64 bit assignment on a 32 bit platform

64 bit assignment on a 32 bit platform

[kum]

Hi,

UINT64 a64; // UINT64 has been typedefed as long long unsigned int
UINT32 a32, b32;
a32 = x; // some value
b32 = y; // some value

a64 = a32 * b32;
a64 += a32 + b32;

Is it necessary to type-cast both a32 and b32 to yield correct 64 bit
results? Without casting, I find that addition (even if the result
overflows) works while the multiplication does not. Is there a
compiler option to make this work without casting? I am using gcc
4.1.1 on an xscale platform.

Thank you all,
kumaresh

Re: 64 bit assignment on a 32 bit platform

[Andrew Haley]

kum writes:
 > Hi,
 > 
 > UINT64 a64; // UINT64 has been typedefed as long long unsigned int
 > UINT32 a32, b32;
 > a32 = x; // some value
 > b32 = y; // some value
 > 
 > a64 = a32 * b32;
 > a64 += a32 + b32;
 > 
 > Is it necessary to type-cast both a32 and b32 to yield correct 64 bit
 > results? Without casting, I find that addition (even if the result
 > overflows) works while the multiplication does not. Is there a
 > compiler option to make this work without casting? I am using gcc
 > 4.1.1 on an xscale platform.

You have to cast at leat one of the operands in a binary operation to
get the widening to anything wider than int.

Andrew.

Re: 64 bit assignment on a 32 bit platform

[kum]

Thanks Andrew

Andrew>You have to cast at least one of the operands in a binary operation to
Andrew> get the widening to anything wider than int.
Is this true if int is 16 bits wide (i.e. 16 bit machine) and the
l-value is 32 bits wide?

Thanks,
kumaresh

Re: 64 bit assignment on a 32 bit platform

[Andrew Haley]

kum writes:
 > Thanks Andrew
 > 
 > Andrew>You have to cast at least one of the operands in a binary operation to
 > Andrew> get the widening to anything wider than int.

 > Is this true if int is 16 bits wide (i.e. 16 bit machine) and the
 > l-value is 32 bits wide?

It's true everywhere, all the time.

Andrew.

Re: 64 bit assignment on a 32 bit platform

[kum]

Thanks again.

On 10/29/07, Andrew Haley <aph-gcc@littlepinkcloud.com> wrote:
> kum writes:
>  > Thanks Andrew
>  >
>  > Andrew>You have to cast at least one of the operands in a binary operation to
>  > Andrew> get the widening to anything wider than int.
>
>  > Is this true if int is 16 bits wide (i.e. 16 bit machine) and the
>  > l-value is 32 bits wide?
>
> It's true everywhere, all the time.
>
> Andrew.
>

Re: 64 bit assignment on a 32 bit platform

[Rask Ingemann Lambertsen]

On Mon, Oct 29, 2007 at 03:37:04PM +0530, kum wrote:
> Hi,
> 
> UINT64 a64; // UINT64 has been typedefed as long long unsigned int
> UINT32 a32, b32;
> a32 = x; // some value
> b32 = y; // some value
> 
> a64 = a32 * b32;
> a64 += a32 + b32;
> 
> Is it necessary to type-cast both a32 and b32 to yield correct 64 bit
> results?

"a64 = a32 * b32" multiplies two 32-bit numbers to give a 32-bit result,
which is then zero-extended to 64 bits. If you want an expanding
multiplication, use "a64 = (UINT64) a32 * (UINT64) b32".

> Without casting, I find that addition (even if the result
> overflows) works while the multiplication does not.

   Which of the two additions are you talking about? Could you perhaps post
some assembly (-O2 -S -dp) of this code?

#include <stdint.h>

uint64_t a;

void addtest1 (uint32_t b, uint32_t c)
{
  a += b + c;
}

> Is there a
> compiler option to make this work without casting? I am using gcc
> 4.1.1 on an xscale platform.

   It would appear to work the way it is supposed to and I don't know of any
option to change that.

-- 
Rask Ingemann Lambertsen
Danish law requires addresses in e-mail to be logged and stored for a year

Re: 64 bit assignment on a 32 bit platform

[kum]

> Without casting, I find that addition (even if the result
> overflows) works while the multiplication does not.
Sorry, addition behaves the same way as multiplication. Thanks for your help.

Thanks all,
kumaresh