variable) temporarily , then assign to b like above process ?

Please help me quick...
Thank You.

[Rameshwar Soni]

Always use a UseAMeaningfulSubjectLine (<-- click) for your thread.

Ranjith Suranga wrote:
Please help me quick..

And EaseUp (<-- click)

And Welcome To Ranch

[Ranji Sura]

Rameshwar Soni wrote:Always use a UseAMeaningfulSubjectLine (<-- click) for your thread.

Ranjith Suranga wrote:
Please help me quick..

And EaseUp (<-- click)

And Welcome To Ranch

Thank you for showing my faults... It is extremely thank you.

[Jeff Verdegan]

Ranjith Suranga wrote:
what I want to know is... when assigning a to b, is it copied directly (8-bits of a) to (8-bits of b) ?
or
is it copied (8-bits of a) to (32-bit int variable) temporarily , then assign to b like above process ?

It's probably not specified. You can look through the JLS if you want to find out for sure.

I think I read somewhere that most mainstream JVM implementations store bytes and ints in the same internal type, so all the JVM would have to do is a sign-extend or mask out the high bytes, if even that.

[Rameshwar Soni]

I think you need not worry about such things, see what Jesper de Jong says here.

Hope it helps and lets see what other people has to say.

[Angus Comber]

Your post got me thinking.

So I knocked up this code.

public class AssignmentTest {
    public static String printbinary(byte val) {
	  int displayMask = 1 << 7;
          String str = new String();
          for(int bit = 1; bit <= 8; ++bit) {
              str += (val & displayMask) == 0 ? '0' : '1';
              val <<= 1; //shift one to left
          }
          return str;
    }

    public static String printbinary(int val) {
	  int displayMask = 1 << 31;
          String str = new String();
          for(int bit = 1; bit <= 32; ++bit) {
              str += (val & displayMask) == 0 ? '0' : '1';
              val <<= 1; //shift one to left
          }
          return str;
    }

  public static void main(String[] args) {   
      byte a, b;
      a = 10;
      b=a;
      System.out.printf("10=%s\n", printbinary(10));
      System.out.printf("a=%s\n", printbinary(a));     
      System.out.printf("b=%s\n", printbinary(b));
  }
}

It prints:
10=00000000000000000000000000001010
a=00001010
b=00001010

Not sure if that answers your question. I would guess that the reference held by a is copied to b.

You may be simplifying the copy operation too much. Probably b=a; means references are copied, not actual values.

Is this homework or something because it is not something that a programmer would generally need to worry about.

[Winston Gutkowski]

Ranjith Suranga wrote:what I want to know is... when assigning a to b, is it copied directly (8-bits of a) to (8-bits of b) ?
or is it copied (8-bits of a) to (32-bit int variable) temporarily , then assign to b like above process?

As Jeff said, it may be specified; but basically, you don't need to know.

What IS specified is exactly how a byte is converted to an int, and vice-versa - see JLS 5.1.2 and 5.1.3.

However, I'd say that all this obsessing with what Java memory looks like is a waste of your time; especially at this stage. In many cases it's not specified; and a major point of Java is that IT manages memory - and in most cases you don't need to know how it does it.

My advice: DON'T use bytes unless you absolutely have to - especially if you're doing arithmetic on them - ints are much easier to work with.

Winston

[Jeff Verdegan]

Angus Comber wrote:
It prints:
10=00000000000000000000000000001010
a=00001010
b=00001010

Not sure if that answers your question.

It doesn't.

Angus Comber wrote:
Not sure if that answers your question. I would guess that the reference held by a is copied to b.

You may be simplifying the copy operation too much. Probably b=a; means references are copied, not actual values.

a and b are primitive variables, not reference variables. They directly hold the numerical values we give them.

[Angus Comber]

a and b are primitive variables, not reference variables. They directly hold the numerical values we give them.

Can you please clarify?

[Jeff Verdegan]

Angus Comber wrote:

a and b are primitive variables, not reference variables. They directly hold the numerical values we give them.

Can you please clarify?

Primitive types--byte, char, short, int, long, float, double, boolean--directly store the values that we care about. There's no reference, and no object. So this statement: "Probably b=a; means references are copied" is incorrect for the a and b we're talking about here.

[Ranji Sura]

It is extremely thank you for all of you to giving answers, ideas, opinions... They really help me to improve. Once again.. thank you for all of you.