Java中的数组是按引用传递还是按值传递? [英] Are arrays in java pass by reference or pass by value?

问题描述

Java中的数组是按引用传递还是按值传递?(我之前也曾问过类似的问题，但似乎没有一个很好的答案，所以希望这次会有所不同.)

Are arrays in java pass by reference or pass by value? (I've similar questions asked before, but none of them seemed to get a good answer, so hopefully this time will be different).

假设我有一个名为 data 的数组，其中包含某种类型的对象.现在让我们假设我将该数组传递并存储在A类中，然后将其传递给B类，而B类更改了该数组的一项.数组的A类版本会改变吗?它是否是原始数组(例如 int )的数组，这有关系吗?ArrayLists呢?

Suppose I have an array called data that contains Objects of some type. Now let us suppose that I pass and store that array in class A and then I pass it to class B and class B changes one of the entries of the array. Will class A's version of the array change? Does it matter if this was an array of primitives (such as int) instead? What about ArrayLists?

谢谢

推荐答案

Java中的所有内容都是按值传递的.但是，如果要传递参考，则为参考的值.

Everything in Java is pass-by-value. However, if you're passing a reference, it's the value of the reference.

由于Java方法无法到达调用者的堆栈中以重新分配变量，因此任何方法调用都不能在那里更改引用(地址)的标识.这就是我们说Java不是按引用传递时的意思.这与C ++(和类似的语言)形成对比，后者在某些情况下允许这样做.

Since Java methods can't reach into the caller's stack to reassign variables, no method call can change the identity of a reference (address) there. This is what we mean when we say Java is not pass-by-reference. This contrasts with C++ (and similar languages), which allows this in some cases.

现在让我们来看一些效果.

Now let's look at some effects.

如果我这样做:

Object[] o = ...
mutateArray(o);

内容之后可能会有所不同，因为所有 mutateArray 所需的内容都是更改其内容的数组的地址.但是， o 的地址将相同.如果我这样做:

the contents can be different afterwards, since all mutateArray needs is the address of an array to change its contents. However, the address of o will be the same. If I do:

String x = "foo";
tryToMutateString(x);

x 的地址此后再次相同.由于字符串是不可变的，因此这意味着它仍将是"foo" .

the address of x is again the same afterwards. Since strings are immutable, this implies that it will also still be "foo".

对对象进行变异就是更改其内容(例如，成功更改 o 的最后一个元素，或尝试将"foo"的最后一个字母更改为"d").这不应与在调用者堆栈中重新分配 x 或 o (不可能)相混淆.

To mutate an object is to change the contents of it (e.g. successfully changing the last element of o, or trying to change the last letter of "foo" to 'd'). This should not be be confused with reassigning x or o in the caller's stack (impossible).

通过共享进行通话的维基百科部分可能会引起更多关注.

The Wikipedia section on call by sharing may shed additional light.

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