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ARM: use MiB for vmalloc sizes

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Rather than using "m" (which is the unit of metres, or milli), and
"MB" in the printk statements, use MiB to make it clear that we are
talking about the power-of-2 megabytes, aka mebibytes.

Reviewed-by: Linus Walleij <linus.walleij@linaro.org>
Reviewed-by: Yanfei Xu <yanfei.xu@windriver.com>
Signed-off-by: Russell King (Oracle) <rmk+kernel@armlinux.org.uk>
This commit is contained in:
Russell King (Oracle) 2021-05-28 11:03:58 +01:00
parent 08b842400f
commit c01914efea
1 changed files with 3 additions and 3 deletions
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6
arch/arm/mm/mmu.c
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@ -1126,7 +1126,7 @@ static unsigned long __initdata vmalloc_size = 240 * SZ_1M;
/*
* vmalloc=size forces the vmalloc area to be exactly 'size'
* bytes. This can be used to increase (or decrease) the vmalloc
* area - the default is 240m.
* area - the default is 240MiB.
*/
static int __init early_vmalloc(char *arg)
{
@ -1135,14 +1135,14 @@ static int __init early_vmalloc(char *arg)
if (vmalloc_reserve < SZ_16M) {
vmalloc_reserve = SZ_16M;
pr_warn("vmalloc area too small, limiting to %luMB\n",
pr_warn("vmalloc area is too small, limiting to %luMiB\n",
vmalloc_reserve >> 20);
}
vmalloc_max = VMALLOC_END - (PAGE_OFFSET + SZ_32M + VMALLOC_OFFSET);
if (vmalloc_reserve > vmalloc_max) {
vmalloc_reserve = vmalloc_max;
pr_warn("vmalloc area is too big, limiting to %luMB\n",
pr_warn("vmalloc area is too big, limiting to %luMiB\n",
vmalloc_reserve >> 20);
}