为什么我不能用两个i32参数调用gen_range？
我有这个代码，但它不能编译：
use rand::Rng;
use std::io;
fn main() {
println!("Guess the number!");
let secret_number = rand::thread_rng().gen_range(0, 101);
println!("The secret number is: {}", secret_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);
}
编译错误：
use rand::Rng;
use std::io;
fn main() {
println!("Guess the number!");
let secret_number = rand::thread_rng().gen_range(0, 101);
println!("The secret number is: {}", secret_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);
}
回答
该gen_range方法需要一个Range参数，而不是两个i32参数，因此更改：
let secret_number = rand::thread_rng().gen_range(0, 101);
到：
let secret_number = rand::thread_rng().gen_range(0..101);
它会编译并工作。注意：方法签名已在crate版本0.8.0中更新，在randcrate 的所有先前版本中，您的代码应按原样工作。
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