Thanks Floh, I was hoping to find the equivalent of 
Windows's _aligned_msize which returns the exact allocated memory but I 
guess I can live with malloc_usable_size for now.

On Thursday, August 6, 2015 at 5:59:46 AM UTC+8, Floh wrote:
>
> AFAIK there is no standard way to query the size you passed in to malloc() 
> by the returned pointer, even malloc_usable_size is not standard.
>
> malloc_usable_size() may return a larger size, but I definitely wouldn't 
> count on that this is always 4 (see here: 
> http://man7.org/linux/man-pages/man3/malloc_usable_size.3.html).
>
> Some platforms or allocator libs have an msize() or _msize() function, but 
> I haven't found this in the dlmalloc sources.
>
> If you absolutely need to store the size passed to alloc inside a memory 
> block then I think you need to implement this on your own (allocate 
> additional space for a header, store size in header, and return pointer 
> after header). This will be dangerous though if you cannot guarantee that 
> all alloc/free calls are properly paired and go through your own wrapper 
> functions.
>
> Cheers,
> -Floh. 
>
> Am Mittwoch, 5. August 2015 12:29:03 UTC+2 schrieb awt:
>>
>> Hi,
>>
>> I need to retrieve the size of a buffer that is dynamically allocated and 
>> this is what I did:
>>
>>
>> size_t size = 64;
>> size_t alignment = 8;
>> memalign(alignment, size);
>>
>> size_t measuredSize = malloc_usable_size(test);
>> std::cout << "measuredSize:" << measuredSize << " size:" << size << 
>> std::endl;
>>
>> However, the size returned by malloc_usable_size is always 4 more than 
>> the amount that I pass in to memalign. In this case, measuredSize will be 
>> 68. I tried using malloc and the result is the same.
>>
>> Is there a way to get the size that I pass into memalign or can I always 
>> assume that malloc_usable_size will always report a higher size of 4 bytes?
>>
>

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