Velocity Reviews > C++ > int packed as hex with memcpy

# int packed as hex with memcpy

Delali Dzirasa
Guest
Posts: n/a

 10-08-2003
I would have a number packed with its hex representation of the integer
below is some sample code of what is being done.

int value = 20; //in hex it is 0x14

..
..
..

{
UCHAR tmp2[2];
tmp2[1] = myVal & 0x00FF;
tmp2[0] = (myVal & 0xFF00) >> 8 ;

memcpy(&myPkt[0], &tmp2, 2); // (where myPkt is a UCHAR*
}

when I check the hex dump I see that 0x0020 was packed instead of the 0x0014
that I want. How can I set the proper flags(?) "if" that is the solution to
pack the hex representation of the integer?

Thanks, your help is greatly appreciated!

Delali

Ron Natalie
Guest
Posts: n/a

 10-08-2003

"Delali Dzirasa" <(E-Mail Removed)> wrote in message news:bm1hbf\$oqo\$(E-Mail Removed)...
> I would have a number packed with its hex representation of the integer
> below is some sample code of what is being done.
>
> int value = 20; //in hex it is 0x14

> {
> UCHAR tmp2[2];

presumably UCHAR is unsigned char?
> tmp2[1] = myVal & 0x00FF;
> tmp2[0] = (myVal & 0xFF00) >> 8 ;
>
> memcpy(&myPkt[0], &tmp2, 2); // (where myPkt is a UCHAR*

memcpy(myPkt, tmp2, sizeof tmp2);

> when I check the hex dump I see that 0x0020 was packed instead of the 0x0014
> that I want. How can I set the proper flags(?) "if" that is the solution to
> pack the hex representation of the integer?

There are no flags and there's no such thing as a hex representation in the code.
The above have stored the value which is both 20 decimal and 14 hex and 24 octal
etc...tmp2[1]. What makes you think otherwise? Are you sure your dumper is really
showing you the bytes in hex?

Peter van Merkerk
Guest
Posts: n/a

 10-08-2003
"Delali Dzirasa" <(E-Mail Removed)> wrote in message
news:bm1hbf\$oqo\$(E-Mail Removed)...
> I would have a number packed with its hex representation of the integer
> below is some sample code of what is being done.
>
> int value = 20; file://in hex it is 0x14
>
> .
> .
> .
>
> {
> UCHAR tmp2[2];
> tmp2[1] = myVal & 0x00FF;
> tmp2[0] = (myVal & 0xFF00) >> 8 ;
>
> memcpy(&myPkt[0], &tmp2, 2); // (where myPkt is a UCHAR*
> }
>
> when I check the hex dump I see that 0x0020 was packed instead of the

0x0014
> that I want. How can I set the proper flags(?) "if" that is the solution

to
> pack the hex representation of the integer?

The code you posted is unfortunately not compilable since AddData() does not
have a return type, and UCHAR and USHORT are not defined in the C++
standard. That being said, I see nothing in your code that would explain the
values in your hex dump. Are you sure that the tool you are using is
displaying the data in hexadecimal format and not decimal format?

--
Peter van Merkerk
peter.van.merkerk(at)dse.nl

Andrey Tarasevich
Guest
Posts: n/a

 10-08-2003
Delali Dzirasa wrote:
> I would have a number packed with its hex representation of the integer
> below is some sample code of what is being done.

You should probably start by explaining what you understand under "hex
representation of the integer" and under "packing a number with its hex
representation" in this particular case. "Hex representation" is in
essence a sequence of characters from '0'..'9', 'A'..'F' set or
something like this. I don't see anything in your code that has anything
to do with obtaining a hex representation of any number, let alone the
"packing".

> int value = 20; //in hex it is 0x14
>
> .
> .
> .
>
> {
> UCHAR tmp2[2];
> tmp2[1] = myVal & 0x00FF;
> tmp2[0] = (myVal & 0xFF00) >> 8 ;
>
> memcpy(&myPkt[0], &tmp2, 2); // (where myPkt is a UCHAR*
> }
>
> when I check the hex dump I see that 0x0020 was packed instead of the 0x0014
> that I want. How can I set the proper flags(?) "if" that is the solution to
> pack the hex representation of the integer?

