如果返回错误结构的构造函数

Question

我正在创建表示图中边的简单通用结构，并且我希望构造函数切换参数，因此较低的将是第一个。

pub type Index = u16;
pub type Cost = u32;

pub struct Edge<T> {
    first: Index,
    second: Index,
    weight: T,
}

impl<T> Edge<T> {
    pub fn new(first: Index, second: Index, weight: T) -> Self {
        return if second > first {
            Edge { second, first, weight }
        } else {
            Edge { first, second, weight }
        }
    }
}

但是我无法使它工作，因为我的测试导致错误。

 #[test]
    fn should_initialize_with_lover_value_first() {
        // given
        let lower: Index = 234;
        let high: Index = 444;
        let cost: Cost = 34;

        // when
        let edge: Edge<Cost> = Edge::new(high, lower, cost);

        // then
        assert_eq!(edge.first, lower, "edge.first = {}, expected = {}", edge.first, lower);
        assert_eq!(edge.second, high, "edge.second = {}, expected = {}", edge.second, high);
    }

我有错误：

thread 'types::generic::edge::tests::should_initialize_with_lover_value_first' panicked at 'assertion failed: `(left == right)`
  left: `444`,
 right: `234`: edge.first = 444, expected = 234', src/types/generic/edge.rs:35:9

我是否错过了某种我不知道的奇怪的生锈行为？

Answer 1

当变量名称与字段名称相同时，Rust 允许您对变量名称进行双关语。所以这：

Edge { second, first, weight }

相当于：

Edge {
    second: second,
    first: first,
    weight: weight,
}

和这个：

Edge { first, second, weight }

相当于：

Edge {
    first: first,
    second: second,
    weight: weight,
}

如您所见，它们彼此等效。

如果你想翻转first和second那么你就可以做到这一点，而不是：

Edge {
    first: second,
    second: first,
    weight,
}