On 25.03.2005 01:42, Alex Baretta wrote:

>>> let f1  x       y = Pervasives.compare x y
>>> let f2 (x: int) y = Pervasives.compare x y
>>> let f3  x       y = if x=y then Pervasives.compare "s1" "s2" else if 
>>> x < y then -1 else 1
>>> let f4 (x: int) y = if x=y then Pervasives.compare "s1" "s2" else if 
>>> x < y then -1 else 1
>>> let f5  x       y = let i = Pervasives.compare x y in if i = 0 then 
>>> Pervasives.compare "s1" "s2" else i
>>> let f6 (x: int) y = let i = Pervasives.compare x y in if i = 0 then 
>>> Pervasives.compare "s1" "s2" else i
>>> let f7  x       y = match Pervasives.compare x y with 0 -> 
>>> Pervasives.compare "s1" "s2" | i -> i
>>> let f8 (x: int) y = match Pervasives.compare x y with 0 -> 
>>> Pervasives.compare "s1" "s2" | i -> i
>>>
>>> let time name f =
>>>   
> 
> 
> I am unsure as to how to interpret these benchmarks...

Yes, these were taken a bit out of context. Basically, the question we 
were discussing was:
> What's the fastest way to compare int*string pairs (or, in general - int*'a), where ints are likely to be distinct whenever the second elements are distinct (basically, the int is a hash)?


The answer is that the "f4" above is the fastest one:

let compair_pair ((i1:int), a1) (i2, a2) =
    if i1 = i2 then
       Pervasives.compare a1 a2
    else if i1 < i2 then
       -1
    else
       1

-- 
Aleksey Nogin

Home Page: http://nogin.org/
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