Using char

[Conan1st]

Hello everyone!

I made a variable under unsigned short but displaying it using char the
value that doesnt change in fact is shown changed...

unsigned short a

a = 321;

printf("a = %d", (char)a);

the display is 'a = 63'...

why is that?

how can I have the right display for a?
--
Posted via http://dbforums.com

[Ben Pfaff]

Conan1st <me*********@dbforums.com> writes:

I made a variable under unsigned short but displaying it using char the
value that doesnt change in fact is shown changed...

unsigned short a

a = 321;

printf("a = %d", (char)a);

Obviously, 321 doesn't fit in `char' on your platform. Use a
wider type.

(Also, %d is for displaying `int's, not `char's, but that's
unlikely to be the real problem.)

[Mac]

On Mon, 27 Oct 2003 23:22:45 +0000, Conan1st wrote:

Hello everyone!

I made a variable under unsigned short but displaying it using char the
value that doesnt change in fact is shown changed...

unsigned short a

a = 321;

printf("a = %d", (char)a);

the display is 'a = 63'...

why is that?

how can I have the right display for a?

printf("a = %hu", a);

When you use a %d with printf, you need to pass an int. %u Is for unsigned
int, and %hu is for a short unsigned int.

If you are determined to use '%d,' then cast 'a' to 'int':

printf("a = %d", (int)a);

Don't forget to add a '\n' at the end of the string, or fflush(stdout),
otherwise you may not ever see the output.

HTH

Mac
--

[Charles Harrison Caudill]

Conan1st <me*********@dbforums.com> wrote:

unsigned short a
a = 321;
printf("a = %d", (char)a); the display is 'a = 63'...
why is that?

Like someone else said, 321 doesn't fit in 8 bits.
a is a short, on x86 it occupies 16 bits. You're
trying to print after casting it to a char. When
you cast it to a char the c compiler puts in logic
mask out all but the lowest byte ( I actually just
told a lie since x86 is little-endian curse them ).
When you print the result as an int it is
effectively doing this:

unsigned short a = 321;
unsigned char b = 0;
b = ( a & 0x00ff );
printf("%d\n", (int)b);

--
Harrison Caudill | .^ www.hypersphere.org
Computer Science & Physics Double Major | | Me*Me=1
Georgia Institute of Technology | '/ I'm just a normal guy

[nobody]

"Conan1st" <me*********@dbforums.com> wrote in message
news:35****************@dbforums.com...

Hello everyone!

I made a variable under unsigned short but displaying it using char the
value that doesnt change in fact is shown changed...

unsigned short a Missing semicolon.

a = 321;

printf("a = %d", (char)a);

the display is 'a = 63'... Are you sure? Unless it is UB (of which I'm not sure), it should output
a = 65
why is that? because 321 % 256 == 65 (or 321 & 0xff == 65, if you prefer).
how can I have the right display for a?

Mac already answered ...

[Dan Pop]

In <35****************@dbforums.com> Conan1st <me*********@dbforums.com> writes:

I made a variable under unsigned short but displaying it using char the
value that doesnt change in fact is shown changed...

unsigned short a
a = 321;
printf("a = %d", (char)a);

the display is 'a = 63'...
why is that?
If you can't engage your brain, C programming is not for you.

Isn't it obvious, from your result, that 321 cannot be represented by the
type char on your implementation?

On the vast majority of hosted implementations, char is an 8-bit type.
One of these 8 bits may or may not be used as a sign bit. How many bits
do you need to represent the value 321?
how can I have the right display for a?

By not converting it to a narrower type before displaying it. Isn't it
obvious? What's less obvious is the right printf invocation, to get the
right answer in a portable way:

printf("a = %u\n", (unsigned)a);

The cast to unsigned is needed because, in a portable programming context,
you don't know to what type a is going to be promoted by the default
argument promotions: int or unsigned int. Since no conversion descriptor
can accept either int or unsigned int, you have to bypass the default
argument promotions by converting the value to a type that is no longer
subject to the default argument promotions.

In standard C, this is an issue only with variadic functions, like
printf. For other functions you can always provide a prototype, so that
the default argument promotions are no longer performed by the compiler,
when the function is called.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Dan Pop]

In <87************@pfaff.stanford.edu> Ben Pfaff <bl*@cs.stanford.edu> writes:

Conan1st <me*********@dbforums.com> writes:

I made a variable under unsigned short but displaying it using char the
value that doesnt change in fact is shown changed...

unsigned short a

a = 321;

printf("a = %d", (char)a);

Obviously, 321 doesn't fit in `char' on your platform. Use a
wider type.

