lkml.org 
[lkml]   [2018]   [Oct]   [27]   [last100]   RSS Feed
Views: [wrap][no wrap]   [headers]  [forward] 
 
Messages in this thread
/
Date
From
SubjectRe: [Outreachy kernel] [RESEND PATCH 2/2] staging: vboxvideo: Use unsigned int instead bool
On Fri, Oct 26, 2018 at 04:04:45PM -0300, Shayenne da Luz Moura wrote:
>This change was suggested by checkpath.pl. Use unsigned int with bitfield
>allocate only one bit to the boolean variable.
>
>CHECK: Avoid using bool structure members because of possible alignment
>issues
>
>Signed-off-by: Shayenne da Luz Moura <shayenneluzmoura@gmail.com>
>---
> drivers/staging/vboxvideo/vbox_drv.h | 14 +++++++-------
> drivers/staging/vboxvideo/vboxvideo_guest.h | 2 +-
> 2 files changed, 8 insertions(+), 8 deletions(-)
>
>diff --git a/drivers/staging/vboxvideo/vbox_drv.h b/drivers/staging/vboxvideo/vbox_drv.h
>index 594f84272957..7d3e329a6b1c 100644
>--- a/drivers/staging/vboxvideo/vbox_drv.h
>+++ b/drivers/staging/vboxvideo/vbox_drv.h
>@@ -81,7 +81,7 @@ struct vbox_private {
> u8 __iomem *vbva_buffers;
> struct gen_pool *guest_pool;
> struct vbva_buf_ctx *vbva_info;
>- bool any_pitch;
>+ unsigned int any_pitch:1;
> u32 num_crtcs;
> /** Amount of available VRAM, including space used for buffers. */
> u32 full_vram_size;

Using bitfields for booleans in these cases is less efficient than just
using "regular" booleans for two reasons:

1. It will use the same amount of space. Due to alignment requirements,
the compiler can't squeeze in anything into the 7 bits that are now
"free". Each member, unless it's another bitfield, must start at a whole
byte.

2. This is actually less efficient (slower) for the compiler to work
with. The smallest granularity we have to access memory is 1 byte; we
can't set individual bits directly in memory. For the original code, the
assembly for 'vbox_private.any_pitch = true' would look something like
this:

movl $0x1,-0x10(%rsp)

As you can see, the compiler can directly write into the variable.
However, when we switch to using bitfields, the compiler must preserve
the original value of the other 7 bits, so it must first read them from
memory, manipulate the value and write it back. The assembly would
look something like this:

movzbl -0x10(%rsp),%eax
or $0x1,%eax
mov %al,-0x10(%rsp)

Which is less efficient than what was previously happening.

--
Thanks,
Sasha

\
 
 \ /
  Last update: 2018-10-26 22:43    [W:0.114 / U:0.608 seconds]
©2003-2020 Jasper Spaans|hosted at Digital Ocean and TransIP|Read the blog|Advertise on this site