Concat 4 8-bit bytes into 1 32-bit word

I am just getting started up with LabView. I have an FPGA that is sending a signed (2's Comp) 32 bit vector to the PC serial port in four 8-bit chunks. I am having trouble figuring out how to recombine the four bytes into the original 32-bit word. It is a simple task in any programming language using concatenation but I am really having problems with this.
Thanks
Jeremy

Two ways come to mind immediately.
1) Go from a string (VISA returns a string from a read) to a byte array then loop and shift each byte the correct amount while OR'ing the result with that from each loop iteration.
2) Use the join numbers vi to combine multiple bytes into a single 32-bit word.
I've attached a vi that demonstrates these two solutions.
Attachments:
MergeBytesExample.vi ‏28 KB

Similar Messages

Importing 16 bit file into 24 bit project

Whats up guy's
When importing a 16 bit file into a 24 bit project
should I convert the file to 24 bits or keep the
original 16 bit depth?
I'm aware that the sample rate is converted to the
projects sample rate when importing files as long as
the corresponding checkbox is enabled in the project
settings - assets pane.
I'm also aware that conversion can be done at a later
state after importing, but I'm curious about the best
or proper approach. Should this be done prior to importing
or should the file be kept in it's original state (16 bits)?
PS - If you could provide the chapter or page of the manual
this information is on that would be extremely helpful.
Thnx in advance

Qstn wrote:
When you say "alter the file's sample data" are you referring to
edits done in the sample editor or sample rate conversion?
both would qualify. there are some corner cases where bit depth does not matter when destructively editing sample data (as I outlined in my initial post).
If you're talking about the sample editor do you mean converting by
saving a copy as 24 bits and making edits to the copy might show tiny
improvements?
yes.
At this moment I don't intend on destructively editing the file but
you never know how processing and mixing might go until your done.
ok. if your file is read and processed in realtime (or via a frozen track) then there's no loss - logic will convert audio samples during playback to a 32 floating point representation as it is read from disk, before it enters the mixer, or is processed by your plugins.
i'm not sure if source bit depth determines bounce in place format- if so, and if you use this feature, then you should convert the source to 24 bit.
the end point is: unless you are desperate for disk space, just convert it to 24 bits. 16 bits is very 'last millennium' in modern production stages - there are a few exceptions to that (e.g., if the files are used by a software sampler, or if the format is required for compatibility with other hardware or software) but... most of the strong arguments for 16 bit died years ago. so, i'd say it's best to consider 24 bits as the minimum during production unless you have very specific reasons to go below that (and you fully understand the tradeoffs). if you want to know the 'whys', that's fine, of course. curiosity and knowledge don't hurt. if you're only looking for a recommendation, that's my recommendation.

Byte into bit array?

I want to implement a little compression programm, and I have to convert the byte array, in which I read a file into a bit array, because I have to be able to change single bits in that array.
1.Is there a wrapper class?
2.How many bits does the class Byte contain?
If it is 8, then i can convert a single byte into 8 bit by using a simple algorithm, right?
The byte simply displays it�s value (0-255)in hex, right?
thanks for your reply!
tim

There is a class named BitSet that seems to be close to what you are looking for (at least in SDK 1.3). But if you can manipulate bytes, it might be more efficient to write your own code.

Is there an easier way to convert bytes into bit(boolean) arrays?

I am currently using this method to convert bytes into bit arrays:
/*convert byte to int such that it is between 0-255 this is the bytes[] array
                    if ((bytes/128)==1)
                         bit[c+0]=true;
                    else
                         bit[c+0]=false;
                    if ((bytes[i]-bitInt[c+0]*128)/64==1)
                         bit[c+1]=true;
                    else
                         bit[c+1]= false;
                    if ((bytes[i]-bitInt[c+0]*128-bitInt[c+1]*64)/32==1)
                         bit[c+2]=true;
                    else
                         bit[c+2]= false;
                    if ((bytes[i]-bitInt[c+0]*128-bitInt[c+1]*64-bitInt[c+2]*32)/16==1)
                         bit[c+3]=true;
                    else
                         bit[c+3]= false;
                    if ((bytes[i]-bitInt[c+0]*128-bitInt[c+1]*64-bitInt[c+2]*32-bitInt[c+3]*16)/8==1)
                         bit[c+4]=true;
                    else
                         bit[c+4]= false;
                    if ((bytes[i]-bitInt[c+0]*128-bitInt[c+1]*64-bitInt[c+2]*32-bitInt[c+3]*16-bitInt[c+4]*8)/4==1)
                         bit[c+5]=true;
                    else
                         bit[c+5]= false;
                    if ((bytes[i]-bitInt[c+0]*128-bitInt[c+1]*64-bitInt[c+2]*32-bitInt[c+3]*16-bitInt[c+4]*8-bitInt[c+5]*4)/2==1)
                         bit[c+6]=true;
                    else
                         bit[c+6]= false;
                    if ((bytes[i]-bitInt[c+0]*128-bitInt[c+1]*64-bitInt[c+2]*32-bitInt[c+3]*16-bitInt[c+4]*8-bitInt[c+5]*4-bitInt[c+6]*2)==1)
                         bit[c+7]=true;
                    else
                         bit[c+7]= false;

