Bits, bytes and megabits

In telecommunications (data rates) 1,000,000 bits in a megabit, apparently.
In computing (file size) 1,048,576 bytes in a megabyte, apparently (1,048,576 bits in a megabit)
There are different names available to distinguish them (e.g. kibibit/kilobit) but apparently, these and similar are often not used. Result? Confusion. http://en.wikipedia.org/wiki/Megabyte
I suspect that quite a few people don't realise this - I didn't. Makes a difference when you are trying to check your upload and download speeds and verify BT's claims about their speeds.

The way I do it is X the results on speedtesters that give results in kbps [something like 74500/15500] by 1024.
The fake results above would be 74500x1024 = 76288000Kbps which = 76.28Mbps [approx 76.3Mbps] which is very near the line max after overheads.cool.
74954kbps = 76752896Kbps = 76.7Mbps
15207 = 15571968Kbps = 15.6Mbps
these are my actual readings.

Similar Messages

Bits, bytes, and all the rest

I need clarification on what some stuff really represents.
My project is to build a Huffman tree (no problem). However, all tutorials and examples that I can find on the net take from a text file with the format <letter> <frequency>, such as:
a 12
b 3
c 4
d 8
Mine has to read any kind of file, such as a text file.
For example, if myDoc.txt contains:
This is my document.
I have to have my program read the bytes from the infile, count the frequency of each byte from 0 through 255. Then the frequencies must be placed on a list of single node Huffman trees, and build the tree from this list.
I think I am having trouble because I just cannot get straight what a byte "looks like". My project says ensure you are not counting the "letters" of the infile, but the "bytes".
So what does a byte look like? When I read in the file as above, what is a "byte" representation of the letter T, i,h, etc?
Any ideas?

Ok, Roshelle....here is a little history lesson that you should have learned or should have been explained to you by your instructor before he/she gave you the assignment to construct a Huffman tree.
A bit is a binary digit which is either 0 or 1 -- it is the only thing that a computer truly understands. Think about it this way, when you turn on the light switch, the light comes on (the computer sees it as a 1), when you turned the switch off, the light goes out (the computer sees it as a 0).
There are 8 bits to a byte -- you can think of it as a mouthful. In a binary number system, 2 to the power of 8 gives you 256. So, computer scientists decided to use this as a basis for assigning a numerical value to each letter of the English alphabets, the digits 0 to 9, some special symbols as well as invisible characters. For example, 0 is equivalent to what is known as null, 1 is CTRL A, 2 is CTRL B, etc...... (this may vary depending on the computer manufacturers).
Now, what was your question again (make sure that you count the byte and not the letters)?
As you can see, when you read data from a file, there may be characters that you don't see such as a carriage return, line feed, tab, etc.....
And like they said, the rest is up to the students!
V.V.

Addition in Java bits, bytes and shift

Hi,
I have a byte array consisting of 8 bit values in each cell.
When I do an addition of two cells - which gives me a 16 bit result I use the following:
Integer a = (int) (arr[value] & 0xff);
a = a << 8;
Integer b = (int) (arr[secondValue] & 0xff);
Integer c = a + b;This seems to work fine. My question is how would I add 3 (24 bit result) or 4 (32 bit result) cells.
Would the following work: 24 bit result
Integer a = (int) (arr[value] & 0xff);
a = a << 16;
Integer b = (int) (arr[secondValue] & 0xff);
Integer c = (int) (arr[thirdValue] & 0xff);
Integer d = a + b + c;I am not sure if I have got the shift (<<16) usage correct or if it should be used in order to obtain the variables b or c.
All help appreciated.

