I'm looking for a freeware c99 compiler for windows. I had intended to use
MS's Visual C++ Express and use its C capability. In the past with my MS
products, I've simply needed to make .c the filetype to invoke the C
compiler. Here's a link http://www.microsoft.com/express/download/#webInstall
The download is 2.6 megs, which is near a reasonable size for a compiler,
but then setup.exe wants to download 87 megs of dot net framework hoggs
that I don't want on my machine.
In the past I've gone with Bloodshed, but I find myself unable to get the
shell to stay open so I can see some output.
The development environment for gcc is austere.
Anyone have another suggestion?
--
Women have simple tastes. They get pleasure out of the conversation of
children in arms and men in love.
H. L. Mencken
Jul 19 '08
173  13749 
santosh said:
Richard Heathfield wrote:
>s0****@gmail.com said:
>>In this particular case it's obvious, isn't it? That you can use
runtime evaluation to calculate the required size.
I don't see why it's advantageous to postpone until runtime a
calculation that can be performed trivially at compilation time.
Would it be as trivial for, say, a long double? I have had cases where
the same long double object produced an output of several lines of
digits for some values and just a few digits for others. Is there a
deterministic formula for doing this at compile-time based only on the
information derivable from float.h?
Assuming you mean a full decimal representation, well, it's pretty simple -
you need at least LDBL_MAX_10_EXP bytes. In practice, however, one doesn't
do this, because in practice one has a good idea of the range of values
that is pertinent to the problem domain, and one can work out exactly how
much storage is required for valid values; invalid values are rejected. I
think I already mentioned this.
<snip>
>(2a) snprintf twice
(2b) snprintf once (to get the size), followed by buffer provision,
followed by sprintf
It is my contention that (2b) is superior to (2a).
How? 2b isn't any more portable than 2a,
Agreed.
and it *might* be infinitesimally faster. No data loss occurs
in either case. So why is 2b superior to 2a?
It's infinitesimally faster - or at least it might be! But more to the
point, it's less to type, so less to get wrong.
<snip>
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
Richard Heathfield wrote:
s0****@gmail.com said:
.... snip ...
>
>if you don't have access to a C99 compiler (unlikely).
Hardly anyone has access to a C99 implementation. Microsoft don't
do one. Borland don't do one. GNU don't do one. That's a huge
proportion of the C marketplace right there.
Oh come on now. Gnu has relatively minor failures in C99
compliance. Admittedly it isn't all the way there, but you only
need one compiler.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.
Richard Heathfield wrote:
CBFalconer said:
>Richard Heathfield wrote:
>>CBFalconer said:
<snip>
>>>In that case I suggest using the %n feature in the format string.
How does that improve matters?
Use the null device to receive the testing run.
Most systems have such a device available.
This doesn't answer the question. How does %n improve matters?
It allows routines in the printf group to report the number of
chars emitted. From this you can resolve the size of buffer
needed, and malloc it. Admittedly finding the system null device
is a nuisance, and non-portable. I published, and you snipped, the
standard sections on the %n feature.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.
arnuld said:
>On Thu, 07 Aug 2008 12:04:42 +0000, Richard Heathfield wrote:
>The maximum number of decimal digits an int can represent, plus one for
the sign and one for the null terminator.
sizeof(int) on my machine is 4 and I tried your expression 4 times:
printf( "max_of_int = %d\n", ( (sizeof(int)* CHAR_BIT) +2 / 3 + 2 ));
That's not mine.
printf( "max_of_int = %d\n", ( ((sizeof(int)* CHAR_BIT)) + (2 / 3) + 2)
Neither is that.
); printf( "max_of_int = %d\n", ( ( ((sizeof(int)* CHAR_BIT) +2) / 3) +
2) );
That's equivalent to mine.
printf( "max_of_int = %d\n", ( ((sizeof(int)* CHAR_BIT) +2) / (3 + 2)
));
That's not mine.
so, 1st printf() is yours and the only match is second. I will takeit
part by part:
sizeof(int)*CHAR_BIT
why are you multiplying the "size of an int" with "how many bits
in a char" ?
To get the number of bits in an int.
