Is there any limit in setting zram device disksize on Linux? - linux

Question

I'm trying to create a zram device on my target device. My target can not allocate memory if the zram disksize is above 100GB, but it's okay with the disksize of 50GB or less.
Is there any limit in setting zram device disksize on Linux? My target device only has 2GB of RAM memory.

Answer 1

I guess you can give a number up to UINT64_MAX - 4095 = 18446744073709547520 on a 64-bit platform.
https://github.com/torvalds/linux/blob/master/drivers/block/zram/zram_drv.h#L101
https://github.com/torvalds/linux/blob/master/drivers/block/zram/zram_drv.c#L1506
https://github.com/torvalds/linux/blob/master/drivers/block/zram/zram_drv.c#L901
So what we have:
... disksize_store(...) {
u64 disksize;
...
// ok, we can give at least UINT64_MAX here.
disksize = unsigned long long memparse(...);
// PAGE_ALIGN, PAGE_SIZE = 1<<12
disksize = PAGE_ALIGN(disksize)
= (((disksize)+((PAGE_SIZE)-1))&(~((typeof(disksize))(PAGE_SIZE)-1)))
= (disksize + ((1<<12)-1))&(~((1<<12)-1))
= (disksize + 4095) & 0xfffffffffffff000
// ^^^^^^^^^^^^^^^ this can overflow
// so max number is UINT64_MAX - 4095 so it doesn't overflow
// otherwise this macro will return 0
...
if (!zram_meta_alloc(..., disksize) {
...
return ...;
}
...
zram->disksize = disksize;
...
}
So let's see into zram_meta_alloc:
... zram_meta_alloc(..., disksize) {
...
num_pages = disksize >> PAGE_SHIFT;
// max num_pages = 0xfffffffffffff = UINT64_MAX >> PAGE_SHIFT
... = vzalloc(num_pages * sizeof(*zram->table));
// ^^^^^^^^^^^^^^^ this can overflow
...
}
vzallloc takes as argument unsigned long. ULONG_MAX should be UINT64_MAX on 64-bit platform. sizeof(*zram->table) is equal to sizeof(unsigned long) + sizeof(unsigned long) + [optional: + sizeof(ktime_t)] + padding (see here). Without padding, assuming 64-bit platform, sizeof(unsigned long) = 8 that should be equal to 8+8[+8] = 16 or 24. But anyway, maximum num_pages is equal to UINT64_MAX >> 12, so to overflow it on 64bit multiplication we would need sizeof(*zram->table) = 2^PAGE_SIZE = 4096, and that shouldn't happen (unless the compiler decides to give over 4000 bytes of padding into the zram->table struct). So we are left with UINT64_MAX - 4095.
So we are left, that the maximum number of disksize is UINT64_MAX-4095. If you give the disksize equal to UINT64_MAX - x, where 0 <= x < 4095, than because of PAGE_ALIGN macro, the disksize will be effectively set to 0. Probably this should be brought up to a kernel developer and they should modify the PAGE_ALIGN macro to support such numbers.
6 days ago to vzalloc calls the call to array_size was added to protect against overflow with this commit.

Answer 2

There is no limit but there is an overhead.
"Note that zram uses about 0.1% of the size of the disk when not in use so a huge zram is wasteful."
https://www.kernel.org/doc/Documentation/blockdev/zram.txt
Also disk_size is a virtual size purely dependent on the input and the compression ratio that receives via chosen alg. Disk-size is the max uncompressed size and general disk parameters.
The only 'actual' control is via mem_limit which is compressed size + disk & zram overheads.
Compression ratio is completely dependent on comp alg chosen from /proc/crypto as zlib & zstd are far more effective but are far slower. It is also very dependent on input as with text zlib & zstd can be over double that what lzo & lz4 will achieve.
If the input is already compressed any alg might garner little to zero compression and without a mem_limit could grab much precious memory from the system.
Mem_limit is the max you are prepared zram to grab from system and a disk-size any more than the compression ratio expected applied to mem_limit is likely a waste.
It will never get used but be part of the 0.1% empty creation overhead.
Maybe try https://github.com/StuartIanNaylor/zram-config