[9fans] portability question

[9fans] portability question

[hugo rivera]

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);


--
Hugo

Re: [9fans] portability question

[Lucio De Re]

On Wed, Jun 16, 2010 at 11:11:09AM +0200, hugo rivera wrote:

>         printf("%lX\n", l);

Would you try %luX?  It may work better?

++L

Re: [9fans] portability question

[hugo rivera]

2010/6/16 Lucio De Re <lucio@proxima.alt.za>:
> On Wed, Jun 16, 2010 at 11:11:09AM +0200, hugo rivera wrote:
>
>>         printf("%lX\n", l);
>
> Would you try %luX?  It may work better?

no, or at least not as I intend. It produces '2868903936X' on 32 bit
linux and '18446744072283488256X' on 64 bit. According to printf(3),
an 'l' should work with both unsigned and signed long quantities.

--
Hugo

Re: [9fans] portability question

[Charles Forsyth]

>Would you try %luX?  It may work better?

he's calling printf, not print

Re: [9fans] portability question

[maht]

On 06/16/10 10:11, hugo rivera 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);
>
>

check the ASM, looks like c is being cast as signed and then sign
extended into a long and then ORed with l.

perhaps this would solve it :

l |= ((unsigned long) c) << 24


But I can't check

Re: [9fans] portability question

[erik quanstrom]

> check the ASM, looks like c is being cast as signed and then sign
> extended into a long and then ORed with l.
>
> perhaps this would solve it :
>
> l |= ((unsigned long) c) << 24

that works.  the extra () are unnecessary.

i think that gcc is using 6.3.1.1 and converting
c to an int (since it fits) so the original expression
is evaulated as
	(int)0xab << (int)24
	-> (int)0xab000000
if there is any sign extension here, it is shifted
away.  then gcc sign-extends this into 8 bytes,
giving the value you see.

- erik

Re: [9fans] portability question

[Bakul Shah]

On Wed, 16 Jun 2010 11:11:09 +0200 hugo rivera <uair00@gmail.com>  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.

Re: [9fans] portability question

[hugo rivera]

Thanks for the feedback.

2010/6/16 Bakul Shah <bakul+plan9@bitblocks.com>:
> On Wed, 16 Jun 2010 11:11:09 +0200 hugo rivera <uair00@gmail.com>  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.
>
>



-- 
Hugo