khublaikhan wrote:
Just to confirm for 32 bit it would be:
// 32-bit
int a = (int)(arr[value] & 0xff);
int b = (int)(arr[secondValue] & 0xff);
int c = (int)(arr[thirdValue] & 0xff);
int d = (int)(arr[fourthValue] & 0xff);
int e = (a<<24) + (b<<16) + (c<<8) + d;
Integer eInt = new Integer(e);
Actually, the more I think about it, you may need to use longs instead of ints for 32-bit (not 16- or 24-bit though). It depends on what your data actually is. If you're expecting your 32-bit number to be an unsigned integer, then you'd better go with long. This is because in Java, ints utilize two's complement, so if the high-order byte you read in for a 32-bit value is very large, the resulting int will be interpreted by Java as a negative number instead of the large positive value you expect.
I'm probably not being very clear. Check out http://en.wikipedia.org/wiki/Two's_complement for the details.
In other words, if you wanted to get 32-bit values in this way, and the computed values are expected to be non-negative integers, use this:
// 32-bit
int a = (int)(arr[value] & 0xff);
int b = (int)(arr[secondValue] & 0xff);
int c = (int)(arr[thirdValue] & 0xff);
int d = (int)(arr[fourthValue] & 0xff);
long e = (a<<24) + (b<<16) + (c<<8) + d;
Long l = new Long(e);If you're actually reading in a 32-bit, two's complement value, then keep using int like you originally posted.

Trying to understand the details of converting an int to a byte[] and back

I found the following code to convert ints into bytes and wanted to understand it better:
public static final byte[] intToByteArray(int value) {
return new byte[]{
(byte)(value >>> 24), (byte)(value >> 16 & 0xff), (byte)(value >> 8 & 0xff), (byte)(value & 0xff) };
}I understand that an int requires 4 bytes and that each byte allows for a power of 8. But, sadly, I can't figure out how to directly translate the code above? Can someone recommend a site that explains the following:
1. >>> and >>
2. Oxff.
3. Masks vs. casts.
thanks so much.

By the way, people often introduce this redundancy (as in your example above):
(byte)(i >> 8 & 0xFF)When this suffices:
(byte)(i >> 8)Since by [JLS 5.1.3|http://java.sun.com/docs/books/jls/third_edition/html/conversions.html#25363] :
A narrowing conversion of a signed integer to an integral type T simply discards all but the n lowest order bits, where n is the number of bits used to represent type T.Casting to a byte is merely keeping only the lower order 8 bits, and (anything & 0xFF) is simply clearing all but the lower order 8 bits. So it's redundant. Or I could just show you the more simple proof: try absolutely every value of int as an input and see if they ever differ:
   public static void main(String[] args) {
      for ( int i = Integer.MIN_VALUE;; i++ ) {
         if ( i % 100000000 == 0 ) {
            //report progress
            System.out.println("i=" + i);
         if ( (byte)(i >> 8 & 0xff) != (byte)(i >> 8) ) {
            System.out.println("ANOMOLY: " + i);
         if ( i == Integer.MAX_VALUE ) {
            break;
   }Needless to say, they don't ever differ. Where masking does matter is when you're widening (say from a byte to an int):
   public static void main(String[] args) {
      byte b = -1;
      int i = b;
      int j = b & 0xFF;
      System.out.println("i: " + i + " j: " + j);
   }That's because bytes are signed, and when you widen a signed negative integer to a wider integer, you get 1s in the higher bits, not 0s due to sign-extension. See [http://java.sun.com/docs/books/jls/third_edition/html/conversions.html#5.1.2]
Edited by: endasil on 3-Dec-2009 4:53 PM

