Java中static变量的初始化顺序

Java中static变量的初始化顺序

package test;

class Bowl {
Bowl(int marker) {
System.out.println("Bowl(" + marker + ")");
}
void f(int marker) {
System.out.println("f(" + marker + ")");
}
}

class Table {
static Bowl b1 = new Bowl(1);
Table() {
System.out.println("Table()");
b2.f(1);
}
void f2(int marker) {
System.out.println("f2(" + marker + ")");
}
static Bowl b2 = new Bowl(2);
}

class Cupboard {
Bowl b3 = new Bowl(3);
static Bowl b4 = new Bowl(4);
Cupboard() {
System.out.println("Cupboard()");
b4.f(2);
}
void f3(int marker) {
System.out.println("f3(" + marker + ")");
}
static Bowl b5 = new Bowl(5);
}

public class StaticInitialization {
public static void main(String[] args) {
System.out.println(
"Creating new Cupboard() in main");
new Cupboard();
System.out.println(
"Creating new Cupboard() in main");
new Cupboard();
t2.f2(1);
t3.f3(1);
}
static Table t2 = new Table();
static Cupboard t3 = new Cupboard();
} ///:~

输出结果：

Bowl(1)

Bowl(2)

Table()

f(1)

Bowl(4)

Bowl(5)

Bowl(3)

Cupboard()

f(2)

Creating new Cupboard() in main

Bowl(3)

Cupboard()

f(2)

Creating new Cupboard() in main

Bowl(3)

Cupboard()

f(2)

f2(1)

f3(1)

分析：：进入main时，发现两个static变量，对于第一个Table型，进入Table类中，发现又有两个static的Bowl类，所以刚开始的顺序为Bowl(1)

Bowl(2) 然后就是Table的构造函数，输出Table() f(1) 接下来是static的cupboard类，进入cupboard类，发现有两个static的bowl类，所以顺序为Bowl(4)

Bowl(5) 对于bowl3,由于它不是static的，所以它的顺序靠后，进入cupboard的构造函数，输出Cupboard() f(2)

接下来就是一个输出语句 又来一个new cupboard，不过它是一般形式，所以进入cupboard类，输出bowl(3)，这是因为它是一般类型的，不是static,对于Static类型的，它只会初始化一次。接下来输出Cupboard(）f(2) 输出一个输出语句。又来一个new cupboard，一般形式，后面的就好理解了。