tile kgdb: fix bug in copy to gdb regs, and optimize memset

David Binderman pointed out that we were doing a full memset()
of the gdb register buffer and then doing a memcpy() to it that
was almost as big.  This commit optimizes that by only doing a
memset() of the registers that are intended to be zero.

While making this change I noticed that we were not copying the
link register (LR, number 55) due to a fencepost error in commit
f419e6f63c ("arch: tile: kernel: kgdb.c: Use memcpy() instead of
pointer copy one by one"), and I've corrected that as well.

Reported-by: David Binderman <dcb314@hotmail.com>
Signed-off-by: Chris Metcalf <cmetcalf@ezchip.com>
This commit is contained in:
Chris Metcalf 2016-01-25 15:05:34 -05:00
parent 81f70ba233
commit 77ef8f5177

View file

@ -126,15 +126,15 @@ void
sleeping_thread_to_gdb_regs(unsigned long *gdb_regs, struct task_struct *task)
{
struct pt_regs *thread_regs;
const int NGPRS = TREG_LAST_GPR + 1;
if (task == NULL)
return;
/* Initialize to zero. */
memset(gdb_regs, 0, NUMREGBYTES);
thread_regs = task_pt_regs(task);
memcpy(gdb_regs, thread_regs, TREG_LAST_GPR * sizeof(unsigned long));
memcpy(gdb_regs, thread_regs, NGPRS * sizeof(unsigned long));
memset(&gdb_regs[NGPRS], 0,
(TILEGX_PC_REGNUM - NGPRS) * sizeof(unsigned long));
gdb_regs[TILEGX_PC_REGNUM] = thread_regs->pc;
gdb_regs[TILEGX_FAULTNUM_REGNUM] = thread_regs->faultnum;
}