I have long been confused by the strange behavior of integers as arguments to functions.
If I pass a variable into a function, I expect the function to be able to modify it.
This expectation applies to variables local to the calling context and to global variables;
it also applied to Strings and Floats and Integers and Chars.
When I’m trying to do something
(like modifying Int arguments inside a function and seeing the change outside it)
and it isn’t working, I fiddle with it until it does.
However, today, I decided that I should actually ask why it didn’t work the way I expect.
Today, I had the embarrassing experience of correcting my fundamentally flawed model of how variables work.
I’m going to skip over explaining the details of my previous model, since I don’t want to encourage its remaining grip on my mind.
Instead, I’m going to work through a very physical metaphor for my new understanding.
The Metaphor: A Table of Boxes and Forms and Colorful Stickers
There are two kinds of values: mutable and immutable.
Immutable values cannot be edited; they are like paper forms filled out in indelible ink.
Numbers (Ints,Floats,BigInts,BigFloats,etc), Chars, Strings, and user-defined immutable types are all immutable.
Mutable values can be edited; they are like boxes with boxes and paper forms inside.
I like to picture them as those organizer-boxes, with dividers it them.
Dictionaries, Arrays, and all other user-defined types are mutable.
There is a table (as in, a piece of furniture) of these boxes and forms; this is all the memory your program is using.
(this metaphor, as you may have noticed, is ignoring the stack/heap distinction and other implementation concerns.
I’m focused on getting correct expectations for what the value of my variables might be after I use them as arguments to a function.)
As your program executes, it fills out new forms, puts new boxes on the table, and throws out boxes and forms that it’s not using.
It also moves things into or out of the boxes.
There is one other component to this setup: stickers.
Any time that code is executing, it is executing within a context.
This context is made of bindings of names (variable names) to the values on the table.
We’ll model these bindings as stickers.
The name of the variable is printed on the sticker,
and the sticker is stuck to the value that the variable is currently bound to.
Each context has its own color of sticker, and ignores stickers that aren’t of its color.
Example 1: x = x + 1
Let’s say we have the variable name
x bound to an Int value,
This value is immutable, so it will be a paper form with
5 written on it in ink.
Because we’re calling this form by the name
x, there is a sticker on the form with the name
x printed on it.
Now that I’ve explained the context, let’s run the line of code
x = x + 1.
First, we’ll take a look at the value on the form labeled
Then, we’ll write the result of
5 + 1 on a new form.
Finally, we’ll move the
x sticker from the form with
5 written on it to the form with
6 written on it.
Example 2: a = 6
a is a 1-dimensional Array.
a as a long, thin box with three dividers in it.
Each space between the dividers is an element of
a is of length 4.
There is a sticker with
a printed on it on the outside of the box.
Each of the spaces in the box has a paper form with a number written on it;
what numbers are written on them is not relevant to this example.
Now, let’s simulate running
a = 6.
(We’re simulating Julia code, so we’re indexing from 1.)
First, we’ll get a paper form and write
6 on it in pen.
Then, we’ll replace the paper in the second box from the left of
a with the new form.
That’s all we need to do; notice that no stickers have been moved.
Example 3: foo(x)
Say we have our integer variable
It’s a form with
6 written on it, with an
x sticker stuck to it.
The sticker is blue.
Now, we have a function:
:::.jl function foo(z::Int) z = z + 1 end
I’m going to ignore irrelevant portions of calling this function,
including that it will return the value of
z + 1.
When we run the line
foo(x), we will move into the context of
Our previous context is the calling context (the one calling
Our current context is inside
foo, so now there is a green
z sticker on our form.
foo, we’ll write
7 on a new form, and move our green
z sticker to it.
Then we return from the function call, and remove all the green stickers.
What is the value of
Well, the blue
x sticker is still stuck to the form with
6 written on it.
This means that
foo did not modify
In fact, passing an Int variable as an argument to a function will never modify that variable.
Inside the context of
foo, we can’t see the blue
x sticker at all — and we definitely can’t move it.
The form is immutable, so you can’t erase and you can’t write anything new on it.
There’s nothing that
foo could possibly do to change the value of
x in the calling context.
Example 4: foo(a)
Picture the box for
a again. It’s got 4 pieces of paper in it; the second one says
There are no labels on any of the four forms; there is a blue
a label on the outside of the box.
Now, we’ll execute
foo is the same function from the previous example.
As we move from the calling context into the
we’ll stick a green
z sticker on the form in the second compartment of the box with the blue
Now, we’ll get out a new form, write
7 on it, and move the green
z sticker to this new form.
Finally, we’ll remove all the green stickers as we return to the calling context.
The value of
a is unchanged.
Example 5: bar(a)
So far, we haven’t managed to mutate any arguments to a function.
This is about to change.
Picture the box for
a. Let’s say that the four forms in have,
respectively, the following numbers written on them:
None of the forms have any labels on them; the only label is the blue
a on the box.
We’ll need a new function for this:
:::.jl function bar(xs::Array) xs = 42 end
Now, let’s simulate:
We’ll move from the blue calling context to the green callee context;
we’ll put a green
xs sticker on the box.
Then we’ll take the piece of paper in the second compartment (with
6 written on it)
and replace it with a new form with
42 written on it.
Now, we’ll return to the calling context, removing the green
Notice that we did not move any stickers.
What is the value of
[1,42,8,19]. The assignment in
bar did affect the value of
It did not affect the binding of
a (the blue sticker),
but it affected which values were inside the mutable box that is the value of
I already had a sort of messy awareness of this whole thing.
I mean, I’ve brushed up against “by value” vs “by reference”;
I’m fine with writing C and using pointer;
I was good on the difference between variables shadowing and modifying a mutable box in OCaml.
But it wasn’t until today that this all clicked into place in my mental model
of how passing variables as arguments to functions works in “normal” languages, like Julia and Java and such.
Suddenly, the division between types that are passed “by reference” and ones that are inexplicably passed “by value” makes complete sense.
I am happy that I noted that I was confused and expressed my confusion despite feeling embarrassed.
After trying several functions in the Julia REPL and determining that the behavior did indeed break my mental model (I felt so confused),
I asked one of my mentors for the summer the relevant question.
Once I convinced him that I was in fact not joking and was honestly surprised and confused,
he was very patient about correcting my misunderstanding.
Being very confused about a fundamental aspect of programming,
misunderstanding such a seemingly basic thing,
was embarrassing to admit to myself;
it was much more embarrassing to reveal that mistake in front of someone as awesome as him.
I still feel embarrassed, but the world also makes a lot more sense, which is worth it.
- Except in certain circumstances in C++.