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*********@db forums.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*********@db forums.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*********@db forums.com> wrote in message
news:35******** ********@dbforu ms.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.c om> Conan1st <me*********@db forums.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.stanfor d.edu> writes:

Conan1st <me*********@db forums.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*********@db forums.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