将数组引用分配给Java中的另一个数组

问题描述

int a[]={1, 2, 3, 4, 5};
int b[]={4, 3, 2, 1, 0};

a=b;

System.out.println("a[0] = "+a[0]);

这显然显示了 a [0] = 4 ，因为为 a 分配了对 b 的引用.

This displays a[0] = 4 as obvious because a is assigned a reference to b.

如果进行了如下修改

int a[]={1, 2, 3, 4, 5};
int b[]={4, 3, 2, 1, 0};

System.out.println("a[(a=b)[0]] = "+a[(a=b)[0]]);  //<-------

然后，它显示 a [(a = b)[0]] = 5 .

为什么不使用此表达式- a [(a = b)[0]] 产生 4 ，即第0个元素 b ，即使它看起来与先前的情况相同?

Why doesn't this expression - a[(a=b)[0]] yield 4, the 0 element of b even though it appears to be the same as the previous case?

推荐答案

第二个表达式在数组索引器表达式内部具有赋值表达式.该表达式的计算结果如下:

The second expression features an assignment expression inside an array indexer expression. The expression evaluates as follows:

• 已选择索引器表达式的目标.那就是原始数组 a
• 首先通过将 b 分配给 a ，然后在索引0处获取 b 的元素来计算索引表达式
• 外部索引器的索引评估为 b [0]
• 将索引应用于 a ，并返回 5
• 将 b 分配给 a 生效.随后对 a [i] 的访问将引用 b ，而不是原始的 a .

• The target of the indexer expression is selected. That's the original array a
• The index expression is evaluated by first assigning b to a, and then taking the element of b at index zero
• The index of the outer indexer is evaluated as b[0]
• The index is applied to a, returning 5
• The assignment of b to a takes effect. Subsequent accesses to a[i] will reference b, not the original a.

本质上，您的单行表达式等同于以下两行代码段:

Essentially, your single-line expression is equivalent to this two-line snippet:

System.out.println("a[(a=b)[0]] = "+a[b[0]]); // Array reference completes first
a=b;                                       // Array assignment is completed last

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