--
Best regards,
Andrey Tarasevich
Brainbench C and C++ Programming MVP

Delali Dzirasa
Guest
Posts: n/a

 10-08-2003
Yes they are sent to a file via another program that I am testing, this
program is acting like a simulator and the other application sits and waits
for data then displays them in log files as to what was sent, so I am not
entirely sure how they are bring printed. I read the log files by opening
then as a binary file and viewing the content. here is a bit of
clarification as to what is happening.

{
//in the first case myVal is 20

UCHAR tmp2[2]; //yes unsigned char
tmp2[1] = myVal & 0x00FF;
tmp2[0] = (myVal & 0xFF00) >> 8 ;
memcpy(&myPkt[0], &tmp2, 2); // (where myPkt is a UCHAR*

myVal = 0x1500;
tmp2[1] = myVal & 0x00FF;
tmp2[0] = (myVal & 0xFF00) >> 8 ;
memcpy(&myPkt[2], &tmp2, 2); // (where myPkt is a UCHAR*

myVal = 0x0430;
tmp2[1] = myVal & 0x00FF;
tmp2[0] = (myVal & 0xFF00) >> 8 ;
memcpy(&myPkt[4], &tmp2, 2); // (where myPkt is a UCHAR*
}

in the binary file I see : 00 20 15 00 04 30

when I need to be seeing: 00 14 15 00 04 30
When I explicitly assign myVal a hex value (0x.......) it works fine.....but
when it is represented in decimal it pack the that decimal as if it were the
original hex value ( ie 0020, and not 0014);

I hope this is a little more clear

Delali

also when I try to change the code: memcpy(&myPkt[2], &tmp2, 2);
as suggested to: memcpy(myPkt[2], tmp2, 2);" I get the following error

"error C2664: 'memcpy' : cannot convert parameter 1 from 'unsigned char' to
'void *'"

"Peter van Merkerk" <(E-Mail Removed)> wrote in message
news:bm1l6t\$hc4hv\$(E-Mail Removed)-berlin.de...
> "Delali Dzirasa" <(E-Mail Removed)> wrote in message
> news:bm1hbf\$oqo\$(E-Mail Removed)...
> > I would have a number packed with its hex representation of the integer
> > below is some sample code of what is being done.
> >
> > int value = 20; file://in hex it is 0x14
> >
> > .
> > .
> > .
> >
> > {
> > UCHAR tmp2[2];
> > tmp2[1] = myVal & 0x00FF;
> > tmp2[0] = (myVal & 0xFF00) >> 8 ;
> >
> > memcpy(&myPkt[0], &tmp2, 2); // (where myPkt is a UCHAR*
> > }
> >
> > when I check the hex dump I see that 0x0020 was packed instead of the

> 0x0014
> > that I want. How can I set the proper flags(?) "if" that is the

solution
> to
> > pack the hex representation of the integer?

>
> The code you posted is unfortunately not compilable since AddData() does

not
> have a return type, and UCHAR and USHORT are not defined in the C++
> standard. That being said, I see nothing in your code that would explain

the
> values in your hex dump. Are you sure that the tool you are using is
> displaying the data in hexadecimal format and not decimal format?
>
> --
> Peter van Merkerk
> peter.van.merkerk(at)dse.nl
>
>

Howard
Guest
Posts: n/a

 10-08-2003

"Delali Dzirasa" <(E-Mail Removed)> wrote in message
news:bm1hbf\$oqo\$(E-Mail Removed)...
> I would have a number packed with its hex representation of the integer
> below is some sample code of what is being done.
>
> int value = 20; //in hex it is 0x14
>
> .
> .
> .
>
> {
> UCHAR tmp2[2];
> tmp2[1] = myVal & 0x00FF;

Suppose you pass in the decimal value 20. That's 0x14 in hex. The above
line masks off the high word (of a character, which is smaller!), so that
you are doing 0x0014 & 0x00ff, which results in 0x0014. Stored in an
unsigned char, that is 0x14, which is just what you started with: decimal
20!

