Convert a String to int?

Question

You can directly convert to an int using the str::parse::<T>() method.

let my_string = "27".to_string();  // `parse()` works with `&str` and `String`!
let my_int = my_string.parse::<i32>().unwrap();

You can either specify the type to parse to with the turbofish operator (::<>) as shown above or via explicit type annotation:

let my_int: i32 = my_string.parse().unwrap();

As mentioned in the comments, parse() returns a Result. This result will be an Err if the string couldn't be parsed as the type specified (for example, the string "peter" can't be parsed as i32).

let my_u8: u8 = "42".parse().unwrap();
let my_u32: u32 = "42".parse().unwrap();

// or, to be safe, match the `Err`
match "foobar".parse::<i32>() {
  Ok(n) => do_something_with(n),
  Err(e) => weep_and_moan(),
}

str::parse::<u32> returns a Result<u32, core::num::ParseIntError> and Result::unwrap "Unwraps a result, yielding the content of an Ok [or] panics if the value is an Err, with a panic message provided by the Err's value."

str::parse is a generic function, hence the type in angle brackets.

If you get your string from stdin().read_line, you have to trim it first.

let my_num: i32 = my_num.trim().parse()
   .expect("please give me correct string number!");

With a recent nightly, you can do this:

let my_int = from_str::<int>(&*my_string);

What's happening here is that String can now be dereferenced into a str. However, the function wants an &str, so we have to borrow again. For reference, I believe this particular pattern (&*) is called "cross-borrowing".