Re: [PATCH] x86: slightly shorten __ticket_spin_trylock() (v2)

[Linus Torvalds <torvalds@linux-foundation.org>]

On Thu, 3 Dec 2009, Jan Beulich wrote:
>  
> -static __always_inline int __ticket_spin_trylock(raw_spinlock_t *lock)
> +static __always_inline u8 __ticket_spin_trylock(raw_spinlock_t *lock)
>  {
>  	int tmp, new;
>  
> @@ -88,12 +88,11 @@ static __always_inline int __ticket_spin
>  		     LOCK_PREFIX "cmpxchgw %w1,%2\n\t"
>  		     "1:"
>  		     "sete %b1\n\t"
> -		     "movzbl %b1,%0\n\t"
>  		     : "=&a" (tmp), "=&q" (new), "+m" (lock->slock)
>  		     :
>  		     : "memory", "cc");
>  
> -	return tmp;
> +	return new;

This is fairly pessimal register allocation.

It used to be that we returned the value in 'tmp', which is %eax, which is 
also the expected return register. Now that we use 'new', it's some random 
other register that is _not_ %eax, which means that while we avoid a 
'movzbl', the regular spin_trylock function call case will now have the 
compiler emitting a 

	movb %X,%al
	ret

at the end just to get the right return register.

Which seems a bit annoying. It looks like we could have done that last 
instruction as just

	sete %b0

instead, and then return 'tmp' instead of 'new', keeping the return value 
in %al and avoiding the unnecessary movement.

> -static __always_inline int __ticket_spin_trylock(raw_spinlock_t *lock)
> +static __always_inline u8 __ticket_spin_trylock(raw_spinlock_t *lock)

Same thing here, afaik.

NOTE! I haven't actually looked at the generated code, and if we actually 
inline it all the way to the caller, it won't matter (and picking another 
register may even help). But while these helpers are marked 
__always_inline, I _thought_ that the way we actually build the final 
'spin_trylock()' function we end up with a real function in the end.

Maybe I'm wrong.

		Linus