Generic functions in Rust - Mini Project: Build & Apply

Using Generic Functions in Rust

📖 Scenario: You are building a small Rust program that can work with different types of data using a single function. This helps avoid repeating code for each data type.

🎯 Goal: Create a generic function in Rust that can take two values of the same type and return the larger one. You will practice writing generic functions and calling them with different types.

📋 What You'll Learn

Create a generic function named largest that takes two parameters of the same generic type T.

Add a trait bound to T so it can be compared using the greater than operator.

Call the largest function with two i32 values and print the result.

Call the largest function with two f64 values and print the result.

💡 Why This Matters

🌍 Real World

Generic functions let programmers write flexible code that works with many data types without repeating themselves.

💼 Career

Understanding generics is important for writing clean, reusable code in Rust and many other programming languages.

1

Create a generic function skeleton

Write a generic function named largest that takes two parameters a and b of the same generic type T. For now, just return a without any comparison.

Rust

fn largest(a: T, b: T) -> T {
    // Your code here
    a
}

Hint

Use fn largest(a: T, b: T) -> T to define a generic function.

2

Add trait bound for comparison

Modify the largest function to add a trait bound so that T can be compared using the greater than operator. Use PartialOrd and Copy traits. Then return the larger of a and b using an if expression.

Rust

fn largest(a: T, b: T) -> T {
    // Your code here
}

Hint

Use T: PartialOrd + Copy to allow comparison and copying.

3

Call the generic function with integers

Call the largest function with two i32 values: 10 and 20. Store the result in a variable named result_int.

Rust

fn main() {
    // Your code here
}

Hint

Call largest(10, 20) and assign to result_int.

4

Call the generic function with floats and print results

Call the largest function with two f64 values: 3.5 and 2.7. Store the result in a variable named result_float. Then print both result_int and result_float using println!.

Rust

fn main() {
    let result_int = largest(10, 20);
    // Your code here
}

Hint

Use println!("Largest integer: {}", result_int); and similarly for result_float.

Practice

1. What does the <T> mean in a Rust function definition like fn example<T>(value: T) {}?

easy

A. It makes the function generic, so it can accept any type.

B. It defines a new variable inside the function.

C. It restricts the function to only work with integers.

D. It marks the function as unsafe.

Solution

1. Step 1: Understand the syntax <T> in Rust

   The <T> after the function name declares a generic type parameter named T.

2. Step 2: Explain the effect of generic type parameter

   This allows the function to accept any type for value, making it flexible and reusable.

3. Final Answer:

   It makes the function generic, so it can accept any type. -> Option A

4. Quick Check:

   Generic type parameter = flexible function input [OK]

Hint: Look for <T> after function name for generics [OK]

Common Mistakes:

• Thinking <T> is a variable declaration
• Assuming it restricts to one type like integer
• Confusing it with unsafe code marker

2. Which of the following is the correct syntax to define a generic function named print_value that takes one argument of any type T?

easy

A. fn print_value(T value) {}

B. fn print_value<T>(value: T) {}

C. fn print_value(value: <T>) {}

D. fn print_value(value: T<>) {}

Solution

1. Step 1: Recall generic function syntax in Rust

   Generic functions use <T> after the function name to declare the generic type.

2. Step 2: Match the syntax with options

   fn print_value<T>(value: T) {} correctly places <T> after the function name and uses value: T as parameter type.

3. Final Answer:

   fn print_value<T>(value: T) {} -> Option B

4. Quick Check:

   Generic syntax = fn name<T>(param: T) [OK]

Hint: Generic type always goes after function name in angle brackets [OK]

Common Mistakes:

• Putting T inside parameter parentheses incorrectly
• Using angle brackets around parameter type
• Missing angle brackets after function name

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

fn show_value<T: std::fmt::Display>(val: T) {
    println!("Value: {}", val);
}

fn main() {
    show_value(10);
    show_value("hello");
}

medium

A. Value: 10\nValue: hello

B. Value: 10\nValue: 10

C. Compilation error due to missing trait bound

D. Runtime error

Solution

1. Step 1: Understand the generic function with trait bound

   The function show_value requires T to implement Display so it can be printed.

2. Step 2: Check calls with integer and string

   Both 10 (integer) and "hello" (string) implement Display, so both calls print correctly.

3. Final Answer:

   Value: 10 Value: hello -> Option A

4. Quick Check:

   Trait bound allows printing = output shown [OK]

Hint: Check trait bounds to know if printing works [OK]

Common Mistakes:

• Assuming generic functions print without trait bounds
• Thinking string prints as number
• Expecting compilation error without trait bound

4. Identify the error in this generic function and how to fix it:

fn compare<T>(a: T, b: T) -> bool {
    a == b
}

medium

A. Function should return i32 instead of bool.

B. Function parameters should be references, not values.

C. Generic type T must be replaced with i32.

D. Missing trait bound T: PartialEq to use == operator.

Solution

1. Step 1: Understand use of == in generic function

   The == operator requires the type T to implement the PartialEq trait.

2. Step 2: Identify missing trait bound

   The function lacks T: PartialEq bound, so compiler will error when using ==.

3. Final Answer:

   Missing trait bound T: PartialEq to use == operator. -> Option D

4. Quick Check:

   Operator needs trait bound = add T: PartialEq [OK]

Hint: Add trait bounds when using operators on generics [OK]

Common Mistakes:

• Ignoring trait bounds for operators
• Changing return type unnecessarily
• Replacing generic with concrete type without reason

5. You want to write a generic function largest that returns the largest element from a slice of any type T. Which of these is the correct way to define it?

hard

A. fn largest<T>(list: &[T]) -> T { let mut largest = list[0]; for &item in list { if item > largest { largest = item; } } largest }

B. fn largest(list: &[i32]) -> i32 { let mut largest = list[0]; for &item in list { if item > largest { largest = item; } } largest }

C. fn largest<T: PartialOrd + Copy>(list: &[T]) -> T { let mut largest = *list[0]; for &item in list { if item > largest { largest = item; } } largest }

D. fn largest<T: PartialEq>(list: &[T]) -> T { list[0] }

Solution

1. Step 1: Identify required traits for comparison and copying

   To compare elements with >, T must implement PartialOrd. To copy elements, Copy trait is needed.

2. Step 2: Check function signature and body

   fn largest<T: PartialOrd + Copy>(list: &[T]) -> T { let mut largest = *list[0]; for &item in list { if item > largest { largest = item; } } largest } correctly adds T: PartialOrd + Copy and uses a loop to find the largest element.

3. Final Answer:

   fn largest<T: PartialOrd + Copy>(list: &[T]) -> T { ... } -> Option C

4. Quick Check:

   Traits PartialOrd + Copy needed for generic largest function [OK]

Hint: Add PartialOrd and Copy traits for generic comparisons [OK]

Common Mistakes:

• Omitting trait bounds causing compile errors
• Returning wrong type or missing copy trait
• Using concrete types instead of generics