java - 比较 Java 的按值传递与 C++ 的按值传递或引用

Question

Java 对对象和原始类型都使用按值传递。因为Java传递的是引用的值，我们可以改变target的值，但不能改变地址。

这与 C++ 相比如何？

Answer 1

区别在于您是否可以影响调用函数中的变量。

让我们暂时把对象放在一边。在纯粹按值传递的 Java 中，您不能更改传递给被调用函数的变量的调用函数值。示例:

void foo() {
    int a = 42;
    bar(a);
    System.out.println("foo says: " + a);
}
void bar(int a) {
    a = 67;
    System.out.println("bar says: " + a);
}

如果调用 foo，您将看到:

bar says: 67foo says: 42

bar could not change foo's a.

That's true in C++ if you pass by value. However, if you pass by reference, you're passing a reference to the calling code's variable. And that means the called code can change it:

void foo() {
    int a = 42;
    bar(a);
    cout << "foo says: " << a;
}
void bar(int& a) {
    a = 67;
    cout << "bar says: " << a;
}

请注意，bar 被定义为接收一个reference (int& a)。如果你调用 bar，你会看到:

bar says: 67foo says: 67

bar was able to change the value of a within foo.

Okay, so let's deal with object references: First off, note that the word "reference" is being used for two completely different things: A reference to the variable in the calling function (that's the pass-by-reference thing), and a reference to an object. When you pass an object reference into a method in Java, the reference is passed by value, just like everything else is.

void foo() {
    List list = new ArrayList();
    List ref2 = list; // (Let's remember that object reference for later...)
    bar(list);
    System.out.println("foo says: " + list.size());
    System.out.println("foo says: Same list? " + (ref2 == list));
}
void bar(List list) {
    // `bar` can modify the state of the object the reference points to
    list.add(new Object());
    System.out.println("bar says (after add): " + list.size());

    // ...but cannot change `foo`'s copy of `list`
    list = new ArrayList();
    System.out.println("bar says (after new): " + list.size());
}

所以你看到了:

bar says (after add): 1bar says (after new): 0foo says: 1foo says: same list? true

bar 可以改变(按值)传入其引用的对象的状态，但不能改变foo对那个对象。 foo 没有看到创建的新列表 bar。