Julia UndefVarError

Question

for i in 1:2
    if i == 2
        print(x)
    end
    if i == 1
        x = 0
    end
end

UndefVarError : x not defined

Why does the code gives that error instead of printing 0 in julia?

While in python the following code print 0?

for i in range(2):
    if i==1:
        print(x)
    if i==0:
        x=0

No language will let you use a variable *before* you declare it. And `x` is undeclared. — , Jun 18 '18 at 15:29
the variable is not used before declaring it and this works in python — Erik Garcia, Jun 18 '18 at 15:35
You're using it on line 3. Where are you declaring it? And no, this won't work in Python. Please open a Python prompt and type in `print(x)`. You'll get `NameError: name 'x' is not defined` — , Jun 18 '18 at 15:36
@Amy: this actually _will_ work in Python, because `x` is set to 0 on the first iteration of the loop and only printed on the second. — DSM, Jun 18 '18 at 15:48
in python x is not defined in the first iteration, it is defined in the second one. And yes this works in python. — Erik Garcia, Jun 18 '18 at 15:52
Julia has local scoping introduced by most code blocks. This is so that the compiler can optimise the functions easily. Check out this in the docs: https://docs.julialang.org/en/v0.6.2/manual/variables-and-scoping/#Local-Scope-1 — niczky12, Jun 18 '18 at 15:57

Bogumił Kamiński · Answer 1 · 2018-06-18T16:15:34.510

The reason is because in the loop a variable gets a new binding each time a loop is executed, see https://docs.julialang.org/en/latest/manual/variables-and-scoping/#For-Loops-and-Comprehensions-1.

In fact while loop changed this behavior between Julia 0.6.3 and Julia 0.7 (in Julia 0.6.3 a new binding was not created). Therefore the following code:

function f()
    i=0
    while i < 2
        i+=1
        if i == 2
            print(x)
        end
        if i == 1
            x = 0
        end
    end
end

Gives the following output.

Julia 0.6.3

julia> function f()
           i=0
           while i < 2
               i+=1
               if i == 2
                   print(x)
               end
               if i == 1
                   x = 0
               end
           end
       end
f (generic function with 1 method)

julia> f()
0

Julia 0.7.0

julia> function f()
           i=0
           while i < 2
               i+=1
               if i == 2
                   print(x)
               end
               if i == 1
                   x = 0
               end
           end
       end
f (generic function with 1 method)

julia> f()
ERROR: UndefVarError: x not defined
Stacktrace:
 [1] f() at .\REPL[2]:6
 [2] top-level scope

For-loop created a new binding already in Julia 0.6.3 at each iteration so it fails both under Julia 0.6.3 and Julia 0.7.0.

EDIT: I have wrapped the examples in a function, but you would get the same result if you executed the while loop in global scope.

niczky12 · Answer 2 · 2018-06-18T17:36:10.333

2

Ignore my comment andgo with Bogumil's answer as that's is the real rwason why your x variable disapears in thesecond iteration.

If you want your code to work like in Python, you can add the global keyword to your assignment of x:

for i in 1:2
    if i == 2
        print(x)
    end
    if i == 1
        global x = 0
    end
end

Note that this is not recommended in most cases as it'll make your code performance suffer. Julia likes local variables that the compiler can optimise away easily.

edited Jun 18 '18 at 17:36

answered Jun 18 '18 at 16:09

niczky12

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The reason is not local scoping but rebinding of a variable in a `for` loop. See the explanation in my answer. Additionally `if` block does not introduce a new scope as is explained in https://docs.julialang.org/en/latest/manual/variables-and-scoping/#scope-of-variables-1. – Bogumił Kamiński Jun 18 '18 at 16:13
You are right. Sorry I mixed things up a bit. I changed my answer to reflect this. – niczky12 Jun 18 '18 at 17:36

Julia UndefVarError

2 Answers2

Julia 0.6.3

Julia 0.7.0