Rogier Wolff wrote:
> On the other hand, having thousands "silent departures from ANSI-C"
> scattered through the kernel is not a good idea. I'm fine with having
> a few __asm__ thingies that generate errors while compiling/linking.
>
> Someone is eventually going to have to port Linux to an architecture
> that doesn't have a gcc port yet.
It breaks with GCC too. How obvious do I have to be?
#include <stdio.h>
int main ()
{
unsigned short a = 0xffff;
a = (a << 1) >> 1;
printf ("a = 0x%x\n", (unsigned) a);
return 0;
}
$ gcc -o test test.c
$ ./test
a = 0xffff
--> high bit /not/ cleared
> Oh, by the way, ANSI defines the behaviour as undefined, as a compiler
> is allowed to recognize that (var <<1) >> 1 evaluates to "var" if done
> with arbitrary precision.
No. For any unsigned type as wide or wider than `unsigned int', that
will clear the top bits. For narrower unsigned types it won't. For
signed types it is implementation defined.
> for example
>
> array [index>>4].firstbyte = 0;
>
> could evaluate (the element size of array happens to be 16 bytes!)
>
> * (char *)array + (index & ~0xf) = 0;
>
> See, the compiler recognized that two shifts over the same width are
> the same.
Your example is wrong. You've shown `(a >> 4) << 4', which is obviously
equal to `a & ~0xf' for any type :-)
-- Jamie
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This archive was generated by hypermail 2b29 : Sun Jan 23 2000 - 21:00:17 EST