2/3 , it gives what, i mean why it is there ?
No idea. That's your invention, not mine.
>It's possible, yes. For example, if CHAR_BIT is 33 (i.e. there are 33
bits in a byte)
and what normal desktop machine have that value, Intel Dual Core, AMD
Athlon 64 3000+ ?
Think "hypothetical example chosen at random".
Laughing part is "I am paid to
write C Code" when I can't understand your conversation..
That's scary, since nobody here is saying anything that is difficult to
understand.
and I call myself a Programmer... huh...
I couldn't possibly comment.
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
CBFalconer said:
Richard Heathfield wrote:
>CBFalconer said:
>>Richard Heathfield wrote:
CBFalconer said:
<snip>
>>>>In that case I suggest using the %n feature in the format string.
How does that improve matters?
Use the null device to receive the testing run.
Most systems have such a device available.
This doesn't answer the question. How does %n improve matters?
It allows routines in the printf group to report the number of
chars emitted.
Which you can get from the return value anyway. How does %n improve
matters? (Third time of asking.)
From this you can resolve the size of buffer
needed, and malloc it.
That may not be necessary, as I have said several times in this thread.
Admittedly finding the system null device
is a nuisance, and non-portable.
Right, so it's inadmissible here.
I published, and you snipped, the
standard sections on the %n feature.
What I asked was *why* you thought %n was an advantage, compared to a
simple capture of the return value. I did not ask you to explain what %n
does. You see, we all *know* what %n does, and in any case your article is
still available for all to see. I snipped your Standard quotes because
they didn't have the slightest bearing on the current discussion. That's
good snip, not bad snip.
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999 s0****@gmail.com said:
<snip>
I see. But if you call snprintf() first to tell you the size, it's
completely irrelevant whether in the second call you use sprintf() or
snprintf();
Yes, the difference is miniscule.
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
In article <48***************@yahoo.com>,
CBFalconer <cb********@maineline.netwrote:
>Richard Heathfield wrote:
>s0****@gmail.com said:
... snip ...
>>
>>if you don't have access to a C99 compiler (unlikely).
Hardly anyone has access to a C99 implementation. Microsoft don't
do one. Borland don't do one. GNU don't do one. That's a huge
proportion of the C marketplace right there.
Oh come on now. Gnu has relatively minor failures in C99
compliance. Admittedly it isn't all the way there, but you only
need one compiler.
It looks very much like someone has hacked CBF's account. This does not
sound at all like the CBF we all knew.
Anyway, the dogma of the group (certain that of RH) is "all or its crap".
If you can find one little thing to nit-pick about, then the
implementation is crap.
Besides that, is the fact that the dogma of the group has also always been
that the compilers listed above (namely, Microsoft [i.e., Windows], gcc
[i.e., Linux] and Borland [i.e., DOS/Windows]) represent a miniscule and
essentially irrelevant part of the C world. That the "big money" is in
embedded systems (not hosted) and that what really matters is all the
(unnamed) compilers for embedded systems (that we never hear about on
this group - funny, that)
Richard Heathfield wrote:
CBFalconer said:
>Richard Heathfield wrote:
>>CBFalconer said:
Richard Heathfield wrote:
CBFalconer said:
<snip>
>In that case I suggest using the %n feature in the format string.
>
How does that improve matters?
Use the null device to receive the testing run.
Most systems have such a device available.
This doesn't answer the question. How does %n improve matters?
<snip>
In this call:
int ret = 0;
int n = 0;
ret = fprintf(temp_file, "%p%n", (void*)ptr, n);
If an output error occurred after the 'p' specifier had been processed,
n will still contain the number of characters written out, while ret
will contain a negative value.
But I think this is not likely at all.
CBFalconer <cb********@yahoo.comwrites:
Richard Heathfield wrote:
>s0****@gmail.com said:
... snip ...
>>
>>if you don't have access to a C99 compiler (unlikely).
Hardly anyone has access to a C99 implementation. Microsoft don't
do one. Borland don't do one. GNU don't do one. That's a huge
proportion of the C marketplace right there.
