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? > -- You received this message because you are subscribed to the Google Groups "emscripten-discuss" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. For more options, visit https://groups.google.com/d/optout.
