Thanks for the explanation, but I'm confused by the actual purpose of
reduce() now.
What's the point of having the v0 parameter at all, if it's not playing the
role it usually plays (of providing a base value in the output type)?
Obviously my example was artificial (it's just simpler than the actual code
in question, but that's why I can't just cast both sides), but why have a
reduce() with v0 parameter if the input types have to match the output type
anyway?
What's the purpose of having v0 at all here?
On Monday, February 10, 2014 5:38:36 PM UTC-5, Steven G. Johnson wrote:
>
>
>
> On Monday, February 10, 2014 4:50:41 PM UTC-5, N. Saphra wrote:
>>
>> The following code doesn't behave as I'd expect:
>>
>> julia> reduce((x,y) -> (x * int(y)), 1, ["1", "2", "3"])
>> ERROR: no method *(ASCIIString,Int64)
>> in anonymous at none:1
>> in r_pairwise at reduce.jl:135
>> in reduce at reduce.jl:183
>>
>> I was expecting reduce() to take my op and do something like *(*(*(1,
>> int("1")), int("2")), int("3")) (I understand the associativity is not
>> guaranteed, but * is associative anyway). Instead it gives this error. Am I
>> misunderstanding something?
>>
>
> It's not working because the correctness of your code depends on
> associativity. For example, if it associates from right to left, then the
> first call to your function will pass ("2","3"), for which "2" * int("3")
> fails.
>
> You are assuming a left-associative sum, which is not in fact what reduce
> uses. As Ivar said, in 0.3 there will be a foldl function that does
> this. Or you can use:
>
> reduce((x,y) -> (int(x) * int(y)), 1, ["1", "2", "3"])
>
> Or as Ivar said you can just use a loop, which will be faster anyway (and
> probably clearer, if not shorter):
>
> result = 1; for s in ["1", "2", "3"] result *= int(s); end; result
>
>
>
