Refun-20241007
🏷 ☺️Refun
今天展示一下Rust 高阶函数示例,以后的代码风格会尽量函数式:
fn is_odd(n: u32) -> bool {
n % 2 == 1
}
fn main() {
println!("Find the sum of all the numbers with odd squares under 1000");
let upper = 1000;
// Imperative approach
// Declare accumulator variable
let mut acc = 0;
// Iterate: 0, 1, 2, ... to infinity
for n in 0.. {
// Square the number
let n_squared = n * n;
if n_squared >= upper {
// Break loop if exceeded the upper limit
break;
} else if is_odd(n_squared) {
// Accumulate value, if it's odd
acc += n_squared;
}
}
println!("imperative style: {}", acc);
// Functional approach
let sum_of_squared_odd_numbers: u32 =
(0..).map(|n| n * n) // All natural numbers squared
.take_while(|&n_squared| n_squared < upper) // Below upper limit
.filter(|&n_squared| is_odd(n_squared)) // That are odd
.sum(); // Sum them
println!("functional style: {}", sum_of_squared_odd_numbers);
}
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