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 >> >