You can loop through and use a bitwise operator instead. Here is an example without the loop.
byte b = 6 ;
if( b & Math.pow( 2, 0 ) == Math.pow( 2, 0 ) ) ;
// the 2^0 bit is on
if( b & Math.pow( 2, 1 ) == Math.pow( 2, 1 ) ) ;
// the 2^1 bit is onetc...
You should get something like 110 when you're done.
If you're wonder what & does (no, its not a boolean &&), it takes the bits in the two numbers you give it and returns a number with all the bits on that are on for each of them.
For example:
10011011 &
11001101 =
10001001
So if we take
110 (6) &
010 (2^1, or 2) =
010 (2 again)
Which means that the number (6) has the 2^1 bit on.

Separating 16 bits of data into one 12 bit and one 4 bit channel

I have 16 bits of binary data and i'm trying to split these into one 12 bit and one 4 bit channel. I know that I could use something like:
Dim Channel1 : Set Channel1 = Block.Channels.Add("Channel1", eI16)
Channel1.Formatter.Bitmask = 4,095 'xxxx 1111 1111 1111
but, i'm not sure how to grab those 4 bits out of the 16 bit channel1 and make channel2. It's like i almost need to say something like:
Dim Channel2 : Set Channel2 = Block.Channels.Add("Channel2", eI16)
Channel2 = Channel1.Formatter.Bitmask = 61,440 '1111 xxxx xxxx xxxx
Can anyone help out???? thanks.

OK...This is still straightforward. Her is what you want to do :
CurrPos = File.Position ' remember where the block starts
Set Block = File.GetBinaryBlock()
' set up all channels which can be read directly
Dim Channel1 : Set Channel1 = Block.Channels.Add("Channel1", eU16)
Dim Channel2 : Set Channel2 = Block.Channels.Add("Channel2", eByte)
Dim ChannelPH : Set ChannelPH = Block.Channels.Add("Dummy", eU16) ' just a placeholder
Dim Channel5 : Set Channel5 = Block.Channels.Add("Channel5", eU16)
Dim Channel6 : Set Channel6 = Block.Channels.Add("Channel6", eByte)
' Now we need a block which has one channel only. This channel will be used to extract the bits.
' The block needs to start where the third schannel of the above block starts :
File.Position = CurrPos + 2+1 ' 2 for the U16 channel, 1 for the byte channel
Set Block = File.GetBinaryBlock()
' Set the scan width. This defines the gaps between the values of a channel
Block.BlockWidth = 2+1+2+2+1 ' U16 + Byte + U16 + U16 + Byte
Dim Channel3 : Set Channel3 = Block.Channels.Add("Channel3", eU16)
' .. apply bitmask and scaling (see my last example)
' Do the same for the second channel
Set Block = File.GetBinaryBlock()
' Set the scan width. This defines the gaps between the values of a channel
Block.BlockWidth = 2+1+2+2+1 ' U16 + Byte + U16 + U16 + Byte
Dim Channel4 : Set Channel4 = Block.Channels.Add("Channel4", eU16)
.. apply bitmask and scaling (see my last example)
Do the same for the second channel
I hope this works as expected. Let me know if you need more help...
Andreas

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.

