Re: Memory alignment

[Raimo Niskanen]

On Fri, Feb 03, 2017 at 04:07:25PM +0100, Boudewijn Dijkstra wrote:
> Op Sat, 28 Jan 2017 06:26:16 +0100 schreef Damian McGuckin  
> :
> > What is the recommended most portable way to force memory alignment for  
> > a datum of any type, assuming one has a pointer say
> >
> > char *x
> >
> > I currently use something like
> >
> > char *xany = aligntonext(x, sizeof(long))
> >
> > where I use my own function 'aligntionext' which is defined below and I  
> > also assume that a 'long' will be the natural word-size of the machine  
> > and that any datum things just needs to align to this boundary. That  
> > said, if the second argument is say 4k, the function will align its  
> > result to a 4k boundary.
> >
> > I was wondering if there is an optimal, better, more acceptable, or more  
> > portable, way.
> >
> 
> Easy and very portable:
> 
> void *
> aligntonext(void *x, size_t size)
> {
>   return (void *)uintptr_t)x + size - 1u) / size) * size);
> }
> 
> Whether it is optimal depends on compiler optimization.

Isn't this stuff macros are made of:

# define aligntonext(ptr, size) \
((void*)uintptr_t)(ptr) + (size) - 1u) / (size)) * (size)))

Or

# define aligntonext(ptr, bits) \
((void*)uintptr_t)(ptr) + (1<<(bits)) - 1u) >> (bits)) << (bits)))

Note that the second argument is evaluated three times in both variants...

-- 

/ Raimo Niskanen, Erlang/OTP, Ericsson AB

Re: Memory alignment

[Boudewijn Dijkstra]

Op Sat, 28 Jan 2017 06:26:16 +0100 schreef Damian McGuckin  
:
What is the recommended most portable way to force memory alignment for  
a datum of any type, assuming one has a pointer say


char *x

I currently use something like

char *xany = aligntonext(x, sizeof(long))

where I use my own function 'aligntionext' which is defined below and I  
also assume that a 'long' will be the natural word-size of the machine  
and that any datum things just needs to align to this boundary. That  
said, if the second argument is say 4k, the function will align its  
result to a 4k boundary.


I was wondering if there is an optimal, better, more acceptable, or more  
portable, way.




Easy and very portable:

void *
aligntonext(void *x, size_t size)
{
return (void *)uintptr_t)x + size - 1u) / size) * size);
}

Whether it is optimal depends on compiler optimization.


--
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Re: Memory alignment

[Damian McGuckin]

On Sat, 28 Jan 2017, Kyoung Jae Seo wrote:


Maybe posix_memalign(3) is API you are looking for.


No. This allocates memory.

I already have the buffer. I am trying to use space within it.

Regards - Damian

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Re: Memory alignment

[Kyoung Jae Seo]

Hi.

On Sat, Jan 28, 2017 at 04:26:16PM +1100, Damian McGuckin wrote:
> What is the recommended most portable way to force memory alignment for a
> datum of any type, assuming one has a pointer say
> 
>   char *x
> 
> I currently use something like
> 
>   char *xany = aligntonext(x, sizeof(long))
> 
> where I use my own function 'aligntionext' which is defined below and I also
> assume that a 'long' will be the natural word-size of the machine and that
> any datum things just needs to align to this boundary. That said, if the
> second argument is say 4k, the function will align its result to a 4k
> boundary.
> 
> I was wondering if there is an optimal, better, more acceptable, or more
> portable, way.

Maybe posix_memalign(3) is API you are looking for.