Convert.ToByte and logical masks, from VB to LabVIEW

Hi guys,
I have to write a VI communicating via RS232 with a microcontroller for a PWM application. A colleague has given me the GUI he has developpd in VB and I would like to integrate it into my LabVIEW programme by converting the VB code into Labview code. Here's the code:
Private Sub LoadButton_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles loadButton.Click
        Dim bufLen As Integer = 12      // Buffer length
        Dim freq As Integer = 0      // frequency
        Dim pWidth As Integer = 0      // pulse width
        Dim dac As Integer = 0       // Value used in oscillator setting for generating pulse frequency
        Dim addr As Integer = 0      // Address of the pulse width in the pulse generator.
        Dim rTime As Integer = 0      // duration of machining/pulse train in ms.
        Dim returnValue As Byte = 0      // A variable for storing the value returned by the
//microcontroller after it receives some data
        Dim buffer(bufLen) As Byte       // creates an array of bytes with 12 cells => buffer size = 8 x12 = 96 bits
// can you explain a bit please I know you're converting the entered values into byte and put them one by one in a specific order. This order is how the
//microcontroller expects them
            addr = (Floor((pWidth - Tinh) / Tinc)) // Formula from hardware, calculates address setting for pulse generator to set pulse width.
            buffer(0) = Convert.ToByte(Floor(3.322 * (Log10(freq / 1039)))) // Caluclates OCT value for use in setting oscillator for pulse freq.
            dac = (Round(2048 - ((2078 * (2 ^ (10 + buffer(0)))) / freq))) // Calculates DAC value for use in setting oscillator for pulse freq.
            buffer(1) = Convert.ToByte((dac And &HF00) >> 8)                         //
// &H is the vb.net to tell it its hex code, F00 gives the top 4 bits from a 12 bit value.
            buffer(2) = Convert.ToByte(dac And &HFF) // For values that are larger than 256 (8bits) the value needs to be split across 2 bytes (16 bits) this gets the //bottom 8 bits. &H is the vb.net to tell it its Hex.
            buffer(3) = Convert.ToByte((addr And &HFF0000) >> 16) // This value may be so large it requires 3 bytes (24bits). This gets the top 8 bits.
            buffer(4) = Convert.ToByte((addr And &HFF00) >> 8) // This gets the middle 8 bits.
            buffer(5) = Convert.ToByte(addr And &HFF)// This gets the bottom 8 bits.
            buffer(6) = Convert.ToByte((rTime And &HFF0000) >> 16) //This value may also be 3 bytes long.
            buffer(7) = Convert.ToByte((rTime And &HFF00) >> 8)
            buffer(8) = Convert.ToByte(rTime And &HFF)
            buffer(9) = Convert.ToByte(2.56 * ocpBox.Value) // The ocp pulse period is formed of 256 steps or counts, so if a percentage is requested, this formula gives the number of steps/counts required to set the pulse width
            buffer(10) = Convert.ToByte(tempBox.Value)
            If (tempCheck.Checked = True) Then
                buffer(11) = 1
            Else
                buffer(11) = 0
            End If
            freq = ((2 ^ buffer(0)) * (2078 / (2 - (dac / 1024))))
            pWidth = Tinh + ((((Convert.ToInt32(buffer(3))) << 16) Or ((Convert.ToInt32(buffer(4))) << 8) Or (Convert.ToInt32(buffer(5)))) * Tinc)
            ' Connect to device
            serialPort1.Write(1) // Sends the number 1. This tells the microcontroller we are about to start sending data. It should respond with a zero if it is ready
//and the connection is good.
The line "serialPort1.Write(buffer, 0, bufLen)" sends the buffered data where the variables are: buffer = the buffered data; 0 = the position in the buffer to start from; bufLen = the position in the buffer to stop sending data.
What's the best way to deal with the Convert.ToBytes and the logical masks And ??
Thanks in advance for your time and consideration,
regards
Alex
Solved!
Go to Solution.

Try this
Beginner? Try LabVIEW Basics
Sharing bits of code? Try Snippets or LAVA Code Capture Tool
Have you tried Quick Drop?, Visit QD Community.

Join two 8 bit integers and send via Serial Port

I am trying to join two 8 bit integers and send the result via the serial port.
I have a constant of hex value A that I need to join with a value from hex 0 - F (this is based on incoming data)
When I use the Join VI, and type cast to change from hex to string for Visa Write, I recieve hex 0A0F.
I have been able to use the hex 0-F with a case structure and then wire the corresponding constant ex A0 - AF.
This makes the program very cumbersome and labour intensive to change. I have 22 commands I have to respond to with the address of 0-F.
Currently, I have a Case structure that is selected with Message ID, then a case that is selected with subtype and then a case for address.
Therefore I have to create a constant inside of each address case for each message as the responses are different.
Thanks for any help
Robin

Gambin,
As I understand it, you want to take the two bytes, put them together,
and output the result as an ASCII string on the serial port. This
is easy. Simply convert each number to an ASCII string,
concatonate the two characters together, and send the resulting string
to the VISA write function. That's it! I have attached a VI
(ver. 7.1) that takes two hex numbers as input, converts them to ASCII,
concatonates the results, and outputs the 'command' string. Fell
free to modify this vi and use it as you see fit. I have left
extra terminals in case you want to add error input/output for data
flow, or whatever. Notice that the display for the concatonated
string is in '/' Codes Display mode. This is because 0A hex is
the newline character in ASCII. You should also check to make
sure that your VISA serial settings are not setup so that the newline
character is the termination character. If it is, the second
character may not be recognised. Hope this helps.
Roy
Attachments:
HextoCommand.vi ‏17 KB