EFI update tu support booting into 64 bit OS

Hello,
my mac (see signature) does not support booting into 64 bit in Snow Leopard. I also have troubles to install any 64 bit Windows.
The tool: http://www.ahatfullofsky.comuv.com/English/Programs/SMS/SMS.html says:
64 bit processor, 32 bit EFI, 32 bit kernel boot mode, 32 bit kernel
and its technically possible to boot into 64 bit kernel mode but it is not recommended by Apple.
It is possible to get a 64 bit EFI update to solve it?
thanks

Hi,
try this
http://www.ahatfullofsky.comuv.com/English/Programs/SMS/SMS.html
It seems that your model is in the list of supported models that can boot the 64 bit kernel.
I had the same issue with my macbook pro with identifier MacBookPro4,1 and booting with 64 bit kernel solved it.
I hope you can solve your problem!
Sorry for my bad English :-D

A late 2007 Mac Pro (2,1) no longer boots into 64-bit Snow Leopard

In Snow Leopard 10.6.7, my late 2007 Mac Pro (2,1/3.0 GHz/Dual Quad-Core Intel Xeon) no longer seems to be booting into 64-bit mode. I've tried holding down the 6 and 4 keys during startup and even used a Terminal sudo command to force the machine to always boot into 64-bit mode. No matter what I try, an About This Mac>More Info>Software check only reports the following:
     64-bit Kernel and Extensions: No
Before I upgraded from Leopard 10.5 to Snow Leopard 10.6, I carefully researched this, and this Mac Pro model specifically was listed by Apple as being capable of 64-bit booting — meaning that it would utilize at least some extra memory. At the time, under OS 10.5 and early versions of 10.6, when I used this Terminal command...
     ioreg -l -p IODeviceTree | grep firmware-abi
...my Mac Pro 2,1 returned this response:
     "firmware-abi" = <"EFI64">
As a result of my tests and research, I upgraded the installed SDRAM from 4MB to the theoretical maximum Apple indicated this model could support (16MB) so that Adobe Photoshop 12 (CS 5) would fully utilize extra installed memory under Snow Leopard. While using the first versions of 10.6 Snow Leopard, the machine worked fine in 64-bit mode. Now, in the wake of the 10.6.6 or the 10.6.7 update (I didn't notice which), I find that this model no longer even reports itself as being 64-bit capable, though it once did.
Now, the aforementioned Terminal query returns this response:
      "firmware-abi" = <"EFI32">
Also, the Library>Preferences folder does even not contain a file called "com.apple.boot.plist," which some owners have reported editing to force 64-bit booting.
The thing is, I spent a fair amount of money upgrading this machine to the maximum amount of SDRAM supported. I purchased it because it was advertised as being 64-bit capable and I upgraded it because Apple originally reported it as being a 64-bit model. Now, I'm not sure if the extra memory I added is being fully utilized, though Photoshop does report seeing 15547MB of available memory.
Have any other owners of these second-generation Intel Mac Pro's encountered this? Does anyone know what's going on? Am I doing something wrong when checking for the 64-bit installation or architecture? Is Snow Leopard 10.6.7 mis-reporting the architecture for this model? Or did Apple silently withdraw 64-bit support for this machine in a recent OS update?

Hatter: Thanks for the feedback and links. Well, as described in one of the threads you linked to, the document below reinforces your assertion that only Mac Pro's 3,1 or later (early 2008) support 64-bit booting in Snow Leopard:
http://support.apple.com/kb/HT3770
However, if you asked me in court, I would swear on a stack of Bibles that sometime in the past, I read an Apple document indicating that the Mac Pro 2,1 supported 64-bit booting and I really thought I recalled booting this particular machine into 64-bit mode under early versions of Snow Leopard.
I guess I'm mistaken. Certainly, this issue can be confusing. When I read them out loud to myself, even perfectly-written, accurate explanations on the subject can sound like a circular, non-sensical George Bush rationalization being delivered by comic Will Ferrell. I must not be correctly recalling the details of booting into 64-bit mode with this machine. And I guess that since Photoshop has its own memory management scheme, and I only have 16 gigabytes of SDRAM installed anyway, it doesn't matter that much. Furthermore, in one of your links, Jason Snell of Macworld writes:
"...If you’re running a Mac powered by an Intel Core 2 Duo processor or an Intel Xeon processor, your Mac is 64-bit capable. And Snow Leopard runs 64-bit-capable applications in 64-bit mode regardless of whether it’s booting into a 64-bit or 32-bit kernel. In fact, the only big advantage of booting into a 64-bit kernel would be the ability to use more than 32 gigabytes of RAM..."

