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signed and unsigned char

Given

signed char str_a[]="Hello, world!\n";
unsigned char str_b[]="Hello, world!\n";

what is the difference, if any, between the following two statements?

printf( "%s", str_a );
printf( "%s", str_b );

If there is a difference, what is the best way to compare *str_a with
0xFF? (On my implementation, unadorned char is signed, and so I'm
using

if( *str_a == (signed char)0xFF ) ...

to quiet compiler warnings.)

--
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.
Nov 14 '05 #1
19 4691
Christopher Benson-Manica wrote:

Given

signed char str_a[]="Hello, world!\n";
unsigned char str_b[]="Hello, world!\n";

what is the difference, if any, between the following two statements?

printf( "%s", str_a );
printf( "%s", str_b );

If there is a difference, what is the best way to compare *str_a with
0xFF?
char type arguments are converted to int.
That result is converted to unsigned char for stdio output.
There shouldn't be a problem there, regardless of sign of char.
(On my implementation, unadorned char is signed, and so I'm
using

if( *str_a == (signed char)0xFF ) ...


if( *(unsigned char*)str_a == -1 )

/* assuming 0xff is meant to be all bits set */

--
pete
Nov 14 '05 #2
On Thu, 12 Feb 2004 17:52:55 +0000 (UTC), Christopher Benson-Manica
<at***@nospam.c yberspace.org> wrote:
Given

signed char str_a[]="Hello, world!\n";
unsigned char str_b[]="Hello, world!\n";

what is the difference, if any, between the following two statements?

printf( "%s", str_a );
printf( "%s", str_b );
Other than that they take different arguments? None that I can see.
There are no conversions to worry about, and once the pointer values
land in printf, there's no way for printf to be able tell the
difference anyway (and no reason for it to care).

If there is a difference, what is the best way to compare *str_a with
0xFF? (On my implementation, unadorned char is signed, and so I'm
using

if( *str_a == (signed char)0xFF ) ...

to quiet compiler warnings.)


I'm not sure how the 2nd question is dependent upon the first... What
you've done is clearly showing your intent, which is a Good Thing. One
alternative is
if (*str_a == -1) ...
but I like yours better.


Leor Zolman
BD Software
le**@bdsoft.com
www.bdsoft.com -- On-Site Training in C/C++, Java, Perl & Unix
C++ users: Download BD Software's free STL Error Message
Decryptor at www.bdsoft.com/tools/stlfilt.html
Nov 14 '05 #3

"Christophe r Benson-Manica" <at***@nospam.c yberspace.org> wrote in message
news:c0******** **@chessie.cirr .com...
Given

signed char str_a[]="Hello, world!\n";
unsigned char str_b[]="Hello, world!\n";

what is the difference, if any, between the following two statements?

printf( "%s", str_a );
printf( "%s", str_b );


'^' != (unsigned char)'^' for example.

Extended ascii chars have negative values (since they are higher than 127).
Nov 14 '05 #4
pete <pf*****@mindsp ring.com> spoke thus:
if( *(unsigned char*)str_a == -1 )


1) Is that better than ... (unsigned char)*str_a ... ?
2) This relies on -1's representation being all bits set, yes?

--
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.
Nov 14 '05 #5
pete wrote:

Christopher Benson-Manica wrote:

Given

signed char str_a[]="Hello, world!\n";
unsigned char str_b[]="Hello, world!\n";

what is the difference, if any, between the following two statements?

printf( "%s", str_a );
printf( "%s", str_b );

If there is a difference, what is the best way to compare *str_a with
0xFF?
char type arguments are converted to int.


True (usually), but irrelevant: the arguments are not
any kind of `char', but are pointers.
That result is converted to unsigned char for stdio output.
Either the Standard doesn't say so, or I've overlooked
the spot where it does.
There shouldn't be a problem there, regardless of sign of char.


There isn't a problem, because the "%s" specifier is defined
to work with any of `char*', `unsigned char*', and `signed char*'.
Something of an oddity, really: Most conversion specifiers are
very strict about the type of the corresponding argument, yet
here's one that accepts arguments of three distinct types.
(On my implementation, unadorned char is signed, and so I'm
using

if( *str_a == (signed char)0xFF ) ...


Undefined behavior, I think. You might do better with

if ( (unsigned char)*str == 0xFF )

--
Er*********@sun .com
Nov 14 '05 #6
Christopher Benson-Manica wrote:

pete <pf*****@mindsp ring.com> spoke thus:
if( *(unsigned char*)str_a == -1 )
1) Is that better than ... (unsigned char)*str_a ... ?


No. Now, I like
if( (unsigned char)*str_a == (unsigned char)-1)
or
if( (unsigned char)*str_a == (unsigned char)0xFF)
best.
2) This relies on -1's representation being all bits set, yes?


If str_a points to an all bits set byte,
then *(unsigned char*)str_a will equal ((unsigned char)-1)

then the question becomes, is
((unsigned char)-1) equal to (-1) ?
and it isn't, so I was wrong.

if( (unsigned char)*str_a == (unsigned char)0xFF )

The conversion of out of range values to signed char,
is implementation defined, so I would avoid it.
The conversion of everything to unsigned char, is well defined.

0xff is of type int.

--
pete
Nov 14 '05 #7
Eric Sosman wrote:
pete wrote:
That result is converted to unsigned char for stdio output.


Either the Standard doesn't say so, or I've overlooked
the spot where it does.


I think pete's right. "Byte output" functions, printf included, are
defined in terms of fputc:

The byte output functions write characters to the stream as if by
successive calls to the fputc function. [7.19.3#12]

and the description of fputc says:

The fputc function writes the character specified by c (converted
to an unsigned char) to the output stream pointed to by stream
[...] [7.19.7.3#2]

Jeremy.
Nov 14 '05 #8
On Thu, 12 Feb 2004 18:16:12 GMT, Leor Zolman <le**@bdsoft.co m> wrote:

if( *str_a == (signed char)0xFF ) ...


oops, of course that's UB... Eric caught it.

Leor Zolman
BD Software
le**@bdsoft.com
www.bdsoft.com -- On-Site Training in C/C++, Java, Perl & Unix
C++ users: Download BD Software's free STL Error Message
Decryptor at www.bdsoft.com/tools/stlfilt.html
Nov 14 '05 #9
"Martin Johansen" <ma******@is.on line.no> writes:
"Christophe r Benson-Manica" <at***@nospam.c yberspace.org> wrote in message
news:c0******** **@chessie.cirr .com...
Given

signed char str_a[]="Hello, world!\n";
unsigned char str_b[]="Hello, world!\n";

what is the difference, if any, between the following two statements?

printf( "%s", str_a );
printf( "%s", str_b );


'^' != (unsigned char)'^' for example.

Extended ascii chars have negative values (since they are higher than 127).


<Slightly OT>
Actually the ASCII value of '^' is 94 (unless my newsreader mangled
whatever extended ASCII character you actually wrote).
</Slightly OT>

All the characters in the string literals above are in the "basic
execution character set". C99 6.2.5p3 says:

If a member of the basic execution character set is stored in a
char object, its value is guaranteed to be positive.

In ASCII, all such characters happen to have values in the range
32..126. In EBCDIC, if I recall correctly, some basic characters have
codes greater than 127; I think this implies that in an implementation
that uses EBCDIC as its execution character set, type char must be
unsigned (assuming CHAR_BIT==8).

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://www.sdsc.edu/~kst>
Schroedinger does Shakespeare: "To be *and* not to be"
Nov 14 '05 #10

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