Four or Two Bytes?

[Hahnemann]

Does anybody know the answer to the following? An unsigned short is 2
bytes long. Why is the following file created as 4 bytes instead of
2?

file = fopen("data.bin ", "wb");
if (file != NULL)
{
unsigned short s = 65535;
printf("Size of unsigned short: %d bytes\n", sizeof(unsigned
short)); // 2 bytes - OK
fwrite(&s, sizeof(unsigned short), sizeof(s), file);
fclose(file); // File written as 4 bytes... hmm
}

- Hahnemann

[pete]

Keith Thompson wrote:

"Joachim Schmitz" <no*********@sc hmitz-digital.dewrite s:

>pete wrote:

[...]

This entire discussion is about as close to having no practical
meaning as I can imagine. An implementation that had as many padding
bits for unsigned short as we're talking about, or even as many value
bits, would be beyond exotic; it would be clinically insane.

We were discussing the DS9K.

--
pete

[Bill Leary]

"Barry Schwarz" <sc******@dqel. comwrote in message
news:68******** *************** *********@4ax.c om...

On Thu, 29 May 2008 22:53:32 -0400, "Bill Leary" <Bi********@msn .com>
wrote:

>>"Hahnemann" <ha************ *@gmail.comwrot e in message
news:2f****** *************** *************@d 1g2000hsg.googl egroups.com...

>>Does anybody know the answer to the following? An unsigned short is 2
bytes long. Why is the following file created as 4 bytes instead of
2?

file = fopen("data.bin ", "wb");
if (file != NULL)
{
unsigned short s = 65535;
printf("Size of unsigned short: %d bytes\n", sizeof(unsigned
short)); // 2 bytes - OK

>> fwrite(&s, sizeof(unsigned short), sizeof(s), file);

fwrite(
pointer to buffer ... ok
size of an element of that buffer ... 2
number of elements to write ... 2
handle for file write should occur to ... ok

So, 2 elements of 2 bytes each should write four bytes.

>> fclose(file); // File written as 4 bytes... hmm

So, fwrite worked correctly. More or less, since it wrote out 4 bytes
from
a 2 byte object. It could just about as easily have crashed when it went
beyond the bounds of the object.

Actually, fwrite invoked undefined behavior and only appeared to work
correctly because the OP was using it on a system where he was allowed
to access memory beyond the end of s. On a well-behaved system (tm),
his program would have been interrupted in an implementation defined
manner.

Sure. But since the specific question was why he got four bytes rather than
two in the target file I emphasized that 2 * 2 is 4, and alluded your point
by commenting "More or less..." and mentioning the crash.

- Bill

[Barry Schwarz]

On Sun, 1 Jun 2008 19:56:33 -0400, "Bill Leary" <Bi********@msn .com>
wrote:

>"Barry Schwarz" <sc******@dqel. comwrote in message
news:68******* *************** **********@4ax. com...

snip

>Actually, fwrite invoked undefined behavior and only appeared to work
correctly because the OP was using it on a system where he was allowed
to access memory beyond the end of s. On a well-behaved system (tm),
his program would have been interrupted in an implementation defined
manner.

Sure. But since the specific question was why he got four bytes rather than
two in the target file I emphasized that 2 * 2 is 4, and alluded your point
by commenting "More or less..." and mentioning the crash.

But once the OP has invoked undefined behavior, there is no guarantee
that 2*2 will still equal 4. The only reason fwrite produced 4 bytes
is that this particular manifestation of undefined behavior produced 4
bytes. The fact that he inadvertently asked for 4 bytes is only
unfortunate coincidence. The next time it executes it could do
something completely different. If the OP were to recompile with
different options, who knows what the code could do.
Remove del for email

[Bill Leary]

"Barry Schwarz" <sc******@dqel. comwrote in message
news:1t******** *************** *********@4ax.c om...

On Sun, 1 Jun 2008 19:56:33 -0400, "Bill Leary" <Bi********@msn .com>
wrote:

>>Sure. But since the specific question was why he got four bytes rather
than
two in the target file I emphasized that 2 * 2 is 4, and alluded your
point
by commenting "More or less..." and mentioning the crash.

But once the OP has invoked undefined behavior, there is no guarantee
that 2*2 will still equal 4. The only reason fwrite produced 4 bytes
is that this particular manifestation of undefined behavior produced 4
bytes. The fact that he inadvertently asked for 4 bytes is only
unfortunate coincidence. The next time it executes it could do
something completely different. If the OP were to recompile with
different options, who knows what the code could do.

You are both technically and theoretically correct.

The root issue, however, was that he was telling the function to write four
bytes when he thought he was telling it to write two bytes.

Explaining the potentially world shattering side effects* of what might have
happened due to his misunderstandin g invoking "undefined behavior" would
hardly have been profitable. What he needed, and asked for, was an
explanation of what actually did happen.

If you want to go on in the "undefined behavior" vein, please do so.
Someone has to fight the Good Fight, and you seem to understand your topic
well and do it with great verve.**

For me, I'm done.

- Bill
______________
* I recall someone in another "undefined behavior" thread saying that the
program was free to do anything at all, including implode the world, if it
so desired.
** Yeah, I'm being a little sarcastic. But just a little. I've just
re-examined a half dozen messages where you made your "undefined behavior"
spiel and in each of them (as far as I could tell) you were dead on correct.
An impressive accomplishment.