为什么Collections.swap会复制输入列表？ [英] Why does Collections.swap copy the input list?

问题描述

在JDK1.6的源代码中，Collections类的swap方法如下所示：

In the source for JDK1.6, the Collections class' swap method is written like this:

public static void swap(List<?> list, int i, int j) {
    final List l = list;
    l.set(i, l.set(j, l.get(i)));
}

创建传递列表的最终副本的原因是什么？为什么他们不直接修改传递的列表？在这种情况下，您还会获得原始类型警告。

What reason is there for creating a final copy of the passed list? Why don't they simply modify the passed list directly? In this case, you also get the raw type warning.

推荐答案

没有列表的副本，只有一个副本对列表的引用。最终关键字并不重要。但是，使用原始类型很重要。如果使用该参数，编译器将报告错误：

There is no copy of the list, there is only a copy of the reference to the list. The final keyword is not important. However, it is important that a raw type is used. If the parameter would be used instead, the compiler would report an error:

public static void swap(List<?> list, int i, int j) {
    // ERROR: The method set(int, capture#3-of ?) in the type List<capture#3-of ?>
    // is not applicable for the arguments (int, capture#4-of ?)
    list.set(i, list.set(j, list.get(i)));
}

这意味着，他们正在使用中间变量来规避泛型的缺点，并摆脱错误信息。

This means, that they are using the intermediate variable to circumvent the shortcomings of generics, and to get rid of the error message.

有趣的问题是：为什么他们不使用通用方法？以下代码有效：

The interesting question is: Why don't they use a generic method? The following code works:

public static <T> void swap(List<T> list, int i, int j) {
    list.set(i, list.set(j, list.get(i)));
}

答案是，此方法在旧代码中生成警告，调用方法原始类型：

The answer is, that this method produces warnings in old code invoking the method with raw types:

List list = ...;
// WARNING: Type safety: Unchecked invocation swap2(List, int, int)
// of the generic method swap2(List<T>, int, int) of type Swap
Collections.swap(list, 0, 1);

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