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About retrieve an arbitrary byte among four bytes of an int.

/*

It seems that when an int with width of four bytes is assigned to a one
byte width char, the first three bytes from left to right are discarded
and the rightest byte is assigned to that char.

So, I can just assign an int to a char to retrieve the rightest byte of
that int, and I also use this method plus right-shift operation to get
the leftest byte of an int.

I should be wrong on it, please correct me with your expertise.

Thank you

lovecreatesbeauty

*/

#include <stdio.h>

int main(void)
{
unsigned int data = 'chum';
unsigned char left = '\0';
unsigned char right = '\0';

printf("data: %#X\n", data);

/* get the leftest byte of an int */
left = data >> (sizeof(int) - 1) * 8;

/* get the rightest byte of an int */
right = data;

printf("left: %#X, %c\n", left, left);
printf("right: %#X, %c\n", right, right);

return 0;
}

/* result: (Win2k, mingw32)
data: 0X6368756D
left: 0X63, c
right: 0X6D, m
*/

Jan 14 '06 #1
6 1950
lovecreatesbeauty wrote:
/*

It seems that when an int with width of four bytes is assigned to a one
byte width char, the first three bytes from left to right are discarded
and the rightest byte is assigned to that char.
That's how it usually works out. From the language's
point view you're assigning values rather than chunks of
the representation, and the value stored in the char is
derived from the value of the int via a conversion.

Your example code uses unsigned ints and chars, for
which things will work as you describe. For signed variants
the question of representation comes into play, and things
could turn out differently on a system that doesn't use
two's complement arithmetic. Such systems are surpassingly
rare nowadays, but ...
So, I can just assign an int to a char to retrieve the rightest byte of
that int, and I also use this method plus right-shift operation to get
the leftest byte of an int.
Yes, for unsigned types. Be warned that negative integers
do not right-shift the same way on all systems.
I should be wrong on it, please correct me with your expertise.

Thank you

lovecreatesbeauty

*/

#include <stdio.h>

int main(void)
{
unsigned int data = 'chum';
This construct isn't portable, and the value assigned
to data is entirely compiler-defined. Either the 'c' or
the 'm' -- or even neither! -- might wind up in the low-
order byte of data. A portable (assuming four-byte int)
way of ensuring the order you expect is to assemble it
from the individual pieces:

unsigned int data = ('c' << 24) + ('h' << 16)
+ ('u' << 8) + 'm';
unsigned char left = '\0';
unsigned char right = '\0';

printf("data: %#X\n", data);

/* get the leftest byte of an int */
left = data >> (sizeof(int) - 1) * 8;
This is fine, assuming eight-bit char. Machines
with other char sizes are rare, but definitely do exist
(they are commoner than machines that don't use two's
complement). To gain more portability, you could #include
the <limits.h> header and use CHAR_BIT instead of 8 (and
you'd make similar adjustments in the initialization of
data, above).
/* get the rightest byte of an int */
right = data;

printf("left: %#X, %c\n", left, left);
printf("right: %#X, %c\n", right, right);

return 0;
}

/* result: (Win2k, mingw32)
data: 0X6368756D
left: 0X63, c
right: 0X6D, m
*/


A little thought will show that shifting by other
amounts can retrieve the 'h' and 'u' characters, too.
In fact, you can retrieve groups of bits that straddle
the char boundaries, if you want. See also "masking."

--
Eric Sosman
es*****@acm-dot-org.invalid

Jan 14 '06 #2
In article <11**********************@g49g2000cwa.googlegroups .com>,
lovecreatesbeauty <lo***************@gmail.com> wrote:
It seems that when an int with width of four bytes is assigned to a one
byte width char, the first three bytes from left to right are discarded
and the rightest byte is assigned to that char.
What are "left" and "right" with respect to int?
The byte order of int in a variable is unspecified, and the
byte order of int in a CPU register might be different.

Your example is for "Win32", which (if I understand correctly) is
an OS implemented only for instruction sets that happen not to use
"bigendian" storage order in memory, and whose CPU registers are not
necessarily the same as their memory storage order.

