Zig doesn't include a garbage collector. The burden of managing memory is on you, the developer. It's a big responsibility as it has a direct impact on the performance, stability and security of your application.
We'll begin by talking about pointers, which is an important topic to discuss in and of itself, but also to start training ourselves to see our program's data from a memory-oriented point of view. If you're already comfortable with pointers, heap allocations and dangling pointers, feel free to skip ahead a couple of parts to heap memory & allocators, which is more Zig-specific.
The following code creates a user with a power of 100, and then calls the levelUp function which increments the user's power by 1. Can you guess the output?
const std = @import("std");
pub fn main() void {
var user = User{
.id = 1,
.power = 100,
};
levelUp(user);
std.debug.print("User {d} has power of {d}\n", .{user.id, user.power});
}
fn levelUp(user: User) void {
user.power += 1;
}
pub const User = struct {
id: u64,
power: i32,
};
That was a unkind trick; the code won't compile: local variable is never mutated. This is in reference to the user variable in main. A variable that is never mutated must be declare const. You might be thinking: but in levelUp we are mutating user, what gives? Let's assume the Zig compiler is mistaken and trick it. We'll force the compiler to see that user is mutated:
const std = @import("std");
pub fn main() void {
var user = User{
.id = 1,
.power = 100,
};
user.power += 0;
Now we get an error in levelUp: cannot assign to constant. We saw in part 1 that function parameters are constants, thus user.power += 1; is not valid. To fix the compile-time error, we could change the levelUp function to:
fn levelUp(user: User) void {
var u = user;
u.power += 1;
}
Which will compile, but our output is User 1 has power of 100, even though the intent of our code is clearly for levelUp to increase the user's power to 101. What's happening?
To understand, it helps to think about data with respect to memory, and variables as labels that associate a type with a specific memory location. For example, in main, we create a User. A simple visualization of this data in memory would be:
user -> ------------ (id)
| 1 |
------------ (power)
| 100 |
------------
There are two important things to note. The first is that our user variable points to the beginning of our structure. The second is that the fields are laid out sequentially. Remember that our user also has a type. That type tells us that id is a 64 bit integer and power is a 32 bit integer. Armed with a reference to the start of our data and the type, the compiler can translate user.power to: access a 32 bit integer located 64 bits from the beginning. That's the power of variables, they reference memory and include the type information necessary to understand and manipulate the memory in a meaningful way.
Here's a slightly different visualization which includes memory addresses. The memory address of the start of this data is a random address I came up with. This is the memory address referenced by the user variable, which is also the value of our first field, id. However, given this initial address, all subsequent addresses have a known relative address. Since id is a 64 bit integer, it takes 8 bytes of memory. Therefore, power has to be at $start_address + 8:
user -> ------------ (id: 1043368d0)
| 1 |
------------ (power: 1043368d8)
| 100 |
------------
To verify this for yourself, I'd like to introduce the addressof operator: &. As the name implies, the addressof operator returns the address of an variable (it can also return the address of a function, isn't that something?!). Keeping the existing User definition, try this main:
pub fn main() void {
const user = User{
.id = 1,
.power = 100,
};
std.debug.print("{*}\n{*}\n{*}\n", .{&user, &user.id, &user.power});
}
This code prints the address of user, user.id and user.power. You might get different results based on your platform and other factors, but hopefully you'll see that the address of user and user.id are the same, while user.power is at an 8 byte offset. I got:
learning.User@1043368d0
u64@1043368d0
i32@1043368d8
The addressof operator returns a pointer to a value. A pointer to a value is a distinct type. The address of a value of type T is a *T. We pronounce that a pointer to T. Therefore, if we take the address of user, we'll get a *User, or a pointer to User:
pub fn main() void {
var user = User{
.id = 1,
.power = 100,
};
user.power += 0;
const user_p = &user;
std.debug.print("{any}\n", .{@TypeOf(user_p)});
}
Our original goal was to increase our user's power by 1, via the levelUp function. We got the code to compile, but when we printed power it was still the original value. It's a bit of a leap, but let's change the code to print the address of user in main and in levelUp:
pub fn main() void {
var user = User{
.id = 1,
.power = 100,
};
user.power += 0;
std.debug.print("main: {*}\n", .{&user});
levelUp(user);
std.debug.print("User {d} has power of {d}\n", .{user.id, user.power});
}
fn levelUp(user: User) void {
std.debug.print("levelUp: {*}\n", .{&user});
var u = user;
u.power += 1;
}
If you run this, you'll get two different addresses. This means that the user being modified in levelUp is different from the user in main. This happens because Zig passes a copy of the value. That might seem like a strange default, but one of the benefits is that the caller of a function can be sure that the function won't modify the parameter (because it can't). In a lot of cases, that's a good thing to have guaranteed. Of course, sometimes, like with levelUp, we want the function to modify a parameter. To achieve this, we need levelUp to act on the actual user in main, not a copy. We can do this by passing the address of our user into the function:
const std = @import("std");
pub fn main() void {
var user = User{
.id = 1,
.power = 100,
};
levelUp(&user);
std.debug.print("User {d} has power of {d}\n", .{user.id, user.power});
}
fn levelUp(user: *User) void {
user.power += 1;
}
pub const User = struct {
id: u64,
power: i32,
};
We had to make two changes. The first is calling levelUp with the address of user, i.e. &user, instead of user. This means that our function no longer receives a User. Instead, it receives a *User, which was our second change.