(Also, %d is for displaying `int's, not `char's, but that's
unlikely to be the real problem.)

Huh? What *exactly* is wrong with passing a char to %d, except for the
pathological case when char gets promoted to unsigned int?

Have they dropped the default argument promotions from the C standard
behind my back?

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[pete]

Mac wrote:

On Mon, 27 Oct 2003 23:22:45 +0000, Conan1st wrote:

unsigned short a printf("a = %d", (char)a); how can I have the right display for a?

printf("a = %hu", a);

That's the best way, in my opinion.

--
pete

[Lew Pitcher]

Ben Pfaff wrote:

Conan1st <me*********@dbforums.com> writes:

I made a variable under unsigned short but displaying it using char the
value that doesnt change in fact is shown changed...

unsigned short a

a = 321;

printf("a = %d", (char)a);

Obviously, 321 doesn't fit in `char' on your platform. Use a
wider type.

(Also, %d is for displaying `int's, not `char's, but that's
unlikely to be the real problem.)

But, Ben, in this case, the char _is_ an int. ;-)

Since printf()'s variadic parameters go through standard function call type
promotion, the char data item is promoted to int as part of the call, before
the call is made.

But you knew that already.

--

Lew Pitcher, IT Consultant, Application Architecture
Enterprise Technology Solutions, TD Bank Financial Group

(Opinions expressed here are my own, not my employer's)

[Jirka Klaue]

Lew Pitcher wrote:

Ben Pfaff wrote:

....

(Also, %d is for displaying `int's, not `char's, but that's
unlikely to be the real problem.)

But, Ben, in this case, the char _is_ an int. ;-)

I think he meant "character", as in printf("%c", a);

Jirka

[Lew Pitcher]

Jirka Klaue wrote:

Lew Pitcher wrote:

Ben Pfaff wrote:

...

(Also, %d is for displaying `int's, not `char's, but that's
unlikely to be the real problem.)

But, Ben, in this case, the char _is_ an int. ;-)

I think he meant "character", as in printf("%c", a);

Assuming that a is a char or an int, then the above is legal and proper. If
a is a char, it will be promoted to int as part of the call (variadic
functions), and if a is an int, it remains an int. In either case, the
resulting int will be explicitly truncated into a character when printed
using the %c format.

In any case, Ben's comment was directed at the OP's code fragment
printf("a = %d", (char)a);

--

Lew Pitcher, IT Consultant, Application Architecture
Enterprise Technology Solutions, TD Bank Financial Group

(Opinions expressed here are my own, not my employer's)

[Jirka Klaue]

Lew Pitcher wrote:

Jirka Klaue wrote:

Lew Pitcher wrote:

Ben Pfaff wrote: ...

(Also, %d is for displaying `int's, not `char's, but that's
unlikely to be the real problem.)

But, Ben, in this case, the char _is_ an int. ;-)

I think he meant "character", as in printf("%c", a);

Assuming that a is a char or an int, then the above is legal and proper.
If a is a char, it will be promoted to int as part of the call (variadic
functions), and if a is an int, it remains an int. In either case, the
resulting int will be explicitly truncated into a character when printed
using the %c format.

What is your understanding of "truncated" in this case?
In any case, Ben's comment was directed at the OP's code fragment
printf("a = %d", (char)a);

Yes, and my comment was directed to you. Did you consider that the
printf("%c", a) is what the OP actually wanted and that Ben wanted
to express exactly that?

Jirka

[Dan Pop]

In <3F***********@mindspring.com> pete <pf******@mindspring.com> writes:

Mac wrote:

On Mon, 27 Oct 2003 23:22:45 +0000, Conan1st wrote:

> unsigned short a > printf("a = %d", (char)a); > how can I have the right display for a?

printf("a = %hu", a);

That's the best way, in my opinion.

Except that it's not clear what is the *actual* type expected by %hu: is
it int or unsigned int?

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Dan Pop]

In <bn**********@mamenchi.zrz.TU-Berlin.DE> Jirka Klaue <jk****@ee.tu-berlin.de> writes:

Lew Pitcher wrote:

Ben Pfaff wrote:

...