Oh come on now. Gnu has relatively minor failures in C99
compliance. Admittedly it isn't all the way there, but you only
need one compiler.
No, if you need to produce code that will run on multiple systems, you
may need multiple compilers. And if you're distributing source code,
you need to ensure that the code will be compatible with your users'
compilers, which you might not have access to.
Given the current state of the market, the best way to ensure that is
to restrict yourself to the intersection of C90 and C99. If you use,
say, gcc for your own system, you can take advantage of the features
that gcc offers -- but what if your users want to compile your code on
a system that doesn't have gcc? Or what if they want to use a
different compiler because it generates faster code for their system
than gcc does?
--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
Keith Thompson wrote:
CBFalconer <cb********@yahoo.comwrites:
[ ... ]
>Oh come on now. Gnu has relatively minor failures in C99
compliance. Admittedly it isn't all the way there, but you only
need one compiler.
No, if you need to produce code that will run on multiple systems, you
may need multiple compilers. And if you're distributing source code,
you need to ensure that the code will be compatible with your users'
compilers, which you might not have access to.
Given the current state of the market, the best way to ensure that is
to restrict yourself to the intersection of C90 and C99.
Which is more or less C90.
<snip>
On 7 Aug 2008 at 3:46, Keith Thompson wrote:
CBFalconer <cb********@yahoo.comwrites:
>If the pointer doesn't change, its representation cannot change,
since the representation has to be capable of external storage,
reading in with scanf, and thus recreating the exact original
pointer.
First off, a conforming implementation could, for example, say that
the "%p" conversion yields a random number of underscores followed by
the pointer representation in hexadecimal. *scanf would just ignore
the underscores. This would be a silly implementation, but I believe
it would be conforming.
It is absolutely astonishing that two grown-up people can conduct a
conversation about nonsense like this, apparently in all seriousness.
Keith Thompson said:
CBFalconer <cb********@yahoo.comwrites:
>Richard Heathfield wrote:
<snip>
>>Hardly anyone has access to a C99 implementation. Microsoft don't
do one. Borland don't do one. GNU don't do one. That's a huge
proportion of the C marketplace right there.
Oh come on now. Gnu has relatively minor failures in C99
compliance. Admittedly it isn't all the way there, but you only
need one compiler.
No, if you need to produce code that will run on multiple systems, you
may need multiple compilers. And if you're distributing source code,
you need to ensure that the code will be compatible with your users'
compilers, which you might not have access to.
Those two sentences summarise my own predicament rather nicely.
Given the current state of the market, the best way to ensure that is
to restrict yourself to the intersection of C90 and C99.
Which is precisely what I do (most of the time).
If you use,
say, gcc for your own system, you can take advantage of the features
that gcc offers -- but what if your users want to compile your code on
a system that doesn't have gcc? Or what if they want to use a
different compiler because it generates faster code for their system
than gcc does?
I am delighted to be able to report that, on occasion, I /can/ take
advantage of compiler-specific features. Alas, it doesn't happen as often
as I'd like, but it /does/ happen... from time to time.
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
Richard Heathfield wrote:
santosh said:
>Richard Heathfield wrote:
.... snip ...
>>
>>(2a) snprintf twice
(2b) snprintf once (to get the size), followed by buffer provision,
followed by sprintf
It is my contention that (2b) is superior to (2a).
How? 2b isn't any more portable than 2a,
Agreed.
>and it *might* be infinitesimally faster. No data loss occurs
in either case. So why is 2b superior to 2a?
It's infinitesimally faster - or at least it might be! But more to
the point, it's less to type, so less to get wrong.
Possibly. However, it probably significantly increases the amount
of code loaded from the library. This may significantly increase
the runtime module size, assuming static library. Two functions
are normally bigger than one.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.
santosh <sa*********@gmail.comwrites:
Keith Thompson wrote:
>CBFalconer <cb********@yahoo.comwrites:
[ ... ]
>>Oh come on now. Gnu has relatively minor failures in C99
compliance. Admittedly it isn't all the way there, but you only
need one compiler.
No, if you need to produce code that will run on multiple systems, you
may need multiple compilers. And if you're distributing source code,
you need to ensure that the code will be compatible with your users'
compilers, which you might not have access to.