Bits & Bytes

Hi,
I'm trying to reach a binary switch file. The file consists of bytes. Each byte consists of 4 bits, e.g., in my first byte position I have 02. I'm trying to read this binary file and write to another file in decoded ASCII/Hex format. I'm facing following problems :
1. If the value is less than 0xF, I'm not able to process the byte.
1.1) The byte is always being read in reverse order, i.e., 20 instead of 02. This problem is also when I read any byte..whether greater than or less than 0xF. But for values greater than 0xF, I'm taking them in a String array and writing it in reverse order.
So, when I read the first bit, the control comes out of the function where I'm doing the processing, into main, and when the control goes back into the funtcion, it again reads the byte read earlier. If I try to take the values in a 2D array, the value is over written.
2. How do I check for EOF in a binary file?
So, how should I handle this. Any answers??? My code is shown below :
Thanks & regards,
Faisal
import java.io.*;
public class fileinputstream1_3
static FileInputStream inFile;
static FileOutputStream outFile;
static FileWriter fileWriter;
static DataInputStream inData;
static PrintWriter outData;
public static void toHex(byte v) throws IOException
double tempVal1 = 0;
double tempVal2 = 0;
double tempVal3 = 0;
double tempVal4 = 0;
int tempVal5 = 0;
int tempVal8 = 0;
int testVal1 = 0;
int j = 0;
String testVal;
String tempVal6 = "";
String tempVal7[] = new String[2];
tempVal7[1] = "";
File f = new File( System.getProperty( "Programs" ), "cdr6.txt" ); // find path to o/p file
fileWriter = new FileWriter( f, true );
outData = new PrintWriter(fileWriter);
// Convert a 4-bit value into a hex digit char
if (v >= 0x0 && v <= 0xF)
     fileWriter.write(((char)(v < 0xA ? v+'0' : v-0xA+'A')));
     else
     if ( v > 0xF )
     tempVal1 = (double) v;
     while ( tempVal1 > 1.0 )
     tempVal1 = tempVal1 / 16.0;
     tempVal2 = (int) tempVal1;
     tempVal3 = tempVal1 - tempVal2;
     tempVal4 = tempVal3 * 16;
     if ( tempVal4 >= 10.0 )
     tempVal5 = (int)tempVal4;
     tempVal6 = Integer.toHexString( 0x10000 | tempVal5).substring(4).toUpperCase();
     else
     tempVal5 = (int)tempVal4;
     if ( tempVal6 == "" )
          tempVal6 = Integer.toString(tempVal5);
          testVal = Double.toString(tempVal4);
          testVal1 = (int)tempVal4;
          tempVal1 = tempVal2;
          tempVal7[0] = Integer.toString(testVal1);
          tempVal7[1] = tempVal6;
     try
     for ( int i = 0; i <= 1; i++ )
     fileWriter.write(tempVal7);
     catch ( IOException ioe )
     System.out.println( "Could not open the files. " +ioe );
     fileWriter.flush();
     fileWriter.close();
     System.out.println("Exiting toHex.");
public static void main(String[] args) throws Exception, IOException
int x;
int size;
File f = new File( System.getProperty( "Programs" ), "cdr1");
try
inFile = new FileInputStream( f );     // open the input binary file
inData = new DataInputStream( inFile ); // associate an input stream to the file
f = new File( System.getProperty( "Programs" ), "cdr3.txt" ); // find path to o/p file
// this is just to reproduce the input file - not much practical use
outFile = new FileOutputStream( f );
outData = new PrintWriter( outFile,true );
catch ( IOException ioe )
System.out.println( "Could not open the files. " +ioe );
try
byte bytearray[] = new byte[size+1];
for (int loop_index = 0; loop_index < bytearray.length ; loop_index++ )
outFile.write(bytearray[loop_index]);
toHex(bytearray[loop_index]);
outFile.write(1);
catch ( IOException ioe)
System.out.println( " IO error " + ioe + " occured.");
     System.out.println("Exiting main...");
     System.in.read();

