In linux, dynamic memory can be allocated by kmalloc(). kmalloc() can allocate maximum 128k byte. How can i increase this allocation?
Is kmalloc is slab allocation?
Regards
Learner
Related
My 32-bit Windows program passes a 1K size to CreateFileMappingA, the virtual memory VMMap shows a size 4K with a gap of 0x10000 (64K) between memory maps.
Overall, there are ~30,000 maps resulting in ~2GB (30,000 x 64K) of virtual memory usage.
I was expecting ~120MB (30,000 x 4K) memory usage, not 2GB leading to 'out of memory' issues!
Question
Are there any tips / tricks or flags to reduce the memory footprint?
We are trying to convert captured screen image buffer to intermediate before encoding need 8MB of Heap Memory for that using malloc in vc++ , but it returns null. We have available memory in RAM. Any idea on this ?
I create a slab cache by kmem_cache_create(... size), then allocate memory from this cache by kmem_cache_alloc().
After I have allocated memory for "size" times, what happen if I call kmem_cache_alloc() to allocat size + 1th memory? Return NULL or extend cache implicitly?
The 'size' argument is not about memory reserved for anything. It is about the size of each allocation as returned by kmem_cache_alloc.
It may be there will be memory shortage in general, in which case, depending on flags pased to kmem_cache_alloc, the kernel may try to free some by e.g. shrinking caches.
I understand that in Linux the mm_struct describes the memory layout of a process. I also understand that the start_brk and brk mark the start and end of the heap section of a process respectively.
Now, this is my problem: I have a process, for which I wrote the source code, that allocates 5.25 GB of heap memory using malloc. However, when I examine the process's mm_sruct using a kernel module I find the value of is equal to 135168. And this is different from what I expected: I expected brk - start_brk to be equal slight above 5.25 GB.
So, what is going on here?
Thanks.
I notice the following in the manpage for malloc(3):
Normally, malloc() allocates memory from the heap, and adjusts the size of the heap as required, using sbrk(2). When allocating blocks of memory larger than MMAP_THRESHOLD bytes, the glibc malloc() implementation allocates the memory as a private anonymous mapping using mmap(2). MMAP_THRESHOLD is 128 kB by default, but is adjustable using mallopt(3). Allocations performed using mmap(2) are unaffected by the RLIMIT_DATA resource limit (see getrlimit(2)).
So it sounds like mmap is used instead of the heap.
When using the VirtualAlloc API to allocate and commit a region of virtual memory with a power of two size of the page boundary such as:
void* address = VirtualAlloc(0, 0x10000, MEM_COMMIT, PAGE_READWRITE); // Get 64KB
The address seems to always be in 64KB alignment, not just the page boundary, which in my case is 4KB.
The question is: Is this alignment reliable and prescribed, or is it just coincidence? The docs state that it is guaranteed to be on a page boundary, but does not address the behavior I'm seeing. I ask because I'd later like to take an arbitrary pointer (provided by a pool allocator that uses this chunk) and determine which 64KB chunk it belongs to by something similar to:
void* chunk = (void*)((uintptr_t)ptr & 0xFFFF0000);
The documentation for VirtualAlloc describes the behavior for 2 scenarios: 1) Reserving memory and 2) Committing memory:
If the memory is being reserved, the specified address is rounded down to the nearest multiple of the allocation granularity.
If the memory is already reserved and is being committed, the address is rounded down to the next page boundary.
In other words, memory is allocated (reserved) in multiples of the allocation granularity and committed in multiples of a page size. If you are reserving and committing memory in a single step, it will be be aligned at a multiple of the allocation granularity. When committing already reserved memory it will be aligned at a page boundary.
To query a system's page size and allocation granularity, call GetSystemInfo. The SYSTEM_INFO structure's dwPageSize and dwAllocationGranularity will hold the page size and allocation granularity, respectively.
This is entirely normal. 64KB is the value of SYSTEM_INFO.dwAllocationGranularity. It is a simple counter-measure against address space fragmentation, 4KB pages are too small. The memory manager will still sub-divide 64KB chunks as needed if you change page protection of individual pages within the chunk.
Use HeapAlloc() to sub-allocate. The heap manager has specific counter-measures against fragmentation.