Given the current state of the market, the best way to ensure that is
to restrict yourself to the intersection of C90 and C99.
Which is more or less C90.
Right. To quantify the "more or less", you have to avoid implicit int
and using any of the new C99 keywords (inline, restrict, are there
others?) as identifiers, and probably a few other things I haven't
thought of.
More generally, you have to avoid anything that conflicts with C99.
For example, writing your own "bool.h" or "complex.h" header would be
a bad idea.
Note that C99 keywords like _Bool aren't an issue, since you can't
portably use them in C90 anyway.
--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
Richard Heathfield wrote:
CBFalconer said:
>Richard Heathfield wrote:
.... snip ...
>>
>>This doesn't answer the question. How does %n improve matters?
It allows routines in the printf group to report the number of
chars emitted.
Which you can get from the return value anyway. How does %n
improve matters? (Third time of asking.)
Didn't realize, in the heat of the news replying, that I could just
use the *printf returned value. I have even been known to use it.
Then the only useful idea in that particular set is using the
(occasionally available) null device. And the only real purpose of
%n is to take length measurements in the middle of the *printf
call.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home.att.net>
Try the download section.
On Thu, 07 Aug 2008 14:57:18 +0000, Richard Heathfield wrote:
>arnuld said:
>(sizeof(int)* CHAR_BIT) +2) / 3 + 2;
That's equivalent to mine.
okay
To get the number of bits in an int.
so, my machine has 32 bytes in an int
> 2/3 , it gives what, i mean why it is there ?
No idea. That's your invention, not mine.
your example: sizeof(int) * CHAR_BIT + 2, its okay till here. 32 bytes +
2 bytes more ( 1 more sign and 1 for null character)
now you divide that expression by 3 and then add 2 to it. Why ?
I couldn't possibly comment.
thanks
-- www.lispmachine.wordpress.com
my email is @ the above blog
check the "About Myself" page
arnuld said:
>On Thu, 07 Aug 2008 14:57:18 +0000, Richard Heathfield wrote:
<snip>
>To get the number of bits in an int.
so, my machine has 32 bytes in an int
That's a lot of bytes.
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
On Fri, 08 Aug 2008 08:14:36 +0000, Richard Heathfield wrote:
That's a lot of bytes.
eeee....... Richard :P
I *really* want to know why you divided the whole thing by 3 and added 2
to the result.
I know the final result is number of decimal digits an int can represent,
which is the maximum value of int on a machine. I am asking is that some
kind of formula like n(n+1)/2 ?
-- www.lispmachine.wordpress.com
my email is @ the above blog
check the "About Myself" page
Hello,
arnuld schrieb:
>On Thu, 07 Aug 2008 14:57:18 +0000, Richard Heathfield wrote:
your example: sizeof(int) * CHAR_BIT + 2, its okay till here. 32 bytes +
2 bytes more ( 1 more sign and 1 for null character)
now you divide that expression by 3 and then add 2 to it. Why ?
Ok, let's take the original expression again:
(sizeof(int) * CHAR_BIT + 2) / 3 + 2
The + 2 at the end is clear and already explained by RH;
sizeof(int)*CHAR_BIT is the number of bits the int occupies. I'll call
it BITS_OF_AN_INT.
So, the only "problamtic" part is: (BITS_OF_AN_INT + 2 ) / 3.
Here, RH is "cheating": In fact, he does not calculate the number of
digits in the decimal system. He calculates the number of digits in
octal!
3 bits build up 1 octal digit, much like 4 bits build up 1 hexadecimal
digit. So, he divides BITS_OF_AN_INT by 3. The "+ 2" makes sure that his
calculation has a ceiling, that is, rounds up.
I hope this clarifies the calculation.
However, IMHO, it is not really obvious. Richard (RH), I hope you
clearly document this whenever you use it.
Regards,
Spiro.
"Anybody who would spend considerable effort emulating a C64 is not
particularly sane to begin with; it is unreasonable to expect them to produce
sane software."