Hi,
I don't understand the question. A byte is alwaysa
byte, and a byte is always 8 bits, and there's no
/KajNot necessarilly. While currently more or less
standardised there is hardware that has wider or
narrower bytes.
Whether Java is supported on those platforms and how
it would handle such situations if it were I don't
know.But a byte is still 8 bits, even if the hardware uses 9 bits internally (let's say that the hardware uses one bit for crc)??
I might make a slight change to; In java a byte is always 8 bits :)

Int (32 bits) - byte (8 bits) using bitwise operators

Hi!
I am storing a 16-bit value into a 32-bit integer and trying to extract the Most Significant 8-byte and Least Significant 8-byte as follows:
int n = 129;
byte msb = (byte)(((byte)(n>>8))&0xff);
byte lsb = (byte)(n&0xff) ;
System.out.println ("msb = "+msb+" lsb = "+lsb);
My msb is "0" and my lbs becomes "-127".
The above code works correctly for positive numbers but for signed values (like >129-255), my lsb comes to a negative number.
How should the above code be to handle the lsb?
Thanks,
x86

My msb is "0" and my lbs becomes "-127".
The above code works correctly for positive numbers
but for signed values (like >129-255), my lsb comes to
a negative number.
How should the above code be to handle the lsb?
Your code works fine. Are you wondering why an int value of 129 shows up as a msb of 0 and a lsb of -127? It's because byte values are signed and a bit pattern of 10000001 is equal to -127.

Working set, virtual bytes and private bytes

Hello everyone,
I am using Windows Server 2003 Performance Counter tool to monitor the memory consumed by my process.
The interested terms are working set, virtual bytes and private bytes. My questions are,
1. If I want to watch the real physical memory consumed by current process, which one should I monitor?
2. If I want to watch the physical memory + swap file consumed by current process, which one should I monitor?
3. Any more clear description of what these terms mean? I read the help from Performance Counter tool, but still confused
which one(s) identifies the real used physical memory, and which one(s) identifies the real used physical memory + page swap file, and which one(s) identifies the required memory (may not be really allocated either in physical memory or in swap page file).
If there are any related learning resource about the concepts, it is appreciated if you could recommend some. :-)
thanks in advance,
George

And for further benefit:
"Virtual bytes" is the total size of the not-free virtual address space of the process. It includes private committed, mapped, physically mapped (AWE), and even reserved address spaces.
"Private committed" is the portion of "virtual bytes" that is private to the process (as opposed to "mapped", which may be shared between processes; this is used by default for code). This normally includes program globals,
the stacks (hence locals), heaps, thread-local storage, etc.
Each process's "private committed" memory is that process's contribution to the system-wide counter called "commit charge". There are also contributions to commit charge that aren't part of any one process, such as the paged pool.
The backing store for "private committed" v.m. is the page file, if you have one; if you have (foolishly in my opinion) deleted your page file, then all private committed v.m., along with things like the paged pool that would normally be backed
by the page file, has to stay in RAM at all times... no matter how long ago it's been referenced (which is why I think running without a page file is foolish).
The "commit limit" is the size of RAM that the OS can use (this will exclude RAM used for things like video buffers) plus the total of the current size(s) of your pagefile(s). Windows will not allow allocations of private committed address space
(e.g. VirtualAlloc calls with the COMMIT option) that would take the commit charge above the limit. If a program tries it, Windows will try to expand the page file (hence increasing the commit limit). If this doesn't succeed (maybe because page file expansion
is disabled, maybe because there's no free disk space to do it) then the attempt to allocate committed address space fails (e.g. the VirtualAlloc call returns with a NULL result, indicating no memory was allocated).
The "Working set" is not quite "the subset of virtual pages that are resident in physical memory", unless by "resident" you mean "the valid bit is set in the page table entry", meaning that no page fault will occur
when they're referenced. But this ignores the pages on the modified and standby page lists. Pages lost from a working set go to one of these lists; if the modified list, they are written to the page file and then moved to the standby list. From there they
may be used for other things, such as to satisfy another process's need for pages, or (in Vista and later) for SuperFetch. But until that happens, they still contains the data from the process they came from and are still associated with that process
and with that processes' virtual pages. Referring to the virtual page for which the physical page is on the standby or modified list will result in a "soft" page fault that is resolved without disk I/O. Much slower than referring to a page that's
in the working set - but much faster than having to go to disk for it.
So the most precise definition of "working set" is "the subset of virtual pages that can be referenced without incurring a page fault."

