The kernel takes a very long time to boot if the memory size is bigger then
32767 MB. The memory size is contained in a structure created by an sclp
call. The kernel accesses the field with a LH instrution which performs a
sign extension of a 16 bit word. In the case of a memory size with bit 2^15
set this results in a very large value and the memory detection just loops for
a long time. In addition if more then 64 GB are used on a 64 bit system the
memory size is read from an incorrect storage location.
Use zero-extention to read the 16 bit memory size and the correct offset to
read the 4 byte memory size on 64 bit.
Signed-off-by: Martin Schwidefsky <schwidefsky@de.ibm.com>
Signed-off-by: Andrew Morton <akpm@osdl.org>
Signed-off-by: Linus Torvalds <torvalds@osdl.org>
l %r2, .Lrcp2-.LPG1(%r13) # try with Read SCP
b .Lservicecall-.LPG1(%r13)
.Lprocsccb:
- lh %r1,.Lscpincr1-PARMAREA(%r4) # use this one if != 0
- chi %r1,0x00
- jne .Lscnd
- l %r1,.Lscpincr2-PARMAREA(%r4) # otherwise use this one
+ lhi %r1,0
+ icm %r1,3,.Lscpincr1-PARMAREA(%r4) # use this one if != 0
+ jnz .Lscnd
+ l %r1,.Lscpincr2-PARMAREA+4(%r4) # otherwise use this one
.Lscnd:
xr %r3,%r3 # same logic
ic %r3,.Lscpa1-PARMAREA(%r4)
l %r2,.Lrcp2-.LPG1(%r13) # try with Read SCP
b .Lservicecall-.LPG1(%r13)
.Lprocsccb:
- lh %r1,.Lscpincr1-PARMAREA(%r4) # use this one if != 0
- chi %r1,0x00
- jne .Lscnd
+ lghi %r1,0
+ icm %r1,3,.Lscpincr1-PARMAREA(%r4) # use this one if != 0
+ jnz .Lscnd
lg %r1,.Lscpincr2-PARMAREA(%r4) # otherwise use this one
.Lscnd:
xr %r3,%r3 # same logic