We no longer need that ugly hack of forcing user to be mutated via user.power += 0;. Initially, we failed to get the code to compile because user was a var; the compiler told us it was never mutated. We thought maybe the compiler was wrong and "tricked" it by forcing a mutation. But, as we now know, the user being mutated in levelUp was a different; the compiler was right.
The code now works as intended. There are still many subtleties with function parameters and our memory model in general, but we're making progress. Now might be a good time to mention that, aside from the specific syntax, none of this is unique to Zig. The model that we're exploring here is the most common, some languages might just hide many of the details, and thus flexibility, from developers.
More than likely, you'd have written levelUp as a method of the User structure:
pub const User = struct {
id: u64,
power: i32,
fn levelUp(user: *User) void {
user.power += 1;
}
};
This begs the question: how do we call a method with a pointer receiver? Maybe we have to do something like: &user.levelUp()? Actually, you just call it normally, i.e. user.levelUp(). Zig knows that the method expects a pointer and passes the value correctly (by reference).
I initially chose a function because it's explicit and thus easier to learn from.
I more than implied that, by default, Zig will pass a copy of a value (called "pass by value"). Shortly we'll see that the reality is a bit more subtle (hint: what about complex values with nested objects?)
Even sticking with simple types, the truth is that Zig can pass parameters however it wants, so long as it can guarantee that the intent of the code is preserved. In our original levelUp, where the parameter was a User, Zig could have passed a copy of the user or a reference to main.user, as long as it could guarantee that the function would not mutate it. (I know we ultimately did want it mutated, but by making the type User, we were telling the compiler that we didn't).
This freedom allows Zig to use the most optimal strategy based on the parameter type. Small types, like User, can be cheaply passed by value (i.e. copied). Larger types might be cheaper to pass by reference. Zig can use any approach, so long as the intent of the code is preserved. To some degree, this is made possible by having constant function parameters.
Now you know one of the reasons function parameters are constants.
We previous looked at what the memory of user within our main function looked like. Now that we've changed levelUp what would its memory look like?:
main:
user -> ------------ (id: 1043368d0) <---
| 1 | |
------------ (power: 1043368d8) |
| 100 | |
------------ |
|
............. empty space |
............. or other data |
|
levelUp: |
user -> ------------- (*User) |
| 1043368d0 |----------------------
-------------
Within levelUp, user is a pointer to a User. Its value is an address. Not just any address, of course, but the address of main.user. It's worth being explicit that the user variable in levelUp represents a concrete value. This value happens to be an address. And, it's not just an address, it's also a type, a *User. It's all very consistent, it doesn't matter if we're talking about pointers or not: variables associate type information with an address. The only special thing about pointers is that, when we use the dot syntax, e.g. user.power, Zig, knowing that user is a pointer, will automatically follow the address.
What's important to understand is that the user variable in levelUp itself exists in memory at some address. Just like we did before, we can see this for ourselves:
fn levelUp(user: *User) void {
std.debug.print("{*}\n{*}\n", .{&user, user});
user.power += 1;
}
The above prints the address the user variable references as well as its value, which is the address of the user in main.
If user is a *User, then what is &user? It's a **User, or a pointer to a pointer to a User. I can do this until one of us runs out of memory!
There are use-cases for multiple levels of indirection, but it isn't anything we need right now. The purpose of this section is to show that pointers aren't special, they're just a value, which is an address, and a type.
Up until now, our User has been simple, containing two integers. It's easy to visualize its memory and, when we talk about "copying", there isn't any ambiguity. But what happens when User becomes more complex and contains a pointer?
pub const User = struct {
id: u64,
power: i32,
name: []const u8,
};
We've added name which is a slice. Recall that a slice is a length and a pointer. If we initialized our user with the name of "Goku", what would it look like in memory?
user -> ------------- (id: 1043368d0)
| 1 |
------------- (power: 1043368d8)
| 100 |
------------- (name.len: 1043368dc)
| 4 |
------------- (name.ptr: 1043368e4)
------| 1182145c0 |
| -------------
|
| ............. empty space
| ............. or other data
|
---> ------------- (1182145c0)
| 'G' |
-------------
| 'o' |
-------------
| 'k' |
-------------
| 'u' |
-------------
The new name field is a slice which is made up of a len and ptr field. These are laid out in sequence along with all the other fields. On a 64 bit platform both len and ptr will be 64 bits, or 8 bytes. The interesting part is the value of name.ptr: it's an address to some other place in memory.
