Re: Align a malloc ptr on a 64bit address

From:
"Chris M. Thomasson" <cristom@charter.net>
Newsgroups:
comp.lang.c++
Date:
Tue, 7 Dec 2010 13:06:24 -0800
Message-ID:
<dlxLo.2313$D86.1065@newsfe11.iad>
"WebShaker" <etienne@tlk.fr> wrote in message
news:4cfd6ec6$0$26306$426a74cc@news.free.fr...

Hi.

I'd like to know if I can be sur that

uint64_t *buffer;
buffer = (uint64_t *)malloc(1024);

will return me a pointer 64 bit aligned.

Else if, how can I do that !!!


Perhaps something like this might be of use to you:

http://pastebin.com/xrjVnkQ2

You can use it like:
____________________________________________________________
#include <cstdio>
#include <align.hpp>

int main()
{
    struct buffer_t
    {
        unsigned char buffer[8192];
    };

    unsigned char raw_buffer[align<buffer_t, 4096>::buf_size];
    buffer_t* const aligned_buffer = align<buffer_t,
4096>::ptr_up(raw_buffer);

    std::printf("raw_buffer == %p\n", (void*)raw_buffer);
    std::printf("aligned_buffer == %p\n", (void*)aligned_buffer);

    return 0;
}
____________________________________________________________

This basically aligned an `buffer_t' object on a 4096 byte boundary. Beware
that the main alignment calculation is non-portable to some "exotic"
architectures...

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