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unsigned char

Does the c standard state that for all c99 implementation char as standard
is equal unsigned char ??? - or do i have to take this into account when
writing my program ???.
Nov 14 '05
10 14432
"Thomas Stegen" <ts*****@cis.st rath.ac.uk> wrote:
Hallvard B Furuseth wrote:
It's not _equal_ to either, though. It's a distinct type, so
char *foo;
unsigned char *bar = foo;
is an error even if char is equivalent to unsigned char.
Note that pointers to unsigned char can point to any byte in
an object. So if foo had been properly initialised the above
would have been legal.


Well, no, it would still be a constraint violation because of the
incompatible types in the initialisation. The type (char *) is
never compatible with the type (unsigned char *), and nor is it
ever compatible with the type (signed char *).

To make it legal you can either use a cast, or an intermediate
variable of type (void *) as conversions to and from (void *)
do not require casts.

Example with cast:
char *foo = 0; /* Properly initialise */
unsigned char *bar = (unsigned char *)foo; /* Convert with cast */

Example without cast:
char *foo = 0; /* Properly initialise */
void *bar = foo; /* Convert to (void *) */
unsigned char *baz = bar; /* Convert from (void *) */
But even
int *p;
int *p2 = p;

is an error since it uses the value of an uninitialised variable.


Though this is not a contraint violation, it is merely undefined
behaviour, so the compiler is not required to detect it and
generate a diagnostic message.

--
Simon.
Nov 14 '05 #11

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