Dangling reference prevention in Rust - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to declare a reference to an integer.

Rust

let x = 10;
let r = &[1];

Drag options to blanks, or click blank then click option'

Ax

B10

C*x

Dr

Attempts:

3 left

💡 Hint

Common Mistakes

Referencing a literal directly causes a temporary value that may not live long enough.

The reference must point to a valid variable, so &x is correct.

2fill in blank

medium

Complete the code to return a reference to the first element of a slice safely.

Rust

fn first_element(slice: &[i32]) -> &i32 {
    &slice[1]0
}

Drag options to blanks, or click blank then click option'

A[0]

B(0)

C.0

D{0}

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses or dot notation to index a slice.

To access the first element of a slice, use square brackets with index 0: slice[0].

3fill in blank

hard

Fix the error in the code to avoid returning a dangling reference.

Rust

fn get_str() -> &str {
    let s = String::from("hello");
    &[1]
}

Drag options to blanks, or click blank then click option'

As

BString

Cs.as_str()

D"hello"

Attempts:

3 left

💡 Hint

Common Mistakes

Returning a reference to a local variable causes dangling references.

Returning &s or s.as_str() creates a dangling reference because s does not live long enough after the function returns. Return a reference to the string literal "hello", which has a 'static lifetime.

4fill in blank

hard

Fill both blanks to create a function that returns the longer of two string slices safely.

Rust

fn longest<'a>(x: &'a str, y: &'a str) -> &'a str {
    if [1] > y.len() {
        [2]
    } else {
        y
    }
}

Drag options to blanks, or click blank then click option'

A>

Bx

Cx.len()

Dy

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing string slices directly instead of their lengths.

Returning the length instead of the string slice.

The function compares the lengths of x and y. The condition must be x.len() > y.len(). If true, return x, else return y.

5fill in blank

hard

Fill all three blanks to create a struct with a reference and implement a method returning that reference.

Rust

struct ImportantExcerpt<'a> {
    part: &'a str,
}

impl<'a> ImportantExcerpt<'a> {
    fn level(&self) -> i32 {
        [1]
    }

    fn announce_and_return_part(&self, announcement: &str) -> &[2] {
        println!("Announcement: {}", announcement);
        [3]
    }
}

Drag options to blanks, or click blank then click option'

A3

Bstr

Cself.part

Dself

Attempts:

3 left

💡 Hint

Common Mistakes

Returning the wrong type or forgetting to return the struct field.

The level method returns an integer, here 3. The announce_and_return_part method returns a string slice &str and returns self.part, which is a reference to a string slice.

Practice

1. What does Rust do to prevent dangling references?

easy

A. It automatically deletes all references after use.

B. It enforces borrowing rules and lifetimes to ensure references are valid.

C. It uses garbage collection to clean up unused references.

D. It allows references to point to any memory without checks.

Solution

1. Step 1: Understand what dangling references are

   Dangling references point to data that no longer exists, causing bugs.

2. Step 2: Identify Rust's prevention method

   Rust uses borrowing rules and lifetimes to ensure references are always valid while used.

3. Final Answer:

   It enforces borrowing rules and lifetimes to ensure references are valid. -> Option B

4. Quick Check:

   Dangling reference prevention = borrowing rules + lifetimes [OK]

Hint: Remember: Rust checks lifetimes to avoid invalid references [OK]

Common Mistakes:

• Thinking Rust uses garbage collection
• Believing references are deleted automatically
• Assuming Rust allows unchecked references

2. Which of the following Rust code snippets correctly prevents a dangling reference?

easy

A. let r;
{
let x = 5;
r = &x;
}
println!("{}", r);

B. let r = &5;
println!("{}", r);

C. let r = &x;
let x = 5;
println!("{}", r);

D. let x = 5;
let r = &x;
println!("{}", r);

Solution

1. Step 1: Analyze each snippet for lifetime validity

   let r; { let x = 5; r = &x; } println!("{}", r); creates a reference to x inside a block, but x is dropped before use, causing dangling reference. let r = &5; println!("{}", r); creates a reference to a temporary value 5, which is dropped while borrowed, causing compile error. let x = 5; let r = &x; println!("{}", r); creates x and r in the same scope, so r is valid. let r = &x; let x = 5; println!("{}", r); tries to reference x before x is declared, so x not in scope, compile error.