> tmp2[0] = (myVal & 0xFF00) >> 8 ;

Here, you have (0x0014 & 0xff00)>>8, which results in (0x0000) >> 8, which
is 0x0000, or simply decimal 0 (zero). So, your two unsigned chars stored
in tmp2 are 0 and 20.

>
> memcpy(&myPkt[0], &tmp2, 2); // (where myPkt is a UCHAR*
> }
>
> when I check the hex dump I see that 0x0020 was packed instead of the

0x0014
> that I want. How can I set the proper flags(?) "if" that is the solution

to
> pack the hex representation of the integer?
>

I think your "hex dump" is not hex at all, but decimal, showing the first
byte as zero, and the second as 20, just like your code told it to do.

I'm not sure why you want to take an unsigned short and store it in two
unsigned characters, but that's hardly "packing", which implies reducing the
space required. What do you need in the output? Characters representing
the hex digits such as ['0','0','1','4']? A pair of unsigned char values,
one for each hex digit, such as [0,0,1,4]? Or do you really need to do this
"packng" at all? I mean, a hex dump of the original decinal value 20 will
show you 0x0014 just like you've been trying to get in the first place.

If you're trying to get the hex digits as separate values, such as
[0,0,1,4], remember that you've got 4 bytes in a short, not 2, and will need
an array of 4 unsigned char's to handle all possible unsigned short values.
0xff00 do.

-Howard

Jakob Bieling
Guest
Posts: n/a

 10-08-2003
"Delali Dzirasa" <(E-Mail Removed)> wrote in message
news:bm1hbf\$oqo\$(E-Mail Removed)...
> I would have a number packed with its hex representation of the integer
> below is some sample code of what is being done.

I do not really understand what you are trying to do. But I think that
you think that an integer can hold a number in decimal representation and
*in addition* in a, as you say, packed hexadecimal representation. That is
not the case. Assign 20 to an int and you will get the *exact* same bit
pattern when assigning 0x14 to it (which is 00010100 in both cases).

In another post you said that if you use the hexadecimal representation,
it works fine. I do not want to call you a liar, but whether you pass '20'
or '0x14' is completely irrelevant and will produce the *exact* same code.
It might be a good idea to post a new *minimal*, compilable code sample,
which people just have to paste into an new cpp file and compile.

hth
--
jb

(replace y with x if you want to reply by e-mail)

Ron Natalie
Guest
Posts: n/a

 10-08-2003

"Delali Dzirasa" <(E-Mail Removed)> wrote in message news:bm1mhs\$lic\$(E-Mail Removed)...

>
> also when I try to change the code: memcpy(&myPkt[2], &tmp2, 2);
> as suggested to: memcpy(myPkt[2], tmp2, 2);" I get the following error

I didn't suggest that.

Ron Natalie
Guest
Posts: n/a

 10-08-2003

"Delali Dzirasa" <(E-Mail Removed)> wrote in message news:bm1mhs\$lic\$(E-Mail Removed)...
> Yes they are sent to a file via another program that I am testing,

Make a complete version of a program that demonstrates the problem AND compiles.
You keep giving us fragments, which look like they ought to be fine. We're not
clairvoyant. The fault is almost certianly in the creation of the value you pass

Howard
Guest
Posts: n/a

 10-08-2003

"Ron Natalie" <(E-Mail Removed)> wrote in message
news:3f846fd8\$0\$36937\$(E-Mail Removed) m...
>
> "Delali Dzirasa" <(E-Mail Removed)> wrote in message

news:bm1mhs\$lic\$(E-Mail Removed)...
> > Yes they are sent to a file via another program that I am testing,

>
> Make a complete version of a program that demonstrates the problem AND

compiles.
> You keep giving us fragments, which look like they ought to be fine.

We're not
> clairvoyant. The fault is almost certianly in the creation of the value

you pass
>
>

I think the fault is actually with the fact the OP is masking off the
high/low words but trying to get bytes:

tmp2[1] = myVal & 0x00FF;
tmp2[0] = (myVal & 0xFF00) >> 8 ;

-Howard