为什么guava joiner实现了一个私有方法iterable（final Object first，final Object second，final Object [] rest）？

private static Iterable<Object> iterable(
        final Object first, final Object second, final Object[] rest) {
    checkNotNull(rest);
    return new AbstractList<Object>() {
        @Override
        public int size() {
            return rest.length + 2;
        }

        @Override
        public Object get(int index) {
            switch (index) {
                case 0:
                    return first;
                case 1:
                    return second;
                default:
                    return rest[index - 2];
            }
        }
    };
}

What is author's purpose?

作者的目的是什么?

I guess he wants to make use of the array generated by compiler, rather than new an ArrayList.

我猜他想利用编译器生成的数组,而不是新的ArrayList。

But still a confusing point, why not write as below?

但仍然是一个令人困惑的问题,为什么不写下面的?

private static Iterable<Object> iterable(final Object[] rest) {
    checkNotNull(rest);
    return new AbstractList<Object>() {
        @Override
        public int size() {
            return rest.length;
        }

        @Override
        public Object get(int index) {
            return rest[index];
        }
    };
}

1 个解决方案

#1

The point here is that this method is called from public methods which look like (source):

这里的要点是这个方法是从公共方法调用的,它看起来像(source):

  public final String join(
      @NullableDecl Object first, @NullableDecl Object second, Object... rest) {
    return join(iterable(first, second, rest));
  }

Using signatures like this is a trick to force you to pass in at least two arguments - after all, if you've not got two arguments, there is nothing to join.

使用这样的签名是一种强制你传递至少两个参数的技巧 - 毕竟,如果你没有两个参数,就没有什么可以加入的。

For example:

Joiner.on(':').join();  // Compiler error.
Joiner.on(':').join("A");  // Compiler error.
Joiner.on(':').join("A", "B");  // OK.
Joiner.on(':').join("A", "B", "C");  // OK.
// etc.

This iterable method just creates an Iterable without having to copy everything into a new array. Doing so would be O(n) in the number of arguments; the approach taken here is O(1).

这个可迭代的方法只需创建一个Iterable,而不必将所有内容复制到一个新数组中。这样做的参数数量为O(n);这里采用的方法是O(1)。

#1