Ok, thanks.
Final quick question, does the syntax John used - A = ASCIIString[fun(i)
for i = 1:3] - have the same convert semantics, or is it different?
On Wednesday, October 29, 2014 2:02:47 PM UTC-4, Stefan Karpinski wrote:
>
> Assignment expressions always return the right hand side, not the left
> hand side, so that chaining assignments don't produce weird results doing
> something like
>
> y = v[i] = x
>
>
> With this rule, this is always equivalent to doing y = x; v[i] = x (and
> the order doesn't matter). If assignment returned the left hand side, then
> it would be equivalent to v[i] = x; y = v[i].
>
> So in the first version, you're getting the result of [fun(i) for i =
> 1:3], not the value that gets assigned to A.
>
> On Wed, Oct 29, 2014 at 1:57 PM, Zenna Tavares <zennat...@gmail.com
> <javascript:>> wrote:
>
>> Ok, but then why does it return the Vector{Any}. The following two
>> functions give different results
>>
>> function makestring(fun)
>> A::Array{ASCIIString} = [fun(i) for i = 1:3]
>> end
>>
>>
>> function makestring(fun)
>> A::Array{ASCIIString} = [fun(i) for i = 1:3]
>> convert(Array{ASCIIString},A)
>> end
>>
>> I thought maybe it was a weird REPL thing, but it's not.
>>
>> On Wednesday, October 29, 2014 1:52:01 PM UTC-4, John Myles White wrote:
>>>
>>> I'm pretty sure this sort of thing always works since type declarations
>>> on variables behave like convert calls:
>>>
>>> julia> function foo()
>>> a::Int64 = 0x01
>>> return a
>>> end
>>> foo (generic function with 1 method)
>>>
>>> julia> foo()
>>> 1
>>>
>>> -- John
>>>
>>> On Oct 29, 2014, at 10:50 AM, Zenna Tavares <zennat...@gmail.com>
>>> wrote:
>>>
>>> > Also, the following runs but still returns a Vector{Any}. How is this
>>> possible?
>>> >
>>> > function makestring(fun)
>>> > A::Array{ASCIIString} = [fun(i) for i = 1:3]
>>> > end
>>> >
>>> >
>>> > On Wednesday, October 29, 2014 1:45:51 PM UTC-4, Zenna Tavares wrote:
>>> > As shown in the following example, I am getting differently typed
>>> arrays depending on where I use the list comprehension.
>>> >
>>> > f(i) = "$i"
>>> > A = [f(i) for i = 1:3]
>>> > function makestring(fun)
>>> > A = [fun(i) for i = 1:3]
>>> > end
>>> > B = makestring(f)
>>> >
>>> > In this example A has type Vector{ASCIIString} while B has type
>>> Vector{Any}. What gives? And is there a workaround such that we get a more
>>> specific type?
>>> >
>>> > I understand there are some open issues related to this, but I am not
>>> sure if this case comes under what's already been discussed.
>>> >
>>> >
>>>
>>>
>
