java - 如果int不继承Object，那为什么 "String.format(String, Object ...)"用int的编译呢？

Question

我读了这篇文章:Is int an object in Java? .

在帖子中，争论的是int 不是从Object 继承的。如果是这样，那么为什么下面的代码编译没有任何错误？鉴于 int 不是 Object 并且 format() 方法的签名是 public static String format(String format, Object.. .args) 如文档所示:javadoc for String !

public class Testing {
    public static void main(String[] args) {
        int integer = 7;
        String str = String.format("%03d", integer);
        System.out.println(str);
    }
}

我也读过“自动装箱”。这到底是什么意思？在编译之前，是否所有原语都被适当的 Object 替换了？如果是这样，那么在使用大型 int (int[]) 数组时，内存使用是否比 Integer ( 整数[])？ double 等也有类似的论点。

欢迎任何见解。

Answer 1

它是由 Autoboxing 引起的.

这是链接的 Java 文档中的一小段，比我能更好地解释它:

Autoboxing is the automatic conversion that the Java compiler makes between the primitive types and their corresponding object wrapper classes. For example, converting an int to an Integer, a double to a Double, and so on. If the conversion goes the other way, this is called unboxing.

Here is the simplest example of autoboxing:

Character ch = 'a';

The rest of the examples in this section use generics. If you are not yet familiar with the syntax of generics, see the Generics (Updated) lesson.

Consider the following code:

List<Integer> li = new ArrayList<>();
for (int i = 1; i < 50; i += 2)
    li.add(i);

Although you add the int values as primitive types, rather than Integer objects, to li, the code compiles. Because li is a list of Integer objects, not a list of int values, you may wonder why the Java compiler does not issue a compile-time error. The compiler does not generate an error because it creates an Integer object from i and adds the object to li. Thus, the compiler converts the previous code to the following at runtime:

List<Integer> li = new ArrayList<>();
for (int i = 1; i < 50; i += 2)
    li.add(Integer.valueOf(i));