a volatile related bug in kernel/timer.c ?

From: KwongYuan Wong
Date: Thu May 17 2012 - 11:28:51 EST


Hi All,

currently, I am working on a private mips-like chip, and I came
across the following senario:

in the function "del_timer" in kernel/timer.c, there is the following code

954 if (timer_pending(timer)) {
955 base = lock_timer_base(timer, &flags);
956 if (timer_pending(timer)) {


suppose timer_pending(timer) check in line 954 is A, and in line 956 is B.

because the timer_pending(timer) check is very simple, so the result
may be saved in a register, and that register is reused
by both A and B. While this should be wrong? the check at B should

reload the value from memory instead of using previous
result kept in register, because lock_timer_base may have side-effect
which change the result of time_pending?

so I guess a barrier() is needed, so that the code should be the following?

if (timer_pending(timer)) {
base = lock_timer_base(timer, &flags);
barrier();
if (timer_pending(timer)) {


in my chip, the generated assembly is like the following:
( the function "lock_timer_base" in inlined also)

1017 del_timer:
1018 .set noreorder
1019 .set nomacro
1020
1021 lw $5,0($4)
1022 addu $3,$0,$0
1023 beq $5,$0,.L121 <=== $5 is the value of the

first "timer_pending(timer)"
1024 nop
1025
1026 lw $3,20($4)
1027 addiu $2,$0,-2

1028 and $6,$3,$2
1029 beq $6,$0,.L122
1030 nop
1031
1032 .L125:
1033 .set push ; .set opportunistic
1034 # 69 "include/asm/irqflags.h" 1
1035 __raw_local_irq_save $7
1036 # 0 "" 2
1037 .set pop
1038 addu $3,$0,$0
1039 beq $5,$0,.L124 <=== in the second check,
it's reused,
but
it should not, $5 should be updated from memory?
1040 nop

I am a compiler engineer, a newbie in kernel, please feel free to
point out if there is anything wrong

thanks very much

---
Warmest, regards,
WANG.Jiong
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