Do not upgrade to an IvyBridge processor if you need to boot into 32-bit mode with Lion

I recently upgraded to a MBP running the Ivybridge processor and tried to boot into 32-bit mode. Although it did boot after a short period (3-5 mins) it totally crashed a locked up. This was precursored by a graphics error and then the mouse cursor could be moved but nothing else could be done and it was only solved by a hard shut down (holding down the power button). Previously I had been running a MBP with the Sandybridge architecture and this booted fine into 32-bit mode and worked without a problem.
I spent a lot of time on the phone with Apple care who said they'd not had any reports of this issue and we went right though my system until they said there was nothing else they could do and I should try a re-install of Lion. I did this onto another hard-drive, booted into it and managed to get the machine to crash again... leading me to understand that it's nothing to do with my setup.
Thinking it might be an issue with my firmware on the mac today I visited an Apple store and tried booting 5 MacBook Pro's and a MacBook Air - all running IvyBridge - into 32-bit mode and managed to cause the same crash on all of them. If the machine is just booted an left with finder it didn't seem to want to crash but when I opened programs, mainly mac mail, it didn't take too much longer for the crash to occur.
If there are any Apple engineer reading this could you try and reproduce the issue and look at fixing this please as a lot of people still use 32-bit and if I'd known that the IvyBridge processors weren't going to work in 32-bit mode I would have stayed with the Sandybridge MBPs.
Many Thanks,
Adam Peters,

That is a pain then. Pretty much all of the OS included applications (finder, Mail, iTunes, Safari,...) are all fully 64-bit programs now (since Lion) so I can see where launching them when booted into 32-bit mode could cause stability issues.
If you boot into 32-bit mode and just run the Tascam software without launching anything else, will it work?
Was it alright in 10.6? And if so, how about keeping 10.6 in a virtual machine or on a separate disc or partition?

How to run i tunes version 10.6.3 on mac os x 10.6, 32 bit mode? it gives mesg to go into 32 bit mode but i cant get into it pls help

how to run i tunes version 10.6.3 on mac os x 10.6, 32 bit mode? it is downloded in 64 bit mode. when i start i tunes it gives mesg to go into 32 bit mode by going into finder, choose the the "get info command" then check the check box "32 bit mode in window" but i am not finding the check box of 32 bit mode. pls help

That crash appears to be casued by the Facebook plug-in.
Create a new account (systempreferences -> accounts or Users & Groups on 10.7 and 10.8), make a new Library in that account, import some shots and see if the problem is repeated there. If it is, then a re-install of the app might be indicated. If it's not, then it's likely the app is okay and the problem is something in the main account.

Can Mac mini boot into 64-bit kernel

http://macperformanceguide.com/SnowLeopard-64bit.html says:
To determine if your Mac has 32-bit or 64-bit firmware, copy/paste the following command into Terminal:
ioreg -l -p IODeviceTree | grep firmware-abi
...output:
| | "firmware-abi" = <EFI64>
As shown above, the firmware-abi has the value EFI64, indicating that my Mac Pro is 64-bit capable. If it reads EFI32, then your Mac cannot be booted into a 64-bit kernel.
...well, my Mac mini displays that EFI64 64-bit info, but when booting with the "64" keys down it nevertheless uses the 32-bit kernel.
So can the Mac mini boot into 64-bit kernel?

What you are missing is that Apple has written into the boot ROM for certain Macs which meet the criteria to not be able to boot with the 64-bit kernel. The Mac mini is not allowed to boot into 64-bit mode by Apple. There is a list of non-pro Macs hard coded into the EFI boot firmware that tells it not to allow these Macs to boot into 64-bit mode. The Mac mini is on the list! Try as you may the Mac mini will not do it without a hack.
If you google this issue you can find a hack to rewrite the boot firmware and allow the mini to boot into 64-bit mode. It can be dangerous to your mini if you make a mistake!
Here is one hack;
http://www.osxbook.com/blog/2009/08/31/is-your-machine-good-enough-for-snow-leop ard-k64/
Dah•veed

