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Kick the Tires: Rust Crash Course Lesson One Exercise Solutions

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Kick the Tires: Rust Crash Course Lesson One Exercise Solutions

In the last post, we gave you some homework. Today, we go over the answers to those questions, explaining the Rust code along the way.

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Below are the solutions to the exercises from the last Rust Crash Course lesson, "Kick the Tires."

This post is part of a series based on teaching Rust at FP Complete. If you're reading this post outside of the blog, you can find links to all posts in the series at the top of the introduction post. You can also subscribe to the RSS feed.

Exercise 1

The following code is broken:

fn main() {
    let val: String = String::from("Hello, World!");
    printer(val);
    printer(val);
}

fn printer(val: String) {
    println!("The value is: {}", val);
}

We got an error message about a move. We'll learn more about moves in the next lesson when we discuss ownership. There are two basic solutions. First, the less-than-ideal one:

Clone

We've moved the original val into the first call to printer, and can't use it again. One workaround is to instead move a clone of val into that call, leaving the original unaffected:

fn main() {
    let val: String = String::from("Hello, World!");
    printer(val.clone());
    printer(val);
}

fn printer(val: String) {
    println!("The value is: {}", val);
}

Notice that I only cloned val the first time, not the second time. We don't need val again after the second call, so it's safe to move it. Using an extra clone is expensive since it requires allocating memory and performing a buffer copy.

Speaking of it being expensive...

Pass by Reference

Instead of moving the value, we can instead pass it into the printer function by reference. Let's first try to achieve that by just modifying printer:

fn main() {
    let val: String = String::from("Hello, World!");
    printer(val);
    printer(val);
}

fn printer(val: String) {
    println!("The value is: {}", val);
}

This doesn't work, because when we call printer, we're still giving it a String and not a reference to a String. Fixing that is pretty easy:

fn main() {
    let val: String = String::from("Hello, World!");
    printer(&val);
    printer(&val);
}

Note that the ampersand means both:

  • A reference to this type.
  • Take a reference to this value.

There's an even better way to write printer:

fn printer(val: &str) {
    println!("The value is: {}", val);
}

By using str instead of String, we can pass in string literals, and do not need to force the allocation of a heap object. We'll get to this in more detail when we discuss strings.

Exercise 2

This is the broken code:

fn main() {
    let i = 1;

    loop {
        println!("i == {}", i);
        if i >= 10 {
            break;
        } else {
            i += 1;
        }
    }
}

The error message we get from the compiler is pretty informative:

cannot assign twice to immutable variable i

In order to fix this, we change the variable from immutable to mutable:

fn main() {
    let mut i = 1;
    ...

Exercise 3

This exercise asked you do see when you could leave out semicolons. Here's a simple rule:

  • The statement is the last statement in a block.
  • The type of the expression is unit.

For example, in this code, removing the semicolon is fine:

fn main() {
    for i in 1..11 {
        println!("i == {}", i);
    }
}

That said, it tends to be somewhat idiomatic to leave semicolons on expressions like these which are purely effectful.

Exercise 4

This exercise was to implement FizzBuzz in Rust. Repeating the rules here:

  • Print the numbers 1 to 100.
  • If the number is a multiple of 3, output fizz instead of the number.
  • If the number is a multiple of 5, output buzz instead of the number.
  • If the number is a multiple of 3 and 5, output fizzbuzz instead of the number.

Here's one possible solution using if/ else fallbacks:

fn main() {
    for i in 1..101 {
        if i % 3 == 0 && i % 5 == 0 {
            println!("fizzbuzz");
        } else if i % 3 == 0 {
            println!("fizz");
        } else if i % 5 == 0 {
            println!("buzz");
        } else {
            println!("{}", i);
        }
    }
}

This has at least one downside: it will need to test i % 3 == 0 and i % 5 == 0 potentially multiple times. Under the surface, the compiler may optimize this away. But still, the repeated modulus calls are just sitting in the code taunting us! We can instead use pattern matching:

fn main() {
    for i in 1..101 {
        match (i % 3 == 0, i % 5 == 0) {
            (true, true) => println!("fizzbuzz"),
            (true, false) => println!("fizz"),
            (false, true) => println!("buzz"),
            (false, false) => println!("{}", i),
        }
    }
}

Or, if you want to have some fun with wildcard matching:

fn main() {
    for i in 1..101 {
        match (i % 3, i % 5) {
            (0, 0) => println!("fizzbuzz"),
            (0, _) => println!("fizz"),
            (_, 0) => println!("buzz"),
            (_, _) => println!("{}", i),
        }
    }
}

I'm not going to tell you which of these is the "best" solution. And there are certainly other implementations that could be attempted. This was meant to give you a feel for some more Rust constructs.

Topics:
web dev ,rust ,tutorial ,rust lang

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