(Also, %d is for displaying `int's, not `char's, but that's
unlikely to be the real problem.)

But, Ben, in this case, the char _is_ an int. ;-)

I think he meant "character", as in printf("%c", a);

The OP was obviously expecting a numeric value, not some random character.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Dan Pop]

In <bn**********@mamenchi.zrz.TU-Berlin.DE> Jirka Klaue <jk****@ee.tu-berlin.de> writes:

Lew Pitcher wrote:

Jirka Klaue wrote:

Lew Pitcher wrote:
Ben Pfaff wrote:
...

> (Also, %d is for displaying `int's, not `char's, but that's
> unlikely to be the real problem.)

But, Ben, in this case, the char _is_ an int. ;-)

I think he meant "character", as in printf("%c", a);

Assuming that a is a char or an int, then the above is legal and proper.
If a is a char, it will be promoted to int as part of the call (variadic
functions), and if a is an int, it remains an int. In either case, the
resulting int will be explicitly truncated into a character when printed
using the %c format.

What is your understanding of "truncated" in this case?

In any case, Ben's comment was directed at the OP's code fragment
printf("a = %d", (char)a);

Yes, and my comment was directed to you. Did you consider that the
printf("%c", a) is what the OP actually wanted and that Ben wanted
to express exactly that?

This hypothesis is not worth considering, since the OP was expecting to
see 321 as the output and didn't understand why he got another value.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Lew Pitcher]

In <bn**********@mamenchi.zrz.TU-Berlin.DE> Jirka Klaue
<jk****@ee.tu-berlin.de> writes:

Lew Pitcher wrote:

Jirka Klaue wrote:

Lew Pitcher wrote:

Ben Pfaff wrote:

...
>(Also, %d is for displaying `int's, not `char's, but that's
>unlikely to be the real problem.)

But, Ben, in this case, the char _is_ an int. ;-)

I think he meant "character", as in printf("%c", a);

Assuming that a is a char or an int, then the above is legal and proper.
If a is a char, it will be promoted to int as part of the call (variadic
functions), and if a is an int, it remains an int. In either case, the
resulting int will be explicitly truncated into a character when printed
using the %c format.

What is your understanding of "truncated" in this case?

truncated: past part of "truncate": to shorten by cutting off the top or end

New Websters Dictionary and Thesaurus

--

Lew Pitcher, IT Consultant, Application Architecture
Enterprise Technology Solutions, TD Bank Financial Group

(Opinions expressed here are my own, not my employer's)

[Jirka Klaue]

Lew Pitcher wrote:

In <bn**********@mamenchi.zrz.TU-Berlin.DE> Jirka Klaue writes:

Lew Pitcher wrote: ....

In
either case, the resulting int will be explicitly truncated into a
character when printed using the %c format.

What is your understanding of "truncated" in this case?

truncated: past part of "truncate": to shorten by cutting off the top or
end

The standard says:

7.19.6.1#8
c If no l length modifier is present, the int argument is converted to an
unsigned char, and the resulting character is written.
...

"Converted" does not necessarily mean "truncated".

Jirka

[pete]

Dan Pop wrote:

In <3F***********@mindspring.com> pete <pf******@mindspring.com> writes:

Mac wrote:

On Mon, 27 Oct 2003 23:22:45 +0000, Conan1st wrote:

> unsigned short a

> printf("a = %d", (char)a);

> how can I have the right display for a?

printf("a = %hu", a);

That's the best way, in my opinion.

Except that it's not clear what is the
*actual* type expected by %hu: is
it int or unsigned int?

I would expect that %hu would *actually* expect
whatever type it is that unsigned short promotes to,
be it int or be it unsigned.

--
pete

[Dan Pop]

In <3F***********@mindspring.com> pete <pf*****@mindspring.com> writes:

Dan Pop wrote:

In <3F***********@mindspring.com> pete <pf******@mindspring.com> writes:

>Mac wrote:
>>
>> On Mon, 27 Oct 2003 23:22:45 +0000, Conan1st wrote:
>
>> > unsigned short a
>
>> > printf("a = %d", (char)a);
>
>> > how can I have the right display for a?
>
>> printf("a = %hu", a);
>
>That's the best way, in my opinion.

