When I was first starting I was horrified by this, when trying to literally translate Fortran to Julia: julia> function f() local x = Vector{Float64}(undef,3) local y = Vector{Float64}(undef,3) y[1] = 1.0 x = y y[1] = 2.0 println(x[1]) …

[image] Julia's loop scope behavior differs from Fortran? the entire “scope” setup of variables in Julia is a horror, plain and simple. Please consider the possibility that it is just unfamiliar, and after a few months or years of programming, you may appreciate Julia’s scoping rules as very w…

[image] lmiq: When I was first starting I was horrified by this, when trying to literally translate Fortran to Julia: julia> function f() local x = Vector{Float64}(undef,3) local y = Vector{Float64}(undef,3) y[1] = 1.0 x = y y[1] = 2.0 …

Told ya. :slight_smile: That is how things work in every dynamically typed language. x = y means that the name x is bound now to the value that was associated with the name y. The “previous” x is lost forever. If you want that the elements of x are associated with the values that are assigned to t…

Assignment is naming. That’s it. Just remember that and you’ll be in good shape. When you say x = Vector{Float64}(undef, 3), you’re saying to Julia that you want to use the name x for that undef vector. When you later say x = y you’re saying to Julia that you have a better use for the name x — i…

[image] CosmoProf: so that if 500 lines later, you update the value of y , then the value of x gets silently updated too?? I can’t believe that. This is exactly what would have happened with the equivalent Python, Java, or C code (for the old pointer arrays). [image] CosmoProf: LeandroM,…

Not just dynamically typed languages, it’s the same thing in C and C++ and Java and C# and Scala, etc. If you have double *x, *y; and later you do x = y what happens? It causes x to point to the same memory as y. If you modify that memory it is visible through both references (because it’s the same …

To be fair, the C++ equivalent would probably be Vector which does copy the array. C++ and their possibility of redefining how attribution works for each different class also seems like an odd duck.

So, what should I expect this function to do? function abc() xOld = 1 xNew = xOld + 1 xOld = xNew println("xOld = ",xOld,", xNew = ",xNew) return end By the previous logic, xNew is bound to xOld+1. But then xOld is bound to xNew, so that means xOld must equal xOld+1. So either …

You can try it and see. Note that this example is different as there is no mutation of any objects, only assignment. The array example involves mutation of arrays.

I’m afraid I don’t understand the difference how x=y is mutation but xOld=xNew is assignment. Does the equal (=) command mean something different for arrays than for numbers?

Julia's assignment behavior differs from Fortran?

General Usage

Topic		Replies	Views
Noteworthy differences from Fortran General Usage documentation , fortran	38	3304	August 12, 2020
Julia's loop scope behavior differs from Fortran? General Usage	38	2963	November 18, 2020
Fortran vs Julia stack allocated arrays Performance array , memory-allocation	18	6682	November 29, 2018
In place assignment of scalars vs vectors Performance	20	1625	June 25, 2021
Trying to understand how to write good code (example) New to Julia	34	3173	February 16, 2017

Julia's assignment behavior differs from Fortran?

Related topics