Types can get much more complex than this with deep nesting. But simple or complex, they all behave the same. Specifically, if we go back to our original code where levelUp took a plain User and Zig provided a copy, how would that look now that we have a nested pointer?
The answer is that only a shallow copy of the value is made. Or, as some put it, only the memory immediately addressable by the variable is copied. It might seem like levelUp would get a half-baked copy of user, possibly with an invalid name. But remember that a pointer, like our user.name.ptr is a value, and that value is an address. A copy of an address is still the same address:
main: user -> ------------- (id: 1043368d0)
| 1 |
------------- (power: 1043368d8)
| 100 |
------------- (name.len: 1043368dc)
| 4 |
------------- (name.ptr: 1043368e4)
| 1182145c0 |-------------------------
levelUp: user -> ------------- (id: 1043368ec) |
| 1 | |
------------- (power: 1043368f4) |
| 100 | |
------------- (name.len: 1043368f8) |
| 4 | |
------------- (name.ptr: 104336900) |
| 1182145c0 |-------------------------
------------- |
|
............. empty space |
............. or other data |
|
------------- (1182145c0) <---
| 'G' |
-------------
| 'o' |
-------------
| 'k' |
-------------
| 'u' |
-------------
From the above, we can see that shallow copying will work. Since a pointer's value is an address, copying the value means we get the same address. This has important implications with respect to mutability. Our function can't mutate the fields directly accessible by main.user since it got a copy, but it does have access to the same name, so can it mutate that? In this specific case, no, name is a const. Plus, our value "Goku" is a string literal which are always immutable. But, with a bit of work, we can see the implication of shallow copying:
const std = @import("std");
pub fn main() void {
var name = [4]u8{'G', 'o', 'k', 'u'};
const user = User{
.id = 1,
.power = 100,
.name = name[0..],
};
levelUp(user);
std.debug.print("{s}\n", .{user.name});
}
fn levelUp(user: User) void {
user.name[2] = '!';
}
pub const User = struct {
id: u64,
power: i32,
name: []u8
};
The above code prints "Go!u". We had to change name's type from []const u8 to []u8 and instead of a string literal, which are always immutable, create an array and slice it. Some might see inconsistency here. Passing by value prevents a function from mutating immediate fields, but not fields with a value behind a pointer. If we did want name to be immutable, we should have declared it as a []const u8 instead of a []u8.
Some languages have a different implementation, but many languages work exactly like this (or very close). While all of this might seem esoteric, it's fundamental to day to day programming. The good news is that you can master this using simple examples and snippets; it doesn't get more complicated as other parts of the system grow in complexity.
Sometimes you need a structure to be recursive. Keeping our existing code, let's add an optional manager of type ?User to our User. While we're at it, we'll create two Users and assign one as the manager to another:
const std = @import("std");
pub fn main() void {
const leto = User{
.id = 1,
.power = 9001,
.manager = null,
};
const duncan = User{
.id = 1,
.power = 9001,
.manager = leto,
};
std.debug.print("{any}\n{any}", .{leto, duncan});
}
pub const User = struct {
id: u64,
power: i32,
manager: ?User,
};
This code won't compile: struct 'learning.User' depends on itself. This fails because every type has to have a known compile-time size.
We didn't run into this problem when we added name even though names can be different lengths. The issue isn't with the size of values, it's with the size of the types themselves. Zig needs this knowledge to do everything we talked about above, like accessing a field based on its offset position. name was a slice, a []const u8, and that has a known size: 16 bytes - 8 bytes for len and 8 bytes for ptr.
You might think this is going to be a problem with any optional or union. But for both optionals and unions, the largest possible size is known and Zig can use that. A recursive structure has no such upper-bound, the structure could recurse once, twice or millions of times. That number would vary from User to User and would not be known at compile time.
We saw the answer with name: use a pointer. Pointers always take usize bytes. On a 64-bit platform, that's 8 bytes. Just like the actual name "Goku" wasn't stored with/along our user, using a pointer means our manager is no longer tied to the user's memory layout.
const std = @import("std");
pub fn main() void {
const leto = User{
.id = 1,
.power = 9001,
.manager = null,
};
const duncan = User{
.id = 1,
.power = 9001,
.manager = &leto,
};
std.debug.print("{any}\n{any}", .{leto, duncan});
}
pub const User = struct {
id: u64,
power: i32,
manager: ?*const User,
};
You might never need a recursive structure, but this isn't about data modeling. It's about understanding pointers and memory models and better understanding what the compiler is up to.
A lot of developers struggle with pointers, there can be something elusive about them. They don't feel concrete like an integer, or string or User. None of this has to be crystal clear for you to move forward. But it is worth mastering, and not just for Zig. These details might be hidden in languages like Ruby, Python and JavaScript, and to a lesser extent C#, Java and Go, but they're still there, impacting how you write code and how that code runs. So take your time, play with examples, add debug print statements to look at variables and their address. The more you explore, the clearer it will get.