llvm dev - Jun 2009 - [LLVMdev] [cfe-dev] LLVM frontend supporting arbitrary bit-width integral datatypes

Duncan Sands wrote:
> Hi Adam,
>
>   
>> One problem, I was trying to solve was, that I need to declare
variables of let's say 5-bit width like  'i5 var',
>> the maximal bit-width may be limited to 64 bits. I need such variables
to represent instruction's operands,
>> example is at the end this message.
>>     
>
> any standard compliant C compiler supports i5, believe it or not.
>
> Try this:
>
> #include <stdio.h>
>
> struct i3 { int i:3; };
>
> int main(void) {
>    struct i3 A, B, C;
>
>    A.i = 2;
>    B.i = 3;
>    C.i = A.i + B.i;
>    printf("%d + %d = %d\n", A.i, B.i, C.i);
>    return 0;
> }
>
> I get:
> 2 + 3 = -3
> which is correct for i3 arithmetic!
> There might be some problems with signed vs unsigned bitfields,
> depending on the compiler.
>   
You're producing a signed overflow, which is simply undefined behavior.
Unsigned overflow is well-defined, even for bitfields, but signed is not.

Sebastian

Hi Sebastian,
> You're producing a signed overflow, which is simply undefined behavior.
> Unsigned overflow is well-defined, even for bitfields, but signed is not.
that's true, here's an unsigned version (outputs 5 + 5 = 2):

#include <stdio.h>

struct i3 { unsigned i:3; };

int main(void) {
   struct i3 A, B, C;

   A.i = 5;
   B.i = 5;
   C.i = A.i + B.i;
   printf("%d + %d = %d\n", A.i, B.i, C.i);
   return 0;
}
Ciao,

Duncan.

On Thu, Jun 25, 2009 at 7:16 AM, Duncan Sands<baldrick at free.fr>
wrote:
> Hi Sebastian,
>
>> You're producing a signed overflow, which is simply undefined
behavior.
>> Unsigned overflow is well-defined, even for bitfields, but signed is
not.
>
> that's true, here's an unsigned version (outputs 5 + 5 = 2):
>
> #include <stdio.h>
>
> struct i3 { unsigned i:3; };
>
> int main(void) {
>   struct i3 A, B, C;
>
>   A.i = 5;
>   B.i = 5;
>   C.i = A.i + B.i;
>   printf("%d + %d = %d\n", A.i, B.i, C.i);
>   return 0;
> }
Note that you're actually doing signed arithmetic here; it just
doesn't happen to matter because "int" is much larger than the
bitfield.

-Eli