Unsigned integer

Hello,
I am trying to wrap a c++ dll ,where most of the variables are (unit8,unit16,uni32 ) unsigned integers.
Since java doesnt have usigned integer type,I am wondering what to do.
1. Pass the usigned int value as a string to the c++ program.Convert from string to uint in the C++ code.
would be one option.
I am hoping that there are other options available to do this more gracefully.
thank you very much,
Regards

I agree that signed and unsigned numbers differ in interpretation of the bits so one can disregard this when passing the data to/from DLL.
When you get an uint32 from the DLL it can be stored in a Java int type:
int i=fromDLL();but it should be converted to long when used:
long ui=(long)i&0xffffffff;to pass "unsigned" to the DLL convert it to int:
toDll((int)ui);You may also need to perform little/big-endian conversion.
Dmitry Hudyakov
Brainbench MVP

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ok but how do I set the lower 32 bits? And what is setting the highest bit even doing because I have no idea how the server will respond if I don't set it. As I said I'm porting this from PHP which does all of this and doesn't explain why so I don't know what it's doing it for. The PHP is
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Unsigned integer problem

Hi,
I have a problem of converting a CRand32 function from C to Java which was originally from IDA decompiler.
There's one thing that's troubling me for quite a long time which is the lack of unsigned types, I could not do operations such as byte shifting correctly and the end result I get is sometimes a negative number or larger than 4billion.
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Edited by: LightPepsi on May 30, 2010 11:40 PM
Edited by: LightPepsi on May 30, 2010 11:40 PM

LightPepsi wrote:
That was the first thing which comes to my mind by changing >> to >>>, however the numbers generated are usually not within the range of an unsigned integer.
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Iteration a signed integer and not an unsigned integer, why?

i was just wondering if there is a reason for keeping the iteration variable/indicator of a loop a signed 32 bit integer.
it is never going to take a negative value.. then why signed; unsigned could have served the purpose
Message Edited by Ujjval Shah on 07-08-2009 09:21 AM
Solved!
Go to Solution.

Remember that array sizes are limited to I32 (the "size" input to initialize array is I32, the output of Array size is I32, etc), so running a FOR loop more than 2^21 times is not even possible (N is also I32).
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LabVIEW Champion . Do more with less code and in less time .

Writing double and unsigned integer to serial port

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Hideya,
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Hello,
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Thanks.
TN
Solved!
Go to Solution.

You just need to do some pointer manipulation to build your float.
But you need to be aware of the size of the data types you're using. Based on your example values, it looks like you are using the IEEE single precision format, which is 4 bytes. In 32 bit operating systems, the unsigned int is also 4 bytes. So you can't combine two unsigned int into an IEEE single without casting them as short ints, which are 2 bytes each. You would need to check if the value in your unsigned int will fit in an unsigned short before casting it.
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printf("Press any key to continue...\n");
GetKey();
Here's a link to a discussion on going the other way, which has a sample program and links to more IEEE format info.
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ACS v5.2 - Unable to update User integer attributes through File Operations

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No warning when assigning to an unsigned char from unsigned int

Is this an error in VC++, or am I missing something? The following code should give two warnings for assigning to a smaller type, but only gives one: (compiled as 32 bit)
unsigned
char c;
unsigned
int i = 24;
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unsigned
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This happens in Visual Studios 2005, 2010, 2012, 2013 and 2015, at least.

Is this an error in VC++, or am I missing something? The following code should give two warnings for assigning to a smaller type, but only gives one: (compiled as 32 bit)
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Warning 2 warning C4244: '=' : conversion from 'unsigned int' to 'unsigned char', possible loss of data
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How to specify in TS unsigned long? unsigned short?