Except that it's not clear what is the
*actual* type expected by %hu: is
it int or unsigned int?

I would expect that %hu would *actually* expect
whatever type it is that unsigned short promotes to,
be it int or be it unsigned.

But the standard doesn't explicitly say so.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[pete]

Dan Pop wrote:

In <3F***********@mindspring.com> pete <pf*****@mindspring.com> writes:

Dan Pop wrote:

In <3F***********@mindspring.com> pete <pf******@mindspring.com> writes:

>Mac wrote:
>>
>> On Mon, 27 Oct 2003 23:22:45 +0000, Conan1st wrote:
>
>> > unsigned short a
>
>> > printf("a = %d", (char)a);
>
>> > how can I have the right display for a?
>
>> printf("a = %hu", a);
>
>That's the best way, in my opinion.

Except that it's not clear what is the
*actual* type expected by %hu: is
it int or unsigned int?

I would expect that %hu would *actually* expect
whatever type it is that unsigned short promotes to,
be it int or be it unsigned.

But the standard doesn't explicitly say so.

I'm starting from the premise that
printf("a = %hu", a);
is defined behavior for the case in question,
and from there I conclude that
something that would cause undefined behavior
(like printf not getting what it's expecting),
isn't happening.

--
pete

[Dan Pop]

In <3F***********@mindspring.com> pete <pf*****@mindspring.com> writes:

Dan Pop wrote:

In <3F***********@mindspring.com> pete <pf*****@mindspring.com> writes:

>Dan Pop wrote:
>>
>> In <3F***********@mindspring.com> pete <pf******@mindspring.com> writes:
>>
>> >Mac wrote:
>> >>
>> >> On Mon, 27 Oct 2003 23:22:45 +0000, Conan1st wrote:
>> >
>> >> > unsigned short a
>> >
>> >> > printf("a = %d", (char)a);
>> >
>> >> > how can I have the right display for a?
>> >
>> >> printf("a = %hu", a);
>> >
>> >That's the best way, in my opinion.
>>
>> Except that it's not clear what is the
>> *actual* type expected by %hu: is
>> it int or unsigned int?
>
>I would expect that %hu would *actually* expect
>whatever type it is that unsigned short promotes to,
>be it int or be it unsigned.

But the standard doesn't explicitly say so.

I'm starting from the premise that
printf("a = %hu", a);
is defined behavior for the case in question,

Where does the standard say that?

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[pete]

Dan Pop wrote:

In <3F***********@mindspring.com> pete <pf*****@mindspring.com> writes:

Dan Pop wrote:

In <3F***********@mindspring.com> pete <pf*****@mindspring.com> writes:

>Dan Pop wrote:
>>
>> In <3F***********@mindspring.com> pete <pf******@mindspring.com> writes:
>>
>> >Mac wrote:
>> >>
>> >> On Mon, 27 Oct 2003 23:22:45 +0000, Conan1st wrote:
>> >
>> >> > unsigned short a
>> >
>> >> > printf("a = %d", (char)a);
>> >
>> >> > how can I have the right display for a?
>> >
>> >> printf("a = %hu", a);
>> >
>> >That's the best way, in my opinion.
>>
>> Except that it's not clear what is the
>> *actual* type expected by %hu: is
>> it int or unsigned int?
>
>I would expect that %hu would *actually* expect
>whatever type it is that unsigned short promotes to,
>be it int or be it unsigned.

But the standard doesn't explicitly say so.

I'm starting from the premise that
printf("a = %hu", a);
is defined behavior for the case in question,

Where does the standard say that?

The standard doesn't actually refer to
an unsigned short variable named "a".

N869
7.19.6.1 The fprintf function
[#6]
h Specifies that a following d, i, o, u, x, or X
conversion specifier applies to a short int or
unsigned short int argument (the argument will
have been promoted according to the integer
promotions, but its value shall be converted to
short int or unsigned short int before
printing); or that a following n conversion
specifier applies to a pointer to a short int
argument.

