Can one of you explain to me what is going on here and how to avoid it? Can you reproduce this behaviour? When I define x as x = zeros(10) x .= .05 x I get Vector{Float64} with 10 elements 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 0.0500 when I assign x to y: y = x x I s…

Julia like most other high level languages (oppose C) think of variables as pointers to data. Thus y=x doesn’t mean “make y hold a copy of x”, but instead means “make y a synonym for x”. Thus changing y, naturally, changes x. If you want a copy, you should instead write y=copy(x).

In addition, you may also find the behavior shown below helpful to keep in mind as a reference. Array slicing (things like a[:] where a is an array) also copies the underlying vector. ######################## Example 1 ######################## julia> a = [1, 2, 3] 3-element Array{Int64,1}: …

What is the advantage of using pointers to same data when we write b = a? When we create a new variable with the value of old, it is usually to modify the data in the new one. Even though I have seen this behavior in other languages also, I’m yet to understand why such a “confusing” behavior is so w…

The behavior is useful because copying data can be expensive. Saying b = a avoids a copy. Moreover, it’s sometimes useful to keep the same array in different forms without copying the underlying data. Again, the behavior of b = a let’s you achieve that. See the following quote from 7 Julia Gotchas …

@bashonubuntu Thanks a lot for the explanation. The example you provided makes the reason more clear. :slight_smile:

b = a is quite rare to write but consider b = [[1,2,3], [4,5,6]] a = b[1] or passing a variable into a function.

Why is Julia messing up my variables?

New to Julia

Juan January 16, 2020, 4:28pm 9

But then why would you need to make b = a and not just keep using a?

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Why is Julia messing up my variables?

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