OK, thanks, makes sense. But how to change the original instance, is
looping the only way?
On Thursday, May 1, 2014 3:12:51 PM UTC+3, Freddy Chua wrote:
>
> b = b .+ 5
>
> creates a new instance of an array, so the original array pointed to by
> "b" is not changed at all.
>
>
>
> On Thursday, May 1, 2014 7:39:14 PM UTC+8, Kaj Wiik wrote:
>>
>> As a new user I was surprised that even if you change the value of
>> function arguments (inside the function) the changes are not always visible
>> outside but in some cases they are.
>>
>> Here's an example:
>>
>> function vappu!(a,b)
>> a[3]=100
>> b = b .+ 5
>> (a,b)
>> end
>>
>> c = [1:5]
>> d = [1:5]
>>
>> vappu!(c,d)
>> ([1,2,100,4,5],[6,7,8,9,10])
>>
>> c
>> 5-element Array{Int64,1}:
>> 1
>> 2
>> 100
>> 4
>> 5
>> d
>> 5-element Array{Int64,1}:
>> 1
>> 2
>> 3
>> 4
>> 5
>>
>>
>> Should I loop over arrays explicitly, what is happening in b = b .+ 5 ?
>>
>> Thanks,
>> Kaj
>>
>