--
pete

[Dan Pop]

In <3F***********@mindspring.com> pete <pf*****@mindspring.com> writes:

Dan Pop wrote:

In <3F***********@mindspring.com> pete <pf*****@mindspring.com> writes:

>Dan Pop wrote:
>>
>> In <3F***********@mindspring.com> pete <pf*****@mindspring.com> writes:
>>
>> >Dan Pop wrote:
>> >>
>> >> In <3F***********@mindspring.com> pete <pf******@mindspring.com> writes:
>> >>
>> >> >Mac wrote:
>> >> >>
>> >> >> On Mon, 27 Oct 2003 23:22:45 +0000, Conan1st wrote:
>> >> >
>> >> >> > unsigned short a
>> >> >
>> >> >> > printf("a = %d", (char)a);
>> >> >
>> >> >> > how can I have the right display for a?
>> >> >
>> >> >> printf("a = %hu", a);
>> >> >
>> >> >That's the best way, in my opinion.
>> >>
>> >> Except that it's not clear what is the
>> >> *actual* type expected by %hu: is
>> >> it int or unsigned int?
>> >
>> >I would expect that %hu would *actually* expect
>> >whatever type it is that unsigned short promotes to,
>> >be it int or be it unsigned.
>>
>> But the standard doesn't explicitly say so.
>
>I'm starting from the premise that
> printf("a = %hu", a);
>is defined behavior for the case in question,

Where does the standard say that?

The standard doesn't actually refer to
an unsigned short variable named "a".

If you read the standard carefully, you'll see that all conversion
descriptors require a well specified argument type and it's not clear
at all that %h[ouxX] is an exception.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Johnny]

>>a = 321;

printf("a = %d", (char)a);

the display is 'a = 63'...

Are you sure? Unless it is UB (of which I'm not sure), it should output
a = 65

Please forgive my stupidity. But what does UB stand for and mean?

Thanks

[Richard Heathfield]

Johnny wrote:

[...] what does UB stand for and mean?

UB means "undefined behaviour", and refers to code whose result is not laid
down by the C standard. Examples include void main, fflush(stdin),
gets(NULL), and a trillion other examples of bad programming.

--
Richard Heathfield : bi****@eton.powernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton

[Minti]

Johnny <aw****@hotmail.com> wrote in message news:<AM*********************@news-server.bigpond.net.au>...

a = 321;

printf("a = %d", (char)a);

the display is 'a = 63'...

Are you sure? Unless it is UB (of which I'm not sure), it should output
a = 65

Please forgive my stupidity. But what does UB stand for and mean?

UB -- > Undefined Behaviour
Here is an exact definition

http://dspace.dial.pipex.com/town/gr...t/welcome.html

Or on a light note here is a definition from standard
<quote>

undefined behavior
behavior, upon use of a nonportable or erroneous program construct, of
erroneous data, or of indeterminately valued objects, for which this
International Standard imposes no requirements
2 NOTE Possible undefined behavior ranges from ignoring the situation
completely with unpredictable
results, to behaving during translation or program execution in a
documented manner characteristic of the
environment (with or without the issuance of a diagnostic message), to
terminating a translation or
execution (with the issuance of a diagnostic message).

3 EXAMPLE An example of undefined behavior is the behavior on integer
overflow.

</quote>
--
Imanpreet Singh Arora
imanomaniac AT acm DOT org

[Malcolm]

"Johnny" <aw****@hotmail.com> wrote in message

Please forgive my stupidity. But what does UB stand for and mean?

UB is "undefined behaviour", like writing to an uninitialised pointer.

The point is that UB means a program is always incorrect. It may produce
correct output, or what the programmer "means", but it is not correct as a C
program.

It shouldn't be confused with implementation-defined behaviour, which is
correct but not portable.

[Minti]

Richard Heathfield <do******@address.co.uk.invalid> wrote in message news:<bo**********@titan.btinternet.com>...

Johnny wrote:

[...] what does UB stand for and mean?

UB means "undefined behaviour", and refers to code whose result is not laid
down by the C standard. Examples include void main, fflush(stdin),
gets(NULL), and a trillion other examples of bad programming.

Just a curious, irrelevant question, if there are trillion examples of
bad programming, how many examples are there of good programming.
Certainly not a trillion.

--
Imanpreet Singh Arora
isingh AT acm DOT org

[Dan Pop]

In <bo**********@titan.btinternet.com> Richard Heathfield <do******@address.co.uk.invalid> writes:

Johnny wrote:

[...] what does UB stand for and mean?

UB means "undefined behaviour",

How do you know? What is preventing UB from meaning "unspecified
behaviour", which, in the context of the C standard, is a completely
different beast?

