Java是“通过引用传递”吗？或“按值传递”？ [英] Is Java "pass-by-reference" or "pass-by-value"?

问题描述

我一直以为Java 传递参考。

I always thought Java was pass-by-reference.

但是，我看过几篇博文（例如，此博客）声称它不是。

However, I've seen a couple of blog posts (for example, this blog) that claim that it isn't.

我认为我不理解他们所做的区别。

I don't think I understand the distinction they're making.

解释是什么？

推荐答案

Java总是通过按值即可。不幸的是，他们决定将对象的位置称为引用。当我们传递一个对象的值时，我们将引用传递给它。这对初学者来说很困惑。

Java is always pass-by-value. Unfortunately, they decided to call the location of an object a "reference". When we pass the value of an object, we are passing the reference to it. This is confusing to beginners.

它是这样的：

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    // we pass the object to foo
    foo(aDog);
    // aDog variable is still pointing to the "Max" dog when foo(...) returns
    aDog.getName().equals("Max"); // true
    aDog.getName().equals("Fifi"); // false 
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // change d inside of foo() to point to a new Dog instance "Fifi"
    d = new Dog("Fifi");
    d.getName().equals("Fifi"); // true
}

在上面的例子中 aDog.getName （）仍将返回Max。函数 foo main 中的值 aDog 未更改>使用 Dog Fifi作为对象引用按值传递。如果它是通过引用传递的，那么 main 中的 aDog.getName（）将返回拨打 foo 后，Fifi。

In the example above aDog.getName() will still return "Max". The value aDog within main is not changed in the function foo with the Dog "Fifi" as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.

同样：

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    foo(aDog);
    // when foo(...) returns, the name of the dog has been changed to "Fifi"
    aDog.getName().equals("Fifi"); // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // this changes the name of d to be "Fifi"
    d.setName("Fifi");
}

在上面的示例中， Fifi 是调用 foo（aDog）后的狗名，因为对象的名称是在 foo（...）。 foo 在 d 上执行的任何操作都是这样的，出于所有实际目的，它们在<$ c $上执行c> aDog 本身（当 d 更改为指向不同的 Dog 实例时除外喜欢 d =新狗（Boxer））。

In the above example, Fifi is the dog's name after call to foo(aDog) because the object's name was set inside of foo(...). Any operations that foo performs on d are such that, for all practical purposes, they are performed on aDog itself (except when d is changed to point to a different Dog instance like d = new Dog("Boxer")).

这篇关于Java是“通过引用传递”吗？或“按值传递”？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！