byte to LONG

[Magix]

I received byte by byte
0x8E byte1
0x3D byte2

0x64 byte3
0x5F byte4

How can I construct to be LONG 0x645F8E3D (shift left >> 16)

Thanks.

[Thomas Stegen]

Magix wrote:

I received byte by byte
0x8E byte1
0x3D byte2

0x64 byte3
0x5F byte4

How can I construct to be LONG 0x645F8E3D (shift left >> 16)

Determine where each by should go and then shift it there and then
bitwise or it with a long.

long result = 0;
unsigned char byte2 = 0x3D;

result |= byte2 << 8;

You can figure out the rest from there I guess.

(I have made the assumption that an unsigned char is 8 bits in the above
and that long is at least twice as large an unsigned char. Not
unreasonable, but not guaranteed. To get around the bytes are not 8 bits
problem you can use CHAR_BIT instead of 8, result |= byte2 << CHAR_BIT;)

--
Thomas.

[pete]

Thomas Stegen wrote:

Magix wrote:

I received byte by byte
0x8E byte1
0x3D byte2

0x64 byte3
0x5F byte4

How can I construct to be LONG 0x645F8E3D (shift left >> 16)

Determine where each by should go and then shift it there and then
bitwise or it with a long.

long result = 0;
unsigned char byte2 = 0x3D;

result |= byte2 << 8;

You can figure out the rest from there I guess.

(I have made the assumption that an unsigned char is 8 bits
in the above
and that long is at least twice as large an unsigned char.

I don't think that you made that assumption.
A portable expression to make a long unsigned value of
0x645F8E3D, out of those byte values, is:

(((long unsigned)byte3 << 24)
+ ((long unsigned)byte4 << 16)
+ (( unsigned)byte1 << 8)
+ byte2)

CHAR_BIT isn't part of it.

--
pete

[Nils O. Selåsdal]

On Sat, 09 Oct 2004 02:22:25 +0000, pete wrote:

A portable expression to make a long unsigned value of

What about where sizeof(long) < 4 ? Or endianess ?

[Chris Torek]

>On Sat, 09 Oct 2004 02:22:25 +0000, pete wrote:

A portable expression to make a long unsigned value of ...

In article <news:pa******* *************** ******@Utel.no>
Nils O. Selåsdal <NO*@Utel.no> wrote:What about where sizeof(long) < 4 ?
Not a problem, because even if sizeof(long)==1 , unsigned long
must still be at least 32 bits long, and hence hold values up
to 0xffffffffUL inclusive.
Or endianess ?

Not a problem either -- the expression depends only on the values
of the inputs, not their representations .
--
In-Real-Life: Chris Torek, Wind River Systems
Salt Lake City, UT, USA (40°39.22'N, 111°50.29'W) [phone]
email: forget about it http://web.torek.net/torek/index.html
Reading email is like searching for food in the garbage, thanks to spammers.

[Keith Thompson]

pete <pf*****@mindsp ring.com> writes:
[...]

A portable expression to make a long unsigned value of
0x645F8E3D, out of those byte values, is:

(((long unsigned)byte3 << 24)
+ ((long unsigned)byte4 << 16)
+ (( unsigned)byte1 << 8)
+ byte2)

CHAR_BIT isn't part of it.

Assuming that the input bytes are 8-bit bytes.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[pete]

Keith Thompson wrote:

pete <pf*****@mindsp ring.com> writes:
[...]

A portable expression to make a long unsigned value of
0x645F8E3D, out of those byte values, is:

(((long unsigned)byte3 << 24)
+ ((long unsigned)byte4 << 16)
+ (( unsigned)byte1 << 8)
+ byte2)

CHAR_BIT isn't part of it.

Assuming that the input bytes are 8-bit bytes.

Absolutely not.
There is no value of CHAR_BIT which could cause the value
of above said expression, to equal anything other than 0x645F8E3DLU
when the values below are used in the expression.
0x8E byte1
0x3D byte2
0x64 byte3
0x5F byte4

--
pete

[CBFalconer]

Keith Thompson wrote:

pete <pf*****@mindsp ring.com> writes:

[...]

A portable expression to make a long unsigned value of
0x645F8E3D, out of those byte values, is:

(((long unsigned)byte3 << 24)
+ ((long unsigned)byte4 << 16)
+ (( unsigned)byte1 << 8)
+ byte2)

CHAR_BIT isn't part of it.

Assuming that the input bytes are 8-bit bytes.

No, assuming they have values in the range 0..255.

--
Chuck F (cb********@yah oo.com) (cb********@wor ldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net> USE worldnet address!

[Keith Thompson]

pete <pf*****@mindsp ring.com> writes:

Keith Thompson wrote:

pete <pf*****@mindsp ring.com> writes:
[...]

> A portable expression to make a long unsigned value of
> 0x645F8E3D, out of those byte values, is:
>
> (((long unsigned)byte3 << 24)
> + ((long unsigned)byte4 << 16)
> + (( unsigned)byte1 << 8)
> + byte2)
>
> CHAR_BIT isn't part of it.

Assuming that the input bytes are 8-bit bytes.

Absolutely not.
There is no value of CHAR_BIT which could cause the value
of above said expression, to equal anything other than 0x645F8E3DLU
when the values below are used in the expression.
0x8E byte1
0x3D byte2
0x64 byte3
0x5F byte4

You're right, my mistake. I paid insufficient attention to the
original question, which asked specifically about constructing the
value 0x645F8E3D from the input bytes 0x8E, 0x3D, 0x64, and 0x5F.

If I wanted to do this kind of thing on a system with CHAR_BIT != 8
I'd probably want to re-examine the underlying assumptions, but the
problem as stated didn't actually make any assumptions about CHAR_BIT
(though it did implicitly assume that the bytes are either unsigned or
wider than 8 bits).

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[JC]

Since you got the four received bytes:

unsigned long l;
char *p;

p = (char *)&l;
*p = byte_2; /* 0x3d, Assuming little endian */
*(p+1) = byte_1; /* 0x8e */
*(p+2) = byte_4; /* ox5f */
*(p+3) = byte_3; /* ox64 */

Now you should contains 0x645f8e3d; well, you get the idea...

"CBFalconer " <cb********@yah oo.com> wrote in message
news:41******** ******@yahoo.co m...

Keith Thompson wrote:

pete <pf*****@mindsp ring.com> writes:

[...]

A portable expression to make a long unsigned value of
0x645F8E3D, out of those byte values, is:

(((long unsigned)byte3 << 24)
+ ((long unsigned)byte4 << 16)
+ (( unsigned)byte1 << 8)
+ byte2)

CHAR_BIT isn't part of it.

Assuming that the input bytes are 8-bit bytes.

No, assuming they have values in the range 0..255.

--
Chuck F (cb********@yah oo.com) (cb********@wor ldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net> USE worldnet address!