Can't boot into 64 bit

I've tried the holiding down 6 4 keys during boot, I've done the terminal comand to set the startup drive to automatically boot into 64 bit mode... nothing works.
At the end of my rope. Thoughts?
Hardware Overview:
Model Name:          MacBook
Model Identifier:          MacBook4,1
Processor Name:          Intel Core 2 Duo
Processor Speed:          2.1 GHz
Number of Processors:          1
Total Number of Cores:          2
L2 Cache:          3 MB
Memory:          4.5 GB
Bus Speed:          800 MHz
Boot ROM Version:          MB41.00C1.B00
SMC Version (system):          1.31f1
Serial Number (system):          ***************
Hardware UUID:          CA090417-06DE-52F7-8BAB-C0CA928EF8DE
Sudden Motion Sensor:
State:          Enabled
System Software Overview:
System Version:          Mac OS X 10.7.3 (11D50b)
Kernel Version:          Darwin 11.3.0
Boot Volume:          Macintosh HD
Boot Mode:          Normal
Computer Name:          *********** MacBook
User Name:          ***************************
Secure Virtual Memory:          Enabled
64-bit Kernel and Extensions:          No
Time since boot:          14 minutes

Ya, it's too bad. I've known for a while that when this laptop reaches its end I'll be going to Windows 7 and a new laptop at half the price.
I'm a little sad because I didn't know the end was so close, but if I can't even run the new OS coming out, well, that's pretty much it.
It's not dead yet, probably still has a year or so left but still. Been with Apple since 2006 and it's been an interesting ride.

Mac Pro not able to boot into 64 bit kernel?

Hello,
I've been trying to boot while using the 6 and 4 buttons held down to no avail. This mac pro was bought in 2007 (work computer). Am I not able to boot into 64 bit mode with Snow Leopard? If so, does that mean I can't take advantage of the 64 bit kernel?
Thanks,
Scott

Man, Yes, when I am rendering all 8 cores in Maya, up to 4 gigs of ram are being used. Why is that so hard to believe? I should be able to render in RAM preview in After Effects a whole lot more than i can now without cacheing.!!!!
Is Maya a 64 bit application? If Maya is not using more than 4GB, than you should be talking to the Maya developers.
You should be able to use Applications -> Utilities -> Activity Monitor and look at the "Kind" field. If it says "intel (64 bits)", then it is a 64 bit application.
How many times do I have to state "YES, I will be using more than 4 gigs of RAM per program"?
Why is it the tech people really don't understand how intensive program dynamics like Maya and Blender are,
You are responding to my post, so I'm going to say, I have been working for major computer manufactures since the early 70's working with "Big Iron" computer systems, not PCs, and I most likely know a lot more about how computers work, then many computer users. Please do not tell me I do not understand. I've been working with 64 bit code since the early 90's.
and yet Apple touts their excellence in 64 bit programs like AFX, yet, I can't use more than 4 gb of RAM?
Are you saying AFX (is that After Effects?) is not a 64 bit application yet? If not, then I suggest you talk to Adobe, not Apple.
Just for fun, I wrote a sample program, which allocated a large amount of memory. I ran it on my iMac with is booted with a 32 bit 10.6.2 Snow Leopard kernel. Here are the results:
PID RESIDENT VIRTUAL PageFlt Reclaim COMMAND
54296 4.74G 9.32G 0K 0K tmp
My program has more than 4GB of physical RAM (4.74GB), and a virtual memory size of 9.32GB.
If I use Activity Monitor, I can see that my program is listed with a Kind of "intel (64 bit)".
So if Maya cannot use more than 4GB of memory, it is a Maya problem. Talk to the Maya developers.
We aren't cows Apple. We are people with limited money trying to make it in the world. Not every mac pro user is a business and we don't want too much control.
You are not talking to Apple, just other Mac users. If you wish to tell something to Apple, I would encourage you to use either:
<http://www.apple.com/feedback/macosx.html>
or
<http://bugreporter.apple.com>
Free ADC (Apple Developer Connection) account needed for BugReporter

URGENT: Covert a true - false value into a bit value

Hello :)
I have a problem, How can I convert a true false value into a bit value, because, I declare a variable value with true or false, that means that I have a 1 or a 0, (thinking about digital information)
for example, one this is my part of my code:
private boolean _x3 = true;
     private boolean _x5 = false;
     private boolean _x6 = true;
     private boolean _x7 = false;
     private boolean _x9 = true;
     private boolean _x10 = false;
     private boolean _x11 = true;
public boolean CheckP1(boolean pP1)
          pP1 = x3^x5 ^_x7 ^_x9 ^_x11;
          return pP1;
This method will send a true or false value, but now I have two doubts, one of them, is, how can get this value, and the second value, how can I conver this value into a bit value
Thank you so much :)
Nataly

Look at a BitSet to hold your boolean values.
Check the API for details.
JJ

Concat 4 8-bit bytes into 1 32-bit word

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