Programming a Guessing Game

guessing_game

use std::io;

fn main() {
    println!("Guess the number!");

    println!("Please input your guess.");

    let mut guess = String::new();
    io::stdin()
        .read_line(&mut guess)
        .expect("Failed to read line");

    println!("You guessed: {}", guess);
}

use std::io;

to obtain user input and then print the result as output, we need to bring the io library into scope.
let mut guess

In RUST, variables are immutable by default.
String::new()

how to call static method in RUST.
&

RUST has reference.
Handling potential failure with the Result type

As mentioned earlier, read_line puts what the user types into the string we’re passing it, but it also returns a value—in this case, an io::Result. Rust has a number of types named Result in its standard library: a generic Result as well as specific versions for submodules, such as io::Result.

Result types are enumerations. For Result the variats are Ok or Err. Err contaions information about why the operation failed.

If I don't call .expect, the program will compile, but I'll get a warning.
Printing values with println! placeholders

Generating a secret number

crate is a collection of Rust code files.

modify Cargo.toml : add rand crate

...
[dependencies]
rand = "0.3.14"

cargo build
# install libraries from crates.io
# Cargo.lock 파일에 적절한 version의 dependency가 깔린다.
# Q. Cargo.toml 파일은 그럼 어쩌지..?

cargo update

cargo doc --open
# 현재 사용하고 있는 라이브러리에 대하여 document를 볼 수 있다...

Crates.io : where people in Rust ecosystem post their open source Rust projects.

shadowing

fn main() {
  let mut test = String::new();
  io::stdin().read_line(&mut test)
  .expect("Failed to read");
  let test: u32 = test.trim().parse().expect("please type a number");
}

test라는 변수명을 사용했는데, 또 사용해도 되나?

Rust에서는 변수의 타입을 변경할 때, shadowing 방식을 지향한다.

number types

i32 : 32-bit number / i64 : 64-bit number

u32 : unsigned 32-bit number

Rust defaults to an i32.

error handling

let mut guess = String::new();
io::stdin()
  .read_line(&mut guess)
  .expect("Failed to read line");

let guess: u32 = match guess.trim().parse() {
  Ok(num) => num,
  Err(_) => continue,
};

expect 에서 match로 바꾸면 이렇게 된다.

Result 타입에 있는 enum인, Ok 와 Err 를 사용하여 위와 같이 하면, 올바른 값을 parsing 할 수 있을 경우에 해당 값을 return하여 guess에 담는다.

RUST Documentation 따라 공부하기 2

Programming a Guessing Game

Generating a secret number