Re: slab redzoning

[Manfred Spraul]

William Lee Irwin III wrote:

> William Lee Irwin III wrote:
>  
>
> > > It returns a false positive when size + 3*BYTES_PER_WORD == 2**n, e.g.
> > > size == 16373. Here, fls(size - 1) == 13, but fls(size - 1 + 12) == 13
> > > while size - 1 + 12 == 16384, where we'd want the check to fail.
> > >      
>
> On Sat, May 22, 2004 at 10:43:47AM +0200, Manfred Spraul wrote:
>  
>
> > No, 16373 must fail: After adding 12 bytes the object size would be 
> > 16385, which would mean an order==3 allocation.
> > And 16372 must succeed: 16384 is still an order==2 allocation.
> > The idea is that there shouldn't be an allocation order increase due to 
> > redzoning, and afaics that doesn't happen, except between 4082 and 4095 
> > bytes.
> >    
>
> Yes. While you've corrected the one-offs in my post (arithmetic is boring,
> we have machines to do that for us now)
I admit, I'm cheating:
I'd copied the test to user space and then tested all values between 32 
and 131072. 16372 passes:
fls(16371) == fls(16383).

I'll send a patch to Andrew to fix the range between 4084 and 4095.

--
   Manfred
#include <stdio.h>

#define BYTES_PER_WORD	4
#define PAGE_SIZE 4096

/*
 * fls: find last bit set.
 */

static int fls(int x)
{
	int r = 32;

	if (!x)
		return 0;
	if (!(x & 0xffff0000u)) {
		x <<= 16;
		r -= 16;
	}
	if (!(x & 0xff000000u)) {
		x <<= 8;
		r -= 8;
	}
	if (!(x & 0xf0000000u)) {
		x <<= 4;
		r -= 4;
	}
	if (!(x & 0xc0000000u)) {
		x <<= 2;
		r -= 2;
	}
	if (!(x & 0x80000000u)) {
		x <<= 1;
		r -= 1;
	}
	return r;
}

int main(void)
{
	int size;

	for (size=32;size<131073;size++) {
		if ((size <= PAGE_SIZE-3*BYTES_PER_WORD || fls(size-1) == fls(size-1+3*BYTES_PER_WORD))) {
			/* printf("%6d: no order change \n", size); */
		} else {
			printf("%6d: order change \n", size);
		}
	}
	return 0;
}