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tighten up the spinlock loop, and some other bikeshedding

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Former-commit-id: 8bd56fadd6e73794415e1f9eae892c772800e559
This commit is contained in:
John Sully 2019-03-01 13:29:21 -05:00
parent 2cb187ce0a
commit c8e0070fd4
1 changed files with 53 additions and 54 deletions
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107
src/fastlock_x64.asm
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@ -19,50 +19,45 @@ fastlock_lock:
; int32_t m_depth
; First get our TID and put it in ecx
push rdi ; we need our struct pointer (also balance the stack for the call)
call gettid ; get our thread ID (TLS is nasty in ASM so don't bother inlining)
mov esi, eax ; back it up in esi
mov rdi, [rsp] ; get our pointer back
push rdi ; we need our struct pointer (also balance the stack for the call)
call gettid ; get our thread ID (TLS is nasty in ASM so don't bother inlining)
mov esi, eax ; back it up in esi
mov rdi, [rsp] ; get our pointer back
cmp [rdi+4], esi ; Is the TID we got back the owner of the lock?
je .LRecursive ; Don't spin in that case
cmp [rdi+4], esi ; Is the TID we got back the owner of the lock?
je .LLocked ; Don't spin in that case
xor eax, eax ; eliminate partial register dependency
mov ax, 1 ; we want to add one
lock xadd [rdi+2], ax ; do the xadd, ax contains the value before the addition
xor eax, eax ; eliminate partial register dependency
inc eax ; we want to add one
lock xadd [rdi+2], ax ; do the xadd, ax contains the value before the addition
; eax now contains the ticket
xor ecx, ecx
ALIGN 16
.Loop:
cmp [rdi], ax ; is our ticket up?
je .LDone ; leave the loop
add ecx, 1000h ; Have we been waiting a long time? (oflow if we have)
; 1000h is set so we overflow on the 1024*1024'th iteration (like the C code)
jc .LYield ; If so, give up our timeslice to someone who's doing real work
pause ; be nice to other hyperthreads
jmp .Loop ; maybe next time we'll get our turn
.LDone:
mov [rdi+4], esi ; lock->m_pidOwner = gettid()
mov dword [rdi+8], 1 ; lock->m_depth = 1
add rsp, 8 ; fix stack
ret
.LYield:
.LLoop:
cmp [rdi], ax ; is our ticket up?
je .LLocked ; leave the loop
pause
add ecx, 1000h ; Have we been waiting a long time? (oflow if we have)
; 1000h is set so we overflow on the 1024*1024'th iteration (like the C code)
jnc .LLoop ; If so, give up our timeslice to someone who's doing real work
; Like the compiler, you're probably thinking: "Hey! I should take these pushs out of the loop"
; But the compiler doesn't know that we rarely hit this, and when we do we know the lock is
; taking a long time to be released anyways. We optimize for the common case of short
; lock intervals. That's why we're using a spinlock in the first place
push rsi
push rax
mov rax, 24 ; sys_sched_yield
syscall ; give up our timeslice we'll be here a while
mov rax, 24 ; sys_sched_yield
syscall ; give up our timeslice we'll be here a while
pop rax
pop rsi
mov rdi, [rsp] ; our struct pointer is on the stack already
xor ecx, ecx ; Reset our loop counter
jmp .Loop ; Get back in the game
.LRecursive:
add dword [rdi+8], 1 ; increment the depth counter
add rsp, 8 ; fix the stack
mov rdi, [rsp] ; our struct pointer is on the stack already
xor ecx, ecx ; Reset our loop counter
jmp .LLoop ; Get back in the game
ALIGN 16
.LLocked:
mov [rdi+4], esi ; lock->m_pidOwner = gettid()
inc dword [rdi+8] ; lock->m_depth++
add rsp, 8 ; fix stack
ret
ALIGN 16
@ -75,32 +70,36 @@ fastlock_trylock:
; int32_t m_depth
; First get our TID and put it in ecx
push rdi ; we need our struct pointer (also balance the stack for the call)
call gettid ; get our thread ID (TLS is nasty in ASM so don't bother inlining)
mov esi, eax ; back it up in esi
pop rdi ; get our pointer back
push rdi ; we need our struct pointer (also balance the stack for the call)
call gettid ; get our thread ID (TLS is nasty in ASM so don't bother inlining)
mov esi, eax ; back it up in esi
pop rdi ; get our pointer back
cmp [rdi+4], esi ; Is the TID we got back the owner of the lock?
je .LRecursive ; Don't spin in that case
cmp [rdi+4], esi ; Is the TID we got back the owner of the lock?
je .LRecursive ; Don't spin in that case
mov eax, [rdi] ; get both active and avail counters
mov ecx, eax ; duplicate in ecx
ror ecx, 16 ; swap upper and lower 16-bits
cmp eax, ecx ; are the upper and lower 16-bits the same?
jnz .LAlreadyLocked ; If not return failure
mov eax, [rdi] ; get both active and avail counters
mov ecx, eax ; duplicate in ecx
ror ecx, 16 ; swap upper and lower 16-bits
cmp eax, ecx ; are the upper and lower 16-bits the same?
jnz .LAlreadyLocked ; If not return failure
; at this point we know eax+ecx have [avail][active] and they are both the same
add ecx, 10000h ; increment avail, ecx is now our wanted value
lock cmpxchg [rdi], ecx ; If rdi still contains the value in eax, put in ecx (inc avail)
jnz .LAlreadyLocked ; If Z is not set then someone locked it while we were preparing
mov eax, 1 ; return SUCCESS!
mov [rdi+4], esi ; lock->m_pidOwner = gettid()
mov dword [rdi+8], eax ; lock->m_depth = 1
ret
.LAlreadyLocked:
xor eax, eax ; return 0 for failure
add ecx, 10000h ; increment avail, ecx is now our wanted value
lock cmpxchg [rdi], ecx ; If rdi still contains the value in eax, put in ecx (inc avail)
jnz .LAlreadyLocked ; If Z is not set then someone locked it while we were preparing
xor eax, eax
inc eax ; return SUCCESS! (eax=1)
mov [rdi+4], esi ; lock->m_pidOwner = gettid()
mov dword [rdi+8], eax ; lock->m_depth = 1
ret
ALIGN 16
.LRecursive:
add dword [rdi+8], 1 ; increment the depth counter
mov eax, 1 ; we successfully got the lock
ret
xor eax, eax
inc eax ; return SUCCESS! (eax=1)
inc dword [rdi+8] ; lock->m_depth++
ret
ALIGN 16
.LAlreadyLocked:
xor eax, eax ; return 0;
ret