Because of this ambiguity, UB should be avoided, unless the context
provides enough information for disambiguation.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Richard Heathfield]

Minti wrote:

Richard Heathfield <do******@address.co.uk.invalid> wrote in message
news:<bo**********@titan.btinternet.com>...

Johnny wrote:

> [...] what does UB stand for and mean?

UB means "undefined behaviour", and refers to code whose result is not
laid down by the C standard. Examples include void main, fflush(stdin),
gets(NULL), and a trillion other examples of bad programming.

Just a curious, irrelevant question, if there are trillion examples of
bad programming, how many examples are there of good programming.
Certainly not a trillion.

It's trivial to write code to generate a trillion examples of good
programming.

Might take up a fair bit of storage, though.
--
Richard Heathfield : bi****@eton.powernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton

[Richard Heathfield]

Dan Pop wrote:

In <bo**********@titan.btinternet.com> Richard Heathfield
<do******@address.co.uk.invalid> writes:

Johnny wrote:

[...] what does UB stand for and mean?
UB means "undefined behaviour",

How do you know? What is preventing UB from meaning "unspecified
behaviour", which, in the context of the C standard, is a completely
different beast?

Nothing, which is why I suggested the alternatives UDB and USB, some time
ago. In message <3D***************@eton.powernet.co.uk> I said: "Others in
this newsgroup call undefined behaviour "UB", but there is nothing to
disambiguate that from "unspecified behaviour", so I think it a poor
abbreviation. UDB and USB, *within the context of this newsgroup*, are not
ambiguous."
Because of this ambiguity, UB should be avoided, unless the context
provides enough information for disambiguation.

UB should /always/ be avoided.

--
Richard Heathfield : bi****@eton.powernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton

[Dan Pop]

In <bo**********@sparta.btinternet.com> Richard Heathfield <do******@address.co.uk.invalid> writes:

Dan Pop wrote:

In <bo**********@titan.btinternet.com> Richard Heathfield
<do******@address.co.uk.invalid> writes:

Johnny wrote:

[...] what does UB stand for and mean?

UB means "undefined behaviour",
How do you know? What is preventing UB from meaning "unspecified
behaviour", which, in the context of the C standard, is a completely
different beast?

Nothing, which is why I suggested the alternatives UDB and USB, some time
ago.

Bad idea, since USB already has a well defined meaning and because we
can't realistically expect people who have just joined the newsgroup
(regardless of their experience as C programmers) to be able to figure
out the meaning of such abbreviations.
In message <3D***************@eton.powernet.co.uk> I said: "Others in
this newsgroup call undefined behaviour "UB", but there is nothing to
disambiguate that from "unspecified behaviour", so I think it a poor
abbreviation. UDB and USB, *within the context of this newsgroup*, are not
ambiguous."

Imagine that right after you joined c.l.c you saw UDBs and USBs all over
the place. Would you have been able to figure out their meaning?

Even if one doesn't fully understand the meanings of "unspecified
behaviour" and "undefined behaviour" (and, especially, the difference
between them), at least one can figure out what the poster is talking
about.

So, if we don't want to turn c.l.c into a closed club for the regulars,
there is no place for such abbreviations here.

Because of this ambiguity, UB should be avoided, unless the context
provides enough information for disambiguation.

UB should /always/ be avoided.

OK, I'll bite: how do you avoid the unspecified behaviour in

printf("Hello world\n");

?

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Minti]

Richard Heathfield <do******@address.co.uk.invalid> wrote in message news:<bo**********@sparta.btinternet.com>...

Minti wrote:

Richard Heathfield <do******@address.co.uk.invalid> wrote in message
news:<bo**********@titan.btinternet.com>...

Johnny wrote:

> [...] what does UB stand for and mean?

UB means "undefined behaviour", and refers to code whose result is not
laid down by the C standard. Examples include void main, fflush(stdin),
gets(NULL), and a trillion other examples of bad programming.

Just a curious, irrelevant question, if there are trillion examples of
bad programming, how many examples are there of good programming.
Certainly not a trillion.

It's trivial to write code to generate a trillion examples of good
programming.

Might take up a fair bit of storage, though.

Maybe it could scan all the USENET c.l.c and get a trillion examples
of good programming.
Might have to filter out fair bit of bad programs though.

--
Imanpreet Singh Arora