Re: malloc() alignment on 32 bit

[Ahmad Fatoum <a.fatoum@pengutronix.de>]

On 19.09.22 13:37, Enrico Scholz wrote:
> Hello,
> 
> on an iMX6ull I stumpled across
> 
> | zstd_decomp_init:536 workspace=8ff1a004+161320
> | ERROR: initcall ubifs_init+0x1/0xc4 failed: Invalid argument
> 
> which is caused by
> 
> | static int zstd_decomp_init(void)
> |	void *wksp = malloc(wksp_size);
> | ...
> | ZSTD_DCtx* ZSTD_initStaticDCtx(void *workspace, size_t workspaceSize)
> |    if ((size_t)workspace & 7) return NULL;  /* 8-aligned */
> 
> 
> Trivial fix would be 'memalign(8, wksp_size)', but is it really ok that
> malloc() for 32 bit has only an alignment of 4?
> 
> Relevant code seems to be in common/tlsf.c
> 
> | enum tlsf_private
> | {
> | #if defined (TLSF_64BIT)
> | 	/* All allocation sizes and addresses are aligned to 8 bytes. */
> | 	ALIGN_SIZE_LOG2 = 3,
> | #else
> | 	/* All allocation sizes and addresses are aligned to 4 bytes. */
> | 	ALIGN_SIZE_LOG2 = 2,
> | #endif
> 
> 'ldrd/strd' require 8 byte alignment which might break with such
> alignment.

I recently learnt too that on 32-bit TLSF only has 4 byte alignment.
I also think that this is too low. 8 byte alignment sounds good IMO.

Cheers,
Ahmad

> 
> 
> Enrico
> 
> 


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