On Fri, Jul 06, 2007 at 06:06:43AM -0600, Wachdorf, Daniel R wrote:
> I am seeing an odd behavior and was wondering if it was a bug. In
> X86_64, Gcc is sign extending a bit shift operation when assigned to an
> unsigned long. Specifically:
>
> #include <stdio.h>
>
> int main(){
>
> unsigned long val;
>
> val = (1 << 31);
>
> printf("Val = %lx\n", val);
>
> return 0;
> }
>
> This results in:
>
> Val = 0xffffffff80000000;
>
> Should the result be 0x80000000? I understand that the bit shift is a
> 32 bit operation, but shouldn't the compiler then up convert that to a
> 64 bit unsigned long?
'1' has signed int type (which on x86_64 is 32-bits), so (1 << 31) is a signed
int type and is negative. If you use 1LU instead of 1, then it is an unsigned
64-bit type, and does not suffer from sign extension:
val = (1LU << 31);
--
Michael Meissner, AMD
90 Central Street, MS 83-29, Boxborough, MA, 01719, USA
[EMAIL PROTECTED]
