Variable shadowing in Rust - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to shadow the variable x with a new value.

Rust

let x = 5;
let x = [1];
println!("{}", x);

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

Ax + 1

Bx - 1

Cx * 2

Dx / 2

Attempts:

3 left

💡 Hint

Common Mistakes

Using the same value instead of a new one.

Trying to modify x without shadowing.

The new variable x shadows the old one by assigning x + 1.

2fill in blank

medium

Complete the code to shadow y with a string value.

Rust

let y = 10;
let y = [1];
println!("{}", y);

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

A20

B"ten"

Cy + 5

Dtrue

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to assign a number instead of a string.

Not using quotes for the string.

Shadowing allows changing the type, so y becomes a string.

3fill in blank

hard

Fix the error by shadowing z with a new value inside the block.

Rust

let z = 3;
{
    let [1] = z + 2;
    println!("{}", z);
}
println!("{}", z);

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

Az_new

Bx

Cy

Dz

Attempts:

3 left

💡 Hint

Common Mistakes

Using a different variable name instead of shadowing.

Trying to modify z without shadowing.

Shadowing z inside the block creates a new z without affecting the outer one.

4fill in blank

hard

Fill both blanks to shadow a with a new value and then shadow it again with a string.

Rust

let a = 1;
let a = [1];
let a = [2];
println!("{}", a);

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

Aa + 1

B"hello"

C"world"

Da * 2

Attempts:

3 left

💡 Hint

Common Mistakes

Trying to shadow with a number then a number again.

Not using quotes for the string shadow.

First shadow a with a + 1, then shadow again with a string "hello".

5fill in blank

hard

Fill all three blanks to shadow val by doubling, then converting to string, then appending an exclamation mark.

Rust

let val = 4;
let val = val [1] 2;
let val = val.to_string();
let val = val [2] [3];
println!("{}", val);

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

A*

B+

C"!"

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using - instead of + for string concatenation.

Trying to multiply strings.

First multiply val by 2, then convert to string, then append "!" using + operator.

Practice

1. What does variable shadowing in Rust allow you to do?

easy

A. Prevent variables from being changed after initialization

B. Create a global variable accessible everywhere

C. Reuse the same variable name to store a new value or type

D. Automatically convert variable types without explicit casting

Solution

1. Step 1: Understand variable shadowing concept

   Variable shadowing means declaring a new variable with the same name as an existing one, which replaces the old variable in the current scope.

2. Step 2: Identify what shadowing allows

   This lets you reuse the same name to update the value or even change the type without needing a new variable name.

3. Final Answer:

   Reuse the same variable name to store a new value or type -> Option C

4. Quick Check:

   Shadowing = reuse variable name [OK]

Hint: Shadowing means reusing variable names for new values [OK]

Common Mistakes:

• Thinking shadowing creates global variables
• Confusing shadowing with immutability
• Assuming shadowing changes variable type automatically

2. Which of the following Rust code snippets correctly demonstrates variable shadowing?

easy

A. let x = 5; let x = x + 1; println!("{}", x);

B. let x = 5; x = x + 1; println!("{}", x);

C. let mut x = 5; x = x + 1; println!("{}", x);

D. let x = 5; mut x = 6; println!("{}", x);

Solution

1. Step 1: Check for correct shadowing syntax

   Shadowing uses let again with the same variable name to create a new variable that hides the old one.

2. Step 2: Analyze each option

   let x = 5; let x = x + 1; println!("{}", x);: let x = 5; let x = x + 1; println!("{}", x); valid shadowing using previous value. let x = 5; x = x + 1; println!("{}", x);: let x = 5; x = x + 1; ... reassignment to immutable, compile error. let mut x = 5; x = x + 1; println!("{}", x);: let mut x = 5; x = x + 1; ... valid mutation but no shadowing (missing second let). let x = 5; mut x = 6; println!("{}", x);: let x = 5; mut x = 6; ... invalid syntax, mut requires let.

3. Final Answer:

   let x = 5; let x = x + 1; println!("{}", x); -> Option A

4. Quick Check:

   Shadowing requires let again [OK]

Hint: Shadowing always uses 'let' again with the same name [OK]

Common Mistakes:

• Trying to reassign immutable variables without 'mut'
• Confusing shadowing with mutable reassignment
• Using 'mut' incorrectly with shadowing

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

let x = 10;
let x = x * 2;
let x = "Value: ".to_string() + &x.to_string();
println!("{}", x);

medium

A. 20

B. Value: x

C. Compilation error

D. Value: 20

Solution

1. Step 1: Follow variable shadowing and type changes

   First, x is 10 (integer). Then shadowed to 20 (integer). Then shadowed again to a String "Value: 20" by concatenating strings.

2. Step 2: Understand the final print output

   The last x is a String "Value: 20", so printing it outputs exactly that.

3. Final Answer:

   Value: 20 -> Option D

4. Quick Check:

   Shadowing can change type; final value is string [OK]

Hint: Shadowing can change variable type and value [OK]

Common Mistakes:

• Assuming variable type cannot change with shadowing
• Expecting numeric output instead of string
• Confusing variable name with literal output

4. Identify the error in this Rust code using variable shadowing:

let x = 5;
let x = x + 1;
let mut x = x * 2;
x = 10;
println!("{}", x);

medium

A. Cannot shadow a variable with 'mut' after immutable shadowing

B. No error; code runs and prints 10

C. Shadowing requires the same type, but types differ here

D. Cannot assign to 'x' because it is not mutable

Solution

1. Step 1: Analyze shadowing and mutability

   First x is immutable 5, then shadowed to 6 (immutable), then shadowed to mutable 12.

2. Step 2: Check assignment and print

   Since the last x is mutable, assigning x = 10; is valid. Printing outputs 10.

3. Final Answer:

   No error; code runs and prints 10 -> Option B

4. Quick Check:

   Shadowing can add mutability; assignment allowed [OK]

Hint: Shadowing can change mutability; last variable controls it [OK]

Common Mistakes:

• Thinking mutability cannot be added by shadowing
• Confusing shadowing with reassignment rules
• Assuming type must stay the same when shadowing

5. Given this Rust code, what is the final value of result?

let x = 3;
let x = x + 2;
{
    let x = x * 2;
    let x = "Number: ".to_string() + &x.to_string();
    println!("{}", x);
}
let result = x + 1;
println!("{}", result);

hard

A. 6

B. 7

C. Number: 10

D. Compilation error due to type mismatch

Solution

1. Step 1: Track shadowing in outer and inner scopes

   Outside the block, x is first 3, then shadowed to 5. Inside the block, x shadows to 10 (5*2), then to a String "Number: 10". This inner shadowing does not affect outer x.

2. Step 2: Calculate final result outside the block

   Outside, x remains 5, so result = x + 1 = 6. The print inside the block shows "Number: 10" but does not change outer x.

3. Final Answer:

   6 -> Option A

4. Quick Check:

   Inner shadowing does not affect outer variable [OK]

Hint: Inner scope shadowing doesn't change outer variable [OK]

Common Mistakes:

• Assuming inner shadowing changes outer variable
• Confusing printed output with variable value
• Expecting type change to persist outside inner scope