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Option 1: exercises/options/options1.rs
题目
rust
// options1.rs
//
// Execute `rustlings hint options1` or use the `hint` watch subcommand for a
// hint.
// I AM NOT DONE
// This function returns how much icecream there is left in the fridge.
// If it's before 10PM, there's 5 pieces left. At 10PM, someone eats them
// all, so there'll be no more left :(
fn maybe_icecream(time_of_day: u16) -> Option<u16> {
// We use the 24-hour system here, so 10PM is a value of 22 and 12AM is a
// value of 0 The Option output should gracefully handle cases where
// time_of_day > 23.
// TODO: Complete the function body - remember to return an Option!
???
}
#[cfg(test)]
mod tests {
use super::*;
#[test]
fn check_icecream() {
assert_eq!(maybe_icecream(9), Some(5));
assert_eq!(maybe_icecream(10), Some(5));
assert_eq!(maybe_icecream(23), Some(0));
assert_eq!(maybe_icecream(22), Some(0));
assert_eq!(maybe_icecream(25), None);
}
#[test]
fn raw_value() {
// TODO: Fix this test. How do you get at the value contained in the
// Option?
let icecreams = maybe_icecream(12);
assert_eq!(icecreams, 5);
}
}
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题目解析
先来看函数maybe_icecream
的注释,了解到这个函数用于返回冰箱里面还剩多少冰淇淋:
- 如果是晚上10点之前,剩下5件
- 晚上10点以后,有人会吃掉它们,所以不会有剩余的。
- 参数
time_of_day
采用24小时制
再来看单元测试的代码:
- 9点返回5,10点返回5,23点返回0,22点返回0
- 由于是24小时的,25不是一个合法的时间,返回了None。
rust
fn maybe_icecream(time_of_day: u16) -> Option<u16> {
// We use the 24-hour system here, so 10PM is a value of 22 and 12AM is a
// value of 0 The Option output should gracefully handle cases where
// time_of_day > 23.
// TODO: Complete the function body - remember to return an Option!
if time_of_day < 22 {
Some(5)
} else if time_of_day <= 23 {
Some(0)
} else {
None
}
}
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对于第二段单元测试的代码,由于maybe_icecream
返回的是Option<u16>
类型,如果要跟5比较,需要unwrap()
rust
#[test]
fn raw_value() {
// TODO: Fix this test. How do you get at the value contained in the
// Option?
let icecreams = maybe_icecream(12);
let icecreams = maybe_icecream(12).unwrap();
assert_eq!(icecreams, 5);
}
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