英文:
Can we access the static member of a class without initializing obects?
问题
public class Test{
static Another a; // 不初始化也可以访问Another类的静态成员
public static void main(String[] args){
System.out.println(a.i);
}
}
class Another{
public static int i=20;
}
这里我可以在不初始化对象的情况下访问类Another的静态成员，系统显示的输出结果是20。
public class Test{
public static void main(String[] args){
Another a; // 但是当我在main方法内部声明时，系统会抛出错误
System.out.println(a.i);
}
}
class Another{
public static int i=20;
}
Test.java:5: error: variable a might not have been initialized
System.out.println(a.i);
^
但是当我将声明移到main代码块内部时，系统会报错。为什么会这样呢？
英文:
public class Test{
static Another a;// without initializing i can able to access static member of Another class
public static void main(String[] args){
System.out.println(a.i);
}
}
class Another{
public static int i=20;
}
Here I can able to the access the static member of class Another without initializing the object and the system is displaying the output as 20.
public class Test{
public static void main(String[] args){
Another a;// but when i declare this inside the main method system is throwing error
System.out.println(a.i);
}
}
class Another{
public static int i=20;
}
Test.java:5: error: variable a might not have been initialized
System.out.println(a.i);
^
But when i move the declaration inside main block system throws an error. why is that?
答案1
得分: 1
以下是翻译好的部分：
public static void main(String[] args){
Another i
System.out.println(a.i);
}
可以替换为
public static void main(String[] args){
System.out.println(Another.i);
}
您之所以出现 **错误** 并不是因为静态变量 `i`，而是因为局部方法变量 `Another a` 没有被初始化 **，由于您没有初始化变量a，函数内部的局部变量在读取其值之前必须被初始化。
英文:
public static void main(String[] args){
Another i
System.out.println(a.i);
}
this can be replaced with
public static void main(String[] args){
System.out.println(Another.i);
}
You are getting error not because the static variable i but **because of local method variable Another a is not initialized ** , Since you have not initialized a, the local variable inside a function must be intialized befor reading its value.
答案2
得分: 0
你应该使用Another.i来访问static变量i，而不是a.i。
如果a有值，a.i是有效的语法，但这会令人困惑（因为任何阅读你代码的人可能错误地认为i是实例变量）。如果在访问时a未初始化，编译器是不允许的。
这两个代码片段的区别在于这里（当a是一个static变量）：
static Another a;
a有一个默认值null。
而在这里（当a是一个局部变量）：
Another a;
它没有默认值。
英文:
You should access the static variable i with Another.i, not a.i.
a.i is valid syntax if a has a value, but it's confusing (since anyone who reads you code may falsely assume i is an instance variable). And if a is not initialized when you access it, the compiler doesn't allow it.
The difference between the two snippets is that here (when a is a static variable):
static Another a;
a has a default value of null.
While here (when a is a local variable):
Another a;
it has no default value.
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