Byte and char !!!

Hi!
What is the diference between an byte and char in Java! Could I use char instead of byte and reverse?
Thanks.

TYPE BYTE BITS SIGN RANGE
byte 1 8 SIGNED -128 to 127
char 2 16 UNSIGNED \u0000 to \uFFFF
Both the date types can be interchanged
Have a look at this
class UText
     public static void main(String[] args)
          byte c1;
          char c2= 'a';
          c1 = (byte) c2;
          c2 = (char) c1;
          System.out.println(c1 +" "+c2);
But It Leads to confusion while interchanging because of its SIGN. And the range of byte is very less.So its better to prefer char.

Having a problem with wireless connection. On my Windows XP 32 bit system and a 6510 printer, initi

On my Dell Latitude D620, Windows XP 32 bit system and my 6510 printer wireless connection (Linksys EA4500) , everything works fine after the computer is booted. After a period of time (usually overnight), I lose connection to the printer (the scanner icon with a red x is shown on my Windows task bar and I can't print. Ran Windows Fix and the HP fix programs. The only thing that solves it is to reboot the computer. A second symptom is that my IE8 stops connecting to the internet at the same time (gives a 'waiting for' response that never changes). Firefox still connects OK when this is happening. So I suspect the problem isn't with the printer, but with my Windows set up. It is a fresh XP SP3 install (hard drive crashed, but it happened before also). I reset IE, checked my proxy, no malware, happens with or without firewalls and security programs.
Hoping that someone else has run into this problem. Been all over Google, and the IE doesn't work / Firefox works is fairly common, but none of the fixes helps.

If you want you can reload Firefox 4.0 or the last "old" version is 3.6.16 and is available here:
http://www.mozilla.com/en-US/firefox/all-older.html
Just reinstall your choice of FF. The bookmarks are in a different directory and are NOT overwritten. Some of the Add-ons that worked in 3.6 DO NOT work with 4.0, pay attention and head any warnings when you start 4.0.
I had to disable extensions: Flashblock and WOT
Firefox SafeMode? - A troubleshooting mode.
1.You can open the Firefox 4.0 SafeMode by holding the Shft key when you use the Firefox desktop or Start menu shortcut.
To exit the Firefox Safe Mode, just close Firefox and wait a few seconds before using the Firefox shortcut to open it again.
For more help, see this: http://support.mozilla.com/en-US/kb/troubleshooting+extensions+and+themes
Hope this helps you out.
Feedback is appreciated especially if this helps you out.

When will Firefox offer a 64-bit version ? I use my 32-bit Firefox,but keep my 64-bit Waterfox and Pale Moon ready, when Firefox crashes on Windows 8.

I am a Firefox Veteran. Been around since the NCSA web browser days and used the first Firefox web browsers. I have never liked or trusted Microsoft Windows, so I've never used Internet Explorer. I had tweaked Windoze Blue Screen Of Death software from the Windows 3.0 days and did minor maintenance on the "Pizza Boxes" to keep them alive until I or my duty station could afford a newer one. Firefox has always performed for me working for "Uncle Sam" or at home. I keep Firefox and the versions: Sea Monkey, Aurora, Nightly on my laptop because I like Firefox and I like to switch from one browser to the other sometimes -- plus, in a Windows crash, I have a backup browser. Windows always "Crashes"; somethings never change. My 32-bit Firefox doesn't seem as reliable on my 64-bit 'Windoze' 8 as it was on Windows 7. My hope is that some of the Firefox Gurus have a 64-bit Firefox up on blocks in the shed out back ready to rock and roll or at least release as experimental to be played with and tested. Thanks guys -- Keep up the good work. Sam

Someday. This is a question asked multiple times but the answer is always the same, there is no real benefit to moving to 64 bit Firefox, and the amount of work required to do something with no benefit makes this a low priority. Eventually it will happen, but just because it uses 64bit doesn't magically make things better.

