Pointer arithmetic

[D'Arcy J.M. Cain]

[I removed comp.lang.c++ from the Newsgroups header as it seems to be 
only a C question]

In article <1991Jan5.001607.5915 at demott.com> Kevin D. Quitt writes:
>const unsigned char *chars   = "some string"
>unsigned char pass[];
>    c   = strchr( chars, toupper( pass[ i ]) ) - chars + 1;
>    gcc would not accept this no matter what I did.

I too have noticed this difference between gcc and Turbo-C.  The problem
seems to be the "unsigned" part of the declaration of chars.  I tried
the following:

#include	<stdio.h>
#include	<string.h>

int		main(void)
{
	const char *chars = "some string";
	int c;

	c = strchr(chars, 'e') - chars + 1;
	return(c);
}

which compiled without a peep with "gcc -O -Wall tst.c -o tst" but
failed when I declared chars as "const unsigned char *chars ...".
(it didn't seem to matter whether or not const was used.)  The error
message is:
tst.c: In function main:
tst.c:6: warning: initialization between incompatible pointer types
tst.c:9: warning: argument passing between incompatible pointer types
tst.c:9: invalid operands to binary -
tst.c:7: warning: `c' may be used uninitialized in this function

So gcc differentiates unsigned char from char.  String literals are
composed of char objects so their addresses can't be assigned to an
unsigned char pointer, strchr() takes a char pointer so an uncasted
unsigned char doesn't match the parameter list and char and unsigned
char can't be subtracted.  So the question is what does ANSI say on
the issue.  K&R2 seems to support gcc in this.  (A2.6, page 194 and
A6.6, page 198.)  In any case, the snippet you provide doesn't exhibit
any reason for using unsigned char over char so perhaps you can just
change the declaration.

-- 
D'Arcy J.M. Cain (darcy at druid)     |
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