[MIPS]clocks_calc_mult_shift() may gen a too big mult value

From: Chen Jie
Date: Mon Oct 31 2011 - 05:01:37 EST

Hi all,

On MIPS, with maxsec=4, clocks_calc_mult_shift() may generate a very
big mult, which may easily cause timekeeper.mult overflow within
timekeeping jobs.

e.g. when clock freq was 250000500(i.e. mips_hpt_frequency=250000500,
and the CPU Freq will be 250000500*2=500001000), mult will be

Attachment is a script that calculates mult values for CPU Freq
between 400050000 and 500050000, with 1KHz step. It outputs mult
values greater than 0xf0000000:
CPU Freq:500001000, mult:0xffffde72, shift:30
CPU Freq:500002000, mult:0xffffbce4, shift:30
CPU Freq:500003000, mult:0xffff9b56, shift:30
CPU Freq:500004000, mult:0xffff79c9, shift:30

The peak value appears around CPU_freq=500001000.

To avoid this, it may need:
1. Supply a bigger maxsec value?
2. In clocks_calc_mult_shift(), pick next mult/shift pair if mult is
too big? Then maxsec will not be strictly obeyed.
3. Change timekeeper.mult to u64?
4. ...

Any idea?

- Chen Jie
#!/bin/env python

def clocks_calc_mult_shift(from_, to_, maxsec):
sftacc = 32;

tmp = maxsec * from_ >> 32;
while tmp:
tmp >>= 1
sftacc -= 1

for sft in xrange(32, 0, -1):
tmp = to_ << sft;
tmp += (from_ / 2)
tmp /= from_
if ((tmp >> sftacc) == 0):

mult = tmp
shift = sft
return mult, shift

for i in xrange(400050000, 500050000, 1000):
mult, shift = clocks_calc_mult_shift(i/2, 1000000000, 4)
if mult > 0xf0000000:
print "CPU Freq:%d, mult:0x%x, shift:%d" % (i, mult, shift)