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Hi Doug,
I'm given the dll and load of h files. The definitions you mentioned are not shown...
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Integer and fractional part of the number

Hello everybody,
I want to divide the floating point number to 2 integer words, first will receive the interger part of number, and the second will receive the fractional part of the number.
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The place to start with this problem is the IEEE-754 numerical
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32 bit floating point is internally:
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the next 8 bits are the exponent bits
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the mantissa value in a 1.22 fixed point binary number, i.e. the most
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You should be able to cast your 32 bit float to a 32 bit unsigned
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Douglas De Clue
LabVIEW programmer
[email protected]
[email protected] (Michael Ross) wrote in message news:...
> This problem needs some limits. Like: only return the decimal portion
> out to 6 places.
>
> First run the number through Remainder and Quotient (divisor of 1).
>
> 751.98743674645641000000 gets remainder of
>
> 0.98743674645641021900
>
> "floor" is 751
>
> Take the remainder and send through Decimal to Fractional String (this
> is where you set your number of decimal places [default is 6]).
>
> returns 0.987437
>
> Then use String Subset and set it to an offset of 2 (cleans off the
> leading zero and decimal point) with the "length" input left at the
> default of "rest" (of string).
>
> 987437 (in string form)
>
> Now use String to Decimal Number
>
> returns 987437 (as a decimal number)
> ====================================
> I think if you don't limit the number of places it is ugly. It is
> ugly going from numbers to strings and then back. But all numbers is a
> bad thing.
>
> The alogirithm for finding the answer mathematically is nasty.
> Especially without a limit on decimal places. You have to figure out
> the number of decimal places and multiply by the inverse of the last
> decimal place. Sounds easy.
> =============================
> Try this: New number: 751.98652
>
> Truncate out the remainder (Remainder and Quotient (divisor of 1)):
>
> 751.98652 ---> 0.98652
>
> Mult by 10 ----> 9.8652
>
> Take the floor (9) and divide by 10 ----> 0.9
>
> subtract the new number from the old. Is the answer non-zero? Not,
> then keep on
>
> 0.98652 - 0.9 = .08652 which not equal to 0
>
> If it is zero, the floor is the answer.
>
> Continue - Not:
>
> This time multiply by 100 ----> 98.652
>
> floor is 98, divide by 100 ----> 0.98
>
> Subtract 0.98 from 0.98652 = .00652 still non zero.
>
> Keep on. 1,000, 10,000...When you get to 100,000
>
> 0.98652 x 100,000 = 98652.00
>
> Get the floor
>
> divide the floor by 100,000 to get the original number (remainder is
> now 0.000), compare to original remainder
>
> 0.98652 - 0.98652 = 0. Yay.
>
> Take the 98652 and go with it. The last place was the one hundred
> thousandth.
>
> I can't take credit for the last, I had uglier stuff going on. I did
> do the string thing.
>
> I swear this is just like some sorry homework problem.
>
> Mike

Represent an 4-byte integer as a hex number???

Hi everyone,
I still have problem with converting an IP address to an POSITIVE number. One of the native method expects unsigned integer pass in to set IP address for a device. But, in java, when I convert a big IP such as 192.0.0.0, it will return a negative integer. Many of people helped me out but I still got the problem since my co-worker did not want to change his code. He wanted me to find an alternative way to do. He suggested me to represent the address in hex format will work ok.
e.g int ipAddr = 0x6C01A7C0; // 192.167.1.108
Then CallCOMObject(ipAddr);
So, my question is do we have a way to convert 192.167.1.108 into 0x6C01A7C0?
Thanks in advance
Hung.

Hi Armalcolm,
Yes, I tried that but the native function throwed exception since the integer I pass in was negative. In the native function, there was an IF statement which would reject all negative integer.
Here is my example
After user enter IP as: 192.167.1.108, I convert (by shifting and adding) to an integer and I got an negative integer (-1062796948). Then I passed it in the native function as:
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The native function throwed exception.
Is that what you mean, Armalcolm?
Alternatively, you could do your shifting and adding into a long, and then and the result with >0xffffffff, passing the result. But there should not be any difference. I could shifting and adding into a long, but I don't know how to represent that long number in hex. Can you help me out with that.
I appreciate for your reply.
Hung.

Unsigned integer

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