Reading input basics in Rust - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this Rust code reading input?

Consider this Rust program that reads a line from the user and prints it back:

use std::io::{self, Write};

fn main() {
    let mut input = String::new();
    print!("Enter your name: ");
    io::stdout().flush().unwrap();
    io::stdin().read_line(&mut input).unwrap();
    println!("Hello, {}!", input.trim());
}

If the user types Alex and presses Enter, what will be printed?

Rust

use std::io::{self, Write};

fn main() {
    let mut input = String::new();
    print!("Enter your name: ");
    io::stdout().flush().unwrap();
    io::stdin().read_line(&mut input).unwrap();
    println!("Hello, {}!", input.trim());
}

A Enter your name: Hello, Alex!

B Hello, Alex!\n

C Hello, Alex!\nEnter your name:

D Enter your name: Hello, Alex!\n

💡 Hint

Remember that print! does not add a newline, so the prompt stays on the same line.

The print! macro prints the prompt without a newline, so the cursor stays on the same line. The user types Alex and presses Enter. The program then prints Hello, Alex! with a newline. So the full output is the prompt and the greeting on the same line, followed by a newline.

❓ Predict Output

intermediate

What error occurs when reading input without mutable String?

Look at this Rust code snippet:

use std::io;

fn main() {
    let input = String::new();
    io::stdin().read_line(&mut input).unwrap();
    println!("You typed: {}", input);
}

What error will the compiler produce?

Rust

use std::io;

fn main() {
    let input = String::new();
    io::stdin().read_line(&mut input).unwrap();
    println!("You typed: {}", input);
}

A error: missing semicolon after `String::new()`

B error: cannot borrow immutable local variable `input` as mutable

C error: expected `&mut String` but found `&String`

D error: cannot move out of borrowed content

💡 Hint

Think about whether the variable can be changed when passed as mutable reference.

The variable input is declared as immutable by default. The read_line function requires a mutable reference to modify the string. So the compiler complains that you cannot borrow an immutable variable as mutable.

❓ Predict Output

advanced

What is the output of this Rust code reading multiple inputs?

Consider this Rust program:

use std::io;

fn main() {
    let mut buffer = String::new();
    io::stdin().read_line(&mut buffer).unwrap();
    let numbers: Vec = buffer
        .trim()
        .split_whitespace()
        .map(|x| x.parse().unwrap())
        .collect();
    println!("Sum: {}", numbers.iter().sum::());
}

If the user inputs 10 20 30 and presses Enter, what will be printed?

Rust

use std::io;

fn main() {
    let mut buffer = String::new();
    io::stdin().read_line(&mut buffer).unwrap();
    let numbers: Vec<i32> = buffer
        .trim()
        .split_whitespace()
        .map(|x| x.parse().unwrap())
        .collect();
    println!("Sum: {}", numbers.iter().sum::<i32>());
}

A Sum: 60

B Sum: 10 20 30

C Sum: 0

D Sum: 102030

💡 Hint

Think about how the input string is split and parsed into numbers before summing.

The input string is split by whitespace into three strings: "10", "20", "30". Each is parsed into an integer and collected into a vector. The sum of these integers is 10 + 20 + 30 = 60.

❓ Predict Output

advanced

What error occurs when parsing invalid input to integer?

Look at this Rust code:

use std::io;

fn main() {
    let mut input = String::new();
    io::stdin().read_line(&mut input).unwrap();
    let num: i32 = input.trim().parse().unwrap();
    println!("Number is {}", num);
}

If the user types abc and presses Enter, what happens?

Rust

use std::io;

fn main() {
    let mut input = String::new();
    io::stdin().read_line(&mut input).unwrap();
    let num: i32 = input.trim().parse().unwrap();
    println!("Number is {}", num);
}

A Program panics with a parse error

B Program prints: Number is 0

C Program waits for more input

D Program prints: Number is abc

💡 Hint

What happens if you try to convert a non-number string to an integer?

The parse() method tries to convert the trimmed input to an integer. Since "abc" is not a valid number, unwrap() causes the program to panic with a parse error.

🧠 Conceptual

expert

How to correctly read multiple lines of input until EOF in Rust?

You want to read multiple lines from standard input until there is no more input (EOF). Which Rust code snippet correctly does this?

A

use std::io;

fn main() {
    let mut input = String::new();
    while io::stdin().read_line(&mut input).unwrap() != 0 {
        println!("Read: {}", input);
        input.clear();
    }
}

B

use std::io;

fn main() {
    let mut input = String::new();
    while io::stdin().read_line(&mut input).unwrap() > 0 {
        println!("Read: {}", input.trim());
        input.clear();
    }
}

C

use std::io::{self, BufRead};

fn main() {
    for line in io::stdin().lock().lines() {
        let line = line.unwrap();
        println!("Read: {}", line);
    }
}

D

use std::io;

fn main() {
    loop {
        let mut input = String::new();
        io::stdin().read_line(&mut input).unwrap();
        if input.is_empty() { break; }
        println!("Read: {}", input.trim());
    }
}

💡 Hint

Think about how to handle input lines safely and efficiently until EOF.

Option C uses io::stdin().lock().lines() which returns an iterator over lines until EOF. It unwraps each line safely and prints it. This is the idiomatic and correct way to read multiple lines until EOF in Rust.

Option C is close but reuses the same string without clearing it properly before reading, which can cause concatenation issues.

Option C checks if the input string is empty, but read_line returns 0 on EOF, not an empty string, so it may loop infinitely.

Option C does not clear the input string, so it will print accumulated input repeatedly.