If you define "left" as Most Significant Bit and "right" as
Least Signficant Bit, and the byte you want to get out is
the one containing the Least Significant Bits, and you recognize
that you are talking about *values* rather than about what might
happen to be stored in memory, and if you are use 'unsigned'
for the int and the char, and your int is 4 bytes, then Yes, a
simple assignment is certain to have that result.

So, I can just assign an int to a char to retrieve the rightest byte of
that int, and I also use this method plus right-shift operation to get
the leftest byte of an int.


You have jumped here from "when an int with of width four bytes" to
"so I can just" make that assignment for the desired result. The
problem with that is that for a random machine, int might not
have a width of -four- bytes. You have taken behaviour that is
true on one machine and generalized it to be true on all machines.
--
"It is important to remember that when it comes to law, computers
never make copies, only human beings make copies. Computers are given
commands, not permission. Only people can be given permission."
-- Brad Templeton
Jan 14 '06 #3

Walter Roberson wrote:
In article <11**********************@g49g2000cwa.googlegroups .com>,
What are "left" and "right" with respect to int?
The byte order of int in a variable is unspecified, and the
byte order of int in a CPU register might be different.

Yes, thank you. I'm a superficial guy with little knowledge.

then Yes, a
simple assignment is certain to have that result.


Thank you again, but it's not a homework.

lovecreatesbeauty

Jan 14 '06 #4
"lovecreatesbeauty" <lo***************@gmail.com> writes:
Walter Roberson wrote:
In article <11**********************@g49g2000cwa.googlegroups .com>,
What are "left" and "right" with respect to int?
The byte order of int in a variable is unspecified, and the
byte order of int in a CPU register might be different.


Yes, thank you. I'm a superficial guy with little knowledge.
then Yes, a
simple assignment is certain to have that result.


Thank you again, but it's not a homework.


He meant "assignment" as in "x=y;", not as in "homework assignment".

--
Keith Thompson (The_Other_Keith) 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.
Jan 14 '06 #5
Eric Sosman <es*****@acm-dot-org.invalid> writes:
lovecreatesbeauty wrote:
/*
It seems that when an int with width of four bytes is assigned to a
one
byte width char, the first three bytes from left to right are discarded
and the rightest byte is assigned to that char.


That's how it usually works out. From the language's
point view you're assigning values rather than chunks of
the representation, and the value stored in the char is
derived from the value of the int via a conversion.

Your example code uses unsigned ints and chars, for
which things will work as you describe. For signed variants
the question of representation comes into play, and things
could turn out differently on a system that doesn't use
two's complement arithmetic. Such systems are surpassingly
rare nowadays, but ...


The semantics of assigning a large value to an unsigned type too small
to hold it are well defined. The result is effectively to discard the
high-order bits (though it's defined in terms of the mathematical
values, not in terms of representation).

Assigning an overly large value to a *signed* type yields an
implementation-defined result (or, in C99, raises an
implementation-defined signal). It typically discards the high-order
bits, but there are other possibilities; for example, saturation
(converting all large values to the maximum value of the target type)
is legal. You should usually just avoid doing that.

--
Keith Thompson (The_Other_Keith) 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.
Jan 14 '06 #6
Walter Roberson wrote:
In article <11**********************@g49g2000cwa.googlegroups .com>,
Your example is for "Win32", which (if I understand correctly) is
an OS implemented only for instruction sets that happen not to use
"bigendian" storage order in memory, and whose CPU registers are not
necessarily the same as their memory storage order.


Thank you. I also tried this code snippet on a HP-UX 11i machine, the
result is:

$ cc +DAportable retrievebytefromint.c
cc: "retrievebytefromint.c", line 5: warning 27: Integer character
constant contains more than one character.
$ a.out
data: 0X6368756D
left: 0X63, c
right: 0X6D, m
$

lovecreatesbeauty

Jan 16 '06 #7

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