交换使用并行分配数组元素 [英] Swapping array elements using parallel assignment

问题描述

这个问题，我打了一下用数组和方法调用的并行任务。因此，这里是一个典型例子，试图交换两个成员的数组，其值：

Intrigued by this question, I have played a bit with parallel assignment with arrays and method calls. So here's an paradigmatic example, trying to swap two members in an array, by their value:

deck = ['A', 'B', 'C']
#=> ["A", "B", "C"]
deck[deck.index("A")], deck[deck.index("B")] = deck[deck.index("B")], deck[deck.index("A")]
#=> ["B", "A"]
deck
#=> ["A", "B", "C"]

阵列没有改变。但是，如果我们改变参数的顺序，它的工作原理：

The array hasn't changed. But if we change the order of arguments, it works:

deck[deck.index("B")], deck[deck.index("A")] = deck[deck.index("A")], deck[deck.index("B")]
#=> ["A", "B"]
deck
#=> ["B", "A", "C"]

我想这与调用分配中的首页方法的顺序做，但不能看清楚。可有人请解释下事物的秩序，为什么第一个例子不交换成员，第二个呢？

I guess it has to do with the order of calling the index methods within the assignment, but not see it clearly. Can someone please explain the order of things underneath, and why the first example doesn't swap the member, and second does?

推荐答案

据预计。它遵循从红宝石如何评估前pressions。

It is expected. It follows from how ruby evaluates expressions.

deck[deck.index("A")], deck[deck.index("B")] = deck[deck.index("B")], deck[deck.index("A")]

隐含

deck[deck.index("A")], deck[deck.index("B")] = 'B', 'A'

的注：此处的字符串'A'和'B'仅用于说明。 Ruby没有在这里创造新的字符串对象的。这本质上是：

deck[deck.index("A")] = 'B' -> deck[0] = 'B' (deck = ['B', 'B', 'C'])
deck[deck.index("B")] = 'A' -> deck[0] = 'A' (deck = ['A', 'B', 'C'])

阵列＃指数收益率时，发现第一个匹配。

现在，

deck[deck.index("B")], deck[deck.index("A")] = deck[deck.index("A")], deck[deck.index("B")]
-> deck[deck.index("B")], deck[deck.index("A")] = 'A', 'B'
-> deck[deck.index("B")] = 'A' -> deck[1] = 'A' (deck = ['A', 'A', 'C'])
-> deck[deck.index("A")] = 'B' -> deck[0] = 'B' (deck = ['B', 'A', 'C'])

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