Dear all,
I open a binary file and want to write 0x00040700 to this file.
how can I set write buffer?
---------------------------------------------------
typedef unsigned char UCHAR;
int iFD=open(szFile Name,O_CREAT|O_ BINARY|O_TRUNC| O_WRONLY,S_IREA D|S_IWRITE);
UCHAR buffer[5]; //???????????
write(iFD,buffe r,5);
---------------------------------------------------
Thanks.
Regards,
cylin.
Nov 14 '05
20  5582 
In <40************ *****@news.indi vidual.net> rl*@hoekstra-uitgeverij.nl (Richard Bos) writes: fclose(outfile) ;
/* For critical applications, you should even check the return value
of fclose(), but I rarely do this. */
Bad idea: most of the things that can go wrong (after fopen succeeded)
happen at fclose time.
To optimise the error checking, when I generate an output file in a
compact succession of operations, I only check fclose and a fflush call
immediately preceding it (which is probably redundant).
OTOH, long running programs that generate output constantly (e.g. one
printf call per main loop iteration) should check each output call,
even if it's on stdout (that gets redirected to a file when the program
is run in non-interactive mode). No point in continuing the execution
if the program can no longer generate output.
Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de
In <c6************ @ID-230325.news.uni-berlin.de> "cylin" <cy***@avant.co m.tw> writes: I use low-level I/O functions because I want to the speed faster than
stardard I/O functions.
Where did you get this silly idea from? The low-level I/O functions have
a relatively high cost and the stardard I/O functions are designed to
minimise the number of low-level I/O function calls, by doing some local
buffering.
As a result of this, it is trivially easy for the ignorant to slow down
his I/O by one order of magnitude, while an expert can only speed up his
I/O by a small percentage, by using the low-level functions instead of the
standard ones and removing one level of buffering, when and where it is
not necessary.
Can't low-level I/O functions do this case?
Why bother?
Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de
> Use shifts and mask operations (bitwise AND) to extract the individual bytes. The details depend on the byte order in which you want to write
to the file.
IMHO one is better off using "htnl()." This will work fine so long as
all you are writing are 32 bit values.
kal <k_*****@yahoo. com> spoke thus: IMHO one is better off using "htnl()." This will work fine so long as
all you are writing are 32 bit values.
I suspect you meant htonl().
--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cybers pace.org | don't, I need to know. Flames welcome. k_*****@yahoo.c om (kal) writes: Use shifts and mask operations (bitwise AND) to extract the individual
bytes. The details depend on the byte order in which you want to write
to the file.
IMHO one is better off using "htnl()."
Wheather such a non-standard method of doing things is preferable when a
standard method exists is certainly debatable.
Martin
--
,--. Martin Dickopp, Dresden, Germany ,= ,-_-. =.
/ ,- ) http://www.zero-based.org/ ((_/)o o(\_))
\ `-' `-'(. .)`-'
`-. Debian, a variant of the GNU operating system. \_/
Martin Dickopp <ex************ ****@zero-based.org> wrote: rl*@hoekstra-uitgeverij.nl (Richard Bos) writes:
if (fwrite(buf, sizeof *buf, sizeof buf/sizeof *buf, outfile) !=
sizeof buf/sizeof *buf) {
puts("Handle a write error here.");
}
fclose(outfile) ;
/* For critical applications, you should even check the return value
of fclose(), but I rarely do this. */
Why do you check the return value of `fwrite' then? Most operating
systems don't write immediately to the underlying device when `fwrite'
is called, but have some buffering mechanism. Therefore, an error is
far more likely to show up in `fclose' than in `fwrite' on such systems.
Good question. Guess I'm just too used to interactive output.
Richard
Christopher Benson-Manica <at***@nospam.c yberspace.org> wrote: kal <k_*****@yahoo. com> spoke thus:
IMHO one is better off using "htnl()." This will work fine so long as
all you are writing are 32 bit values.
I suspect you meant htonl().
And _I_ suspect that function is not ISO.
Richard
Richard Bos <rl*@hoekstra-uitgeverij.nl> spoke thus: And _I_ suspect that function is not ISO.
Going off-topic doesn't have to mean making things up :)
--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cybers pace.org | don't, I need to know. Flames welcome.
Christopher Benson-Manica <at***@nospam.c yberspace.org> wrote IMHO one is better off using "htnl()."
I suspect you meant htonl().
Yes, it is "htonl()." Thank you very much. It was my mistake.
Are there any equivalent functions in the standard C library?
Is there even the concept of byte ordering in C?
In <a5************ **************@ posting.google. com> k_*****@yahoo.c om (kal) writes: Christopher Benson-Manica <at***@nospam.c yberspace.org> wrote
> IMHO one is better off using "htnl()."
I suspect you meant htonl().
Yes, it is "htonl()." Thank you very much. It was my mistake.
Are there any equivalent functions in the standard C library?
Nope.
Is there even the concept of byte ordering in C?
Nope. In theory the bits could be scattered all over the place, rather
than being nicely grouped in bytes, i.e. the physical representation of
a short could be:
b0 b7 b5 sign b4 b3 b1 b10 | b2 b14 b8 b9 b6 b12 b13 b12
---------------------------+----------------------------
first byte second byte
as long as all the operations generate the specified results (and unsigned
short replaces the sign bit by bit 15). Of course, this is not going
to happen in practice, but the standard is perfectly happy with it.
Some very popular (at the time) old architectures had glitches at the
byte ordering level (neither big endian nor little endian for 32-bit
integers) but they are gone now.
Dan
--
Dan Pop
DESY Zeuthen, RZ group
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