Languages

Memory Management

Driptanil DattaSoftware Developer

Mar 202510 min read

1. Garbage Collection:

Automatically manages memory
Available in Java, Python, etc.

No control over memory allocation
Larger program size
Slows down the program

2. Manual Memory Management:

Programmer manages memory
smaller program size
Available in C, C++, etc.

Can lead to memory leaks and dangling pointers
Error-prone

3. Ownership:

Rust's way of managing memory
Ensures memory safety without garbage collector
No memory leaks or dangling pointers
Control over memory allocation
Faster than garbage collection

Slower write time (fighting with the borrow checker)

Types of Memory

Stack

fixed-size data stored on the stack
size is known at compile time
stack frames are pushed and popped (no need to free memory)
very fast access

Heap

dynamic-size data stored on the heap
size is calculated at runtime and can grow or shrink
slower access

Ownership Rules

Each value in Rust has a variable that is its owner

There can only be one owner at a time

When the owner goes out of scope, the value will be dropped

https://doc.rust-lang.org/book/ch04-01-what-is-ownership.html (opens in a new tab)

Scope

{ // s is not valid here, it’s not yet declared
    let s = String::from("hello"); // s is valid from this point forward
 
    // do stuff with s
} // s goes out of scope, deallocates from memory

When a variable goes out of scope, Rust calls drop function and cleans up the memory

Move

Rust defaults to moving instead of copying

fn main() {
    let s1 = String::from("hello");
    let s2 = s1;
 
    println!("{}, world!", s1);
}

Execution Output

Error: borrow of moved value: `s1`

Execution Output

$ cargo run

35 | let s1 = String::from("hello");
| -- move occurs because `s1` has type `String`, which does not implement the `Copy` trait
36 | let s2 = s1;

Move (not shallow copy)
s1 is moved to s2 and s1 is no longer valid

Copy

fn main() {
    let x = 5;
    let y = x; // x is copied
 
    println!("x = {}, y = {}", x, y);
}

Rust has a copy trait for data types that are stored on the stack

Clone

fn main() {
    let s1 = String::from("hello");
    let s2 = s1.clone();
 
    println!("s1 = {}, s2 = {}", s1, s2);
}

clone method that can be used to deep copy

Ownership and Functions

fn main() {
    let s = String::from("hello");
 
    takes_ownership(s);
 
    println!("{}", s);
}
 
fn takes_ownership(some_string: String) {
    println!("{}", some_string);
}

Execution Output

Error: borrow of moved value: `s`

some_string takes ownership of s and s is no longer valid

fn main() {
  let x = 5;
 
  makes_copy(x);
 
  println!("{}", x);
}
 
fn makes_copy(some_integer: i32) {
  println!("{}", some_integer);
}

some_integer is a copy of x and x is still valid as it is stored on the stack

fn main() {
  let s1 = gives_ownership();
 
  println!("{}", s1);
}
 
fn gives_ownership() -> String {
  let some_string = String::from("hello");
 
  some_string
}

ownership is transferred to s1

fn main() {
  let s1 = String::from("hello");
 
  let s2 = takes_and_gives_back(s1);
 
  println!("{}", s2);
}
 
fn takes_and_gives_back(a_string: String) -> String {
  a_string
}

ownership is transferred to takes_and_gives_back and then back to s2

References

fn main() {
    let s1 = String::from("hello");
 
    let (s2, len) = calculate_length(s1);
 
    println!("The length of '{}' is {}", s2, len);
}
 
fn calculate_length(s: String) -> (String, usize) {
    let length = s.len();
 
    (s, length)
}

fn main() {
    let s1 = String::from("hello");
 
    let len = calculate_length(&s1);
 
    println!("The length of '{}' is {}", s1, len);
}
 
fn calculate_length(s: &String) -> usize {
    s.len()
}

& is a reference to the value
references don't have ownership of the underlying value
& is immutable by default

Mutable References

fn main() {
    let mut s = String::from("hello");
 
    change(&mut s);
 
    println!("{}", s);
}
 
fn change(some_string: &mut String) {
    some_string.push_str(", world");
}

&mut is mutable reference, only one mutable reference is allowed at a time, in particular scope

fn main() {
    let mut s = String::from("hello");
 
    let r1 = &mut s;
    let r2 = &mut s;
 
    println!("{}, {}", r1, r2);
}

Execution Output

Error: cannot borrow `s` as mutable more than once at a time

fn main() {
    let mut s = String::from("hello");
 
    let r1 = &s;
    let r3 = &mut s;
 
    println!("{}, {}", r1, r3);
}

Execution Output

Error: cannot borrow `s` as mutable because it is also borrowed as immutable

Mutable and immutable references cannot coexist
However, immutable references are allowed to coexist

fn main() {
    let mut s = String::from("hello");
 
    let r1 = &s;
    let r2 = &s;
 
    println!("{}, {}", r1, r2);
 
    let r3 = &mut s;
 
    println!("{}", r3);
}

Immutable references go out of scope after println!
Mutable reference is allowed after immutable references go out of scope

Dangling References

fn main() {
    let reference_to_nothing = dangle();
}
 
fn dangle() -> &String {
    let s = String::from("hello");
 
    &s
}

Execution Output

Error: missing lifetime specifier

this function's return type contains a borrowed value, but there is no value for it to be borrowed from

Rules of References

At any given time, you can have either one mutable reference or any number of immutable references

References must always be valid

Slices

fn main() {
    let s = String::from("hello");
 
    let slice = &s[0..2];
 
    println!("{}", slice);
}

Slices are references a contiguous sequence of elements in a collection, instead of referencing the whole collection

fn main() {
    let s = String::from("hello world");
 
    let word = first_word(&s); // 5
 
    s.clear();
}
 
fn first_word(s: &String) -> int {
    let bytes = s.as_bytes();
 
    for (i, &item) in bytes.iter().enumerate() {
        if item == b' ' {
            return i;
        }
    }
 
    s.len()
}

fn main() {
    let mut s = String::from("hello world");
 
    let hello = &s[..5];
    let world = &s[6..];
    let full = &s[..];
 
    let word = first_word(&s); // hello
 
    s.clear();
 
    println!("{}", word);
}
 
fn first_word(s: &String) -> &str {
    let bytes = s.as_bytes();
 
    for (i, &item) in bytes.iter().enumerate() {
        if item == b' ' {
            return &s[..i];
        }
    }
 
    &s[..]
}

Execution Output

Error: cannot borrow `s` as mutable because it is also borrowed as immutable

first_word returns a slice of s and s is borrowed immutably
s.clear tries to borrow s mutably causing an error

fn main() {
    let mut s = String::from("hello world");
 
    let s2 = "hello world"; // string literal of type `&s`
 
    let word = first_word(s2);
}
 
fn first_word(s: &str) -> &str {
    let bytes = s.as_bytes();
 
    for (i, &item) in bytes.iter().enumerate() {
        if item == b' ' {
            return &s[..i];
        }
    }
 
    &s[..]
}

s2 is a string literal of type &str and first_word accepts &str

Slices on Different Types

fn main() {
    let a = [1, 2, 3, 4, 5];
 
    let slice = &a[1..3];
 
    println!("{:?}", slice);
}

2. Concepts 4. Structs