Vista 64 bit OS and iTunes

I am getting an error message when initializing iTunes.
iTunes was not properly installed. If you wish to import or burn CD's
you need to reinstall iTunes.
I uninstalled, cleaned the reg, and reloaded a few times.
Diagnostic DVD/CD test results "Upper Filter Status Failed" (no CD Rom Driver found)
All seems to be working ok otherwise and CD's seem to be importing OK
Bud
Message was edited by: idraliv

iTunes doesn't officially support 64 bit Windows and the 64 bit Gear drivers you need for burning arn't supplied. You can download them from here:
http://www.gearsoftware.com/support/drivers.cfm
A number of people also had problems with the creation of the CD configuration folder, but as your iTunes is working, I guess that doesn't apply to you.

Really bad banding and artifacts in 16-bit layered greyscale file. Bad banding is retained when converted to 8-bit with No flattening. Flatten 16-bit image and banding and artifacts disappear even with no dither or noise layer. Is this a known bug?

Has anyone else experienced this?
I thought it was a monitor problem at first, but I'm using a 10-bit per channel Eizo. It seems to be a Photoshop bug based on many layers interacting with each other. When I flatten everything is fine. But I need to work on this image with layers and decent fidelity. Interestingly when I convert to 8-bit without flattening (and utilising the dither function to potentially reduce banding even further), it uses the terrible artifact laden/banding to do the conversion even though when I zoom into 1:1 in my 16-bit document it looks fine. But 50% or 25% zoom I suddenly get these awful artifacts.
Here's some screenshots to clarify. Please note I've used Photoshop for 24 years so I'm no slouch but this is the first time I've seen this. The problem is I need to do some subtle airbrush and texture work on this and it's almost unusable unless I flatten (please note, it does the same in the original 8-bit file and just to say I Gaussian-Blurred every layer with a 30px radius after converting to 16-bit so the there's no longer any 8-bit information in there or reasons for banding/artifacts). I can only think it is some of bug in Photoshop's layering engine.
Has anyone seen this before - dealt with this before?
Thanks in advance.
Not sure these embedded images will work.
8% zoomed - see all the strange banding and triangular artifacts
<a href="http://imgur.com/izrGuia"><img src="http://i.imgur.com/izrGuia.png" title="source: imgur.com" /></a>
Flattened view - all smooth:
<a href="http://imgur.com/Pn35IAK"><img src="http://i.imgur.com/Pn35IAK.png" title="source: imgur.com" /></a>
50% zoom, still there:
<a href="http://imgur.com/Z207hFd"><img src="http://i.imgur.com/Z207hFd.png" title="source: imgur.com" /></a>
100% artifacts disappear:
<a href="http://imgur.com/6aGOz0V"><img src="http://i.imgur.com/6aGOz0V.png" title="source: imgur.com" /></a>
100% 16-bit layered
<a href="http://imgur.com/0XJfe5e"><img src="http://i.imgur.com/0XJfe5e.png" title="source: imgur.com" /></a>
and finally 8-bit layered converted from 16-bit with dither.
<a href="http://imgur.com/PSxiu43"><img src="http://i.imgur.com/PSxiu43.png" title="source: imgur.com" /></a>
help!

I can't speak to why, perhaps someone more knowledgeable than I can speak to that. But if it only happens at certain views then it's purely a display issue; it won't happen in print or when you downsample to display size. Such banding issues have always disappeared for me when I output to 8 bit.
I'd assume that it's because of your monitor; it can't display all the colors and when zoomed out there's too wide of a range in a small area so banding occurs. But that's just my guess.

I just bought a new computer. My old computer was a 32bit system but my new one is a 64bit. I own ps cs6. My question is"will I be able to deactivate ps on my old system, and download the 64 bit version and install and activate it or do I have to buy t

I just bought a new computer. My old computer was a 32bit system but my new one is a 64bit. I own ps cs6. My question is"will I be able to deactivate ps on my old system, and download the 64 bit version and install and activate it or do I have to buy the 64bit version in order to move forward.

Nope, you can install it on your new computer with no issues. The disc contains both the 32 and 64 bit versions.
Benjamin

Bits, bytes and megabits

Similar Messages

Maybe you are looking for