2. Step 2: Identify the snippet that correctly prevents dangling references

   let x = 5; let r = &x; println!("{}", r); is the correct example where the reference r points to x which lives long enough for r to be used safely.

3. Final Answer:

   let x = 5; let r = &x; println!("{}", r); -> Option D

4. Quick Check:

   Reference lifetime valid in same scope = let x = 5; let r = &x; println!("{}", r); [OK]

Hint: Keep referenced data alive in the same scope as the reference [OK]

Common Mistakes:

• Referencing variables that go out of scope too early
• Declaring references before the data they point to
• Assuming temporary values always live long enough

3. What will be the output of this Rust code?

fn main() {
    let r;
    {
        let x = 10;
        r = &x;
    }
    println!("{}", r);
}

medium

A. Compilation error due to dangling reference

B. 10

C. Undefined behavior at runtime

D. 0

Solution

1. Step 1: Understand variable lifetimes in the code

   Variable x is declared inside the inner block and dropped when the block ends. Reference r points to x but is used outside the block.

2. Step 2: Identify Rust's behavior on this code

   Rust's compiler detects that r would be a dangling reference because x does not live long enough. It will produce a compilation error.

3. Final Answer:

   Compilation error due to dangling reference -> Option A

4. Quick Check:

   Dangling reference causes compile error in Rust [OK]

Hint: References must not outlive the data they point to [OK]

Common Mistakes:

• Expecting code to run and print 10
• Thinking Rust allows unsafe dangling references
• Confusing runtime errors with compile-time errors

4. Identify the error in this Rust code and how to fix it:

fn get_ref() -> &i32 {
    let x = 42;
    &x
}

medium

A. Error: missing lifetime annotation. Fix by adding explicit lifetime.

B. No error, code is correct.

C. Error: returning reference to local variable. Fix by returning the value instead.

D. Error: x is not mutable. Fix by declaring x as mutable.

Solution

1. Step 1: Analyze the function's return type and variable scope

   The function returns a reference to an integer, but x is a local variable that is dropped when the function ends.

2. Step 2: Identify the problem and fix

   Returning a reference to a local variable causes a dangling reference. The fix is to return the value itself (i32) instead of a reference.

3. Final Answer:

   Error: returning reference to local variable. Fix by returning the value instead. -> Option C

4. Quick Check:

   Don't return references to local variables [OK]

Hint: Never return references to variables that go out of scope [OK]

Common Mistakes:

• Adding lifetimes without fixing the root cause
• Making variables mutable unnecessarily
• Ignoring that local variables are dropped after function ends

5. You want to return a reference to a string slice from a function without causing dangling references. Which approach is correct?

fn first_word(s: &String) -> &str {
    // ???
}

hard

A. Return a slice of s using lifetimes: &s[..1]

B. Create a new String inside the function and return a reference to it

C. Return a reference to a local string variable inside the function

D. Return the whole String by value instead of a reference

Solution

1. Step 1: Understand the input and output lifetimes

   The function takes a reference to a String and returns a string slice (&str) referencing part of that String.

2. Step 2: Choose the safe way to return a slice without dangling references

   Returning a slice of the input String (&s[..1]) is safe because the slice's lifetime is tied to the input reference's lifetime.

3. Final Answer:

   Return a slice of s using lifetimes: &s[..1] -> Option A

4. Quick Check:

   Return slices tied to input lifetimes to avoid dangling references [OK]

Hint: Return slices of input references, not local variables [OK]

Common Mistakes:

• Returning references to local variables
• Creating new owned Strings and returning references to them
• Ignoring lifetime relationships between input and output