("asuffield" in http://forums.thedailywtf.com/forums/p/7199/134136.aspx)
--
Spiro R. Trikaliotis http://opencbm.sf.net/ http://www.trikaliotis.net/ http://www.viceteam.org/
Hello,
I wrote:
Here, RH is "cheating": In fact, he does not calculate the number of
digits in the decimal system. He calculates the number of digits in
octal!
Of course, one has to remember that the octal representation needs at
least as many digits as the decimal representation. Without this fact,
this calculation would be pointless.
Regards,
Spiro.
"Anybody who would spend considerable effort emulating a C64 is not
particularly sane to begin with; it is unreasonable to expect them to produce
sane software."
("asuffield" in http://forums.thedailywtf.com/forums/p/7199/134136.aspx)
--
Spiro R. Trikaliotis http://opencbm.sf.net/ http://www.trikaliotis.net/ http://www.viceteam.org/
Spiro Trikaliotis said:
Hello,
arnuld schrieb:
>>On Thu, 07 Aug 2008 14:57:18 +0000, Richard Heathfield wrote:
>your example: sizeof(int) * CHAR_BIT + 2, its okay till here. 32 bytes
+ 2 bytes more ( 1 more sign and 1 for null character)
now you divide that expression by 3 and then add 2 to it. Why ?
Ok, let's take the original expression again:
(sizeof(int) * CHAR_BIT + 2) / 3 + 2
The + 2 at the end is clear and already explained by RH;
Right. Actually, I explained the entire calculation, for a variety of
number bases, on 22 July of this year (i.e. about two weeks ago). Here is
the message ID: <5q******************************@bt.com>
Perhaps I should do a Web page (sigh).
<snip>
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
On Wed, 06 Aug 2008 16:17:16 +0000, Kenny McCormack wrote:
In article <g7**********@registered.motzarella.org>, santosh
<sa*********@gmail.comwrote:
>>CBFalconer wrote:
>>santosh wrote:
Richard Heathfield wrote:
... snip ...
open a shell, find the program, run the program from the shell.
That's the ideal. But assuming that 'pereges' cannot do that (since
if he could, he obviously wouldn't be misusing getch like this),
getchar provides the behaviour he needs (block just before return
from main) without rendering the code nonportable.
He can. He just hasn't learned how.
Perhaps, but teaching him to open a shell is off-topic for this group,
as is getch(), hence the advise to use getchar().
There you have it gentlemen. What more evidence do you need?
Thank you Judas for the victim;
Stay a while; you'll watch it bleed.
>
(For those of you slow on the uptake: The "it" is proof that the regs
and wannabees [santosh] would far, far rather give totally worthless and
incomprehensible [to newbies] advice, than give a useful "off-topic"
answer.)
I think santosh prefers the C programming language to a syntax equivalent
of the american playground classic: smear the queer.
--
Howard
On Thu, 07 Aug 2008 11:00:52 -0400, CBFalconer <cb********@yahoo.com>
wrote:
Richard Heathfield wrote:
>(2b) snprintf once (to get the size), followed by buffer provision,
followed by sprintf [vs snprintf twice]
It's infinitesimally faster - or at least it might be! But more to
the point, it's less to type, so less to get wrong.
Possibly. However, it probably significantly increases the amount
of code loaded from the library. This may significantly increase
the runtime module size, assuming static library. Two functions
are normally bigger than one.
In general, but very unlikely in this case. Because of their
near-total duplication of extensive (compared to other std-lib
features) functionality, all of the *printf routines (and similarly
*scanf) are usually (near-trivial) wrappers around, or sometimes just
entrypoints into, a single 'engine'.
- formerly david.thompson1 || achar(64) || worldnet.att.net This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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The next Access Europe User Group meeting will be on Wednesday 1 May 2024 starting at 18:00 UK time (6PM UTC+1) and finishing by 19:30 (7.30PM).
In this session, we are pleased to welcome a new...
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by: conductexam |
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I have .net C# application in which I am extracting data from word file and save it in database particularly. To store word all data as it is I am converting the whole word file firstly in HTML and...
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by: adsilva |
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A Windows Forms form does not have the event Unload, like VB6. What one acts like?
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