let secret_num = rand::thread_rng().gen_range(1, 101);
But In the official documentation, the code is
let secret_num = rand::thread_rng().gen_range(1..=100);

Why is it returning an range instead of an integer.
Full Code
use std::io;
use std::cmp::Ordering;
use rand::Rng;
use colored::*;
fn main() {
let secret_num = rand::thread_rng().gen_range(1..=100);
println!("The secret number is {}", secret_num);
println!("Guess the Number!");
loop {
println!("Please input you guess: ");
let mut guess = String::new();
io::stdin()
.read_line(&mut guess)
.expect("Failed to read line");
println!("You guessed: {}", guess);
let guess: u32 = match guess.trim().parse() {
Ok(num) => num,
Err(_) => continue,
};
match guess.cmp(&secret_num) {
Ordering::Less => println!("{}", "Too small!".red()),
Ordering::Greater => println!("{}", "Too big!".red()),
Ordering::Equal => {
println!("{}", "You win!".green());
break;
},
}
}
}
You're looking at documentation for [email protected] while you're using some older version.
The API differs in that the former takes a Range
fn gen_range<T, R>(&mut self, range: R) -> T
where
T: SampleUniform,
R: SampleRange<T>,
{
//...
}
but the older version you're using expects a low and high bound.
fn gen_range<T: SampleUniform, B1, B2>(&mut self, low: B1, high: B2) -> T
where
B1: SampleBorrow<T> + Sized,
B2: SampleBorrow<T> + Sized,
{
//...
}
In your case rust-analyzer or whatever you use infers T to be a Range because the first argument low is of type B1: SampleBorrow<T> + Sized, (something that can be borrowed and returns a reference to T aka Range here). If you read the further error it should also tell you that it's still missing a second parameter high.
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