Ah, I see the error in my thinking now. Knowing what the ! signifies makes
it make a lot more sense.
Thanks guys for putting up with a newbie. This was probably one of those 1
in the morning questions that I should have waited to look at the next day
before asking for help; it seems obvious now.
-- Sean
On Thursday, December 11, 2014 3:26:12 AM UTC-6, Mike Innes wrote:
>
> Think of append!(X, Y) as equivalent to X = vcat(X, Y). You called append!
> twice, so X gets Y appended twice.
>
> julia> X = [1,2]; Y = [3,4];
>
> julia> X = vcat(X,Y)
> [1, 2, 3, 4]
>
> In your example you went ahead and did this again:
>
> julia> X = (X = vcat(X, Y))
> [1, 2, 3, 4, 3, 4]
>
> But if you reset X, Y via the first statement and *then* call X =
> append!(X, Y), it works as you would expect.
>
> julia> X = [1,2]; Y = [3,4];
>
> julia> X = append!(X, Y) # same as X = (X = vcat(X, Y))
> [1, 2, 3, 4]
>
> On 11 December 2014 at 07:51, Alex Ames <alexande...@gmail.com
> <javascript:>> wrote:
>
>> Functions that end with an exclamation point modify their arguments, but
>> they can return values just like any other function. For example:
>>
>> julia> x = [1,2]; y = [3, 4]
>> 2-element Array{Int64,1}:
>> 3
>> 4
>>
>> julia> append!(x,y)
>> 4-element Array{Int64,1}:
>> 1
>> 2
>> 3
>> 4
>>
>> julia> z = append!(x,y)
>> 6-element Array{Int64,1}:
>> 1
>> 2
>> 3
>> 4
>> 3
>> 4
>>
>> julia> z
>> 6-element Array{Int64,1}:
>> 1
>> 2
>> 3
>> 4
>> 3
>> 4
>>
>> julia> x
>> 6-element Array{Int64,1}:
>> 1
>> 2
>> 3
>> 4
>> 3
>> 4
>>
>> The append! function takes two arrays, appends the second to the first,
>> then returns the values now contained by the first array. No recursion
>> craziness required.
>>
>> On Thursday, December 11, 2014 1:11:50 AM UTC-6, Sean McBane wrote:
>>>
>>> Ivar is correct; I was running in the Windows command prompt and
>>> couldn't copy and paste so I copied it by hand and made an error.
>>>
>>> Ok, so I understand that append!(X,Y) is modifying X in place. But I
>>> still do not get where the output for the second case, where the result of
>>> append!(X,Y) is assigned back into X is what it is. It would make sense to
>>> me if this resulted in a recursion with Y forever getting appended to X,
>>> but as it is I don't understand.
>>>
>>> Thanks.
>>>
>>> -- Sean
>>>
>>> On Thursday, December 11, 2014 12:42:45 AM UTC-6, Ivar Nesje wrote:
>>>>
>>>> I assume the first line should be
>>>>
>>>> > X = [1,2]; Y = [3,4];
>>>>
>>>> Then the results you get makes sense. The thing is that julia has
>>>> mutable arrays, and the ! at the end of append! indicates that it is a
>>>> function that mutates it's argument.
>>>
>>>
>
