On Wed, 16 Jun 2010 11:11:09 +0200 hugo rivera <[email protected]> wrote:
> Can someone clarify why the program included outputs 'AB000000' (as I
> expect) on 32 bit systems and 'FFFFFFFFAB000000' on 64 bit systems?
> where all those 1's came from? what's the portable way of doing this?
> sorry for newbie questions like this.
>
>
> unsigned long l;
> unsigned char c;
>
> l = 0L;
> c = 0xAB;
> l |= c << 24;
> printf("%lX\n", l);
For use of C on non-plan9 machines I would recommend
downloading the last draft of the C9x standard as a ready
reference. Google for n843.
As per Section 6.5.7 of C9x, both operands of << must be of
"integer type" and the result type is that of the left
operand. Since sizeof c < sizeof(int), it is promoted. Now
6.3.1.2 says "if an int can represent all values of the
original type, the value is converted to an int". So c is
first converted to an int which means c << 24 is an integer
and -ve. Since an int is smaller than a long (in your case)
it is promoted to a long. Changing the |= statement to
l |= (unsigned)c << 24;
should give you what you want.
