As I read from various Java book and tutorials, variables declared in a interface are constants and can't be overridden.
I made a simple code to test it
interface A_INTERFACE
{
int var=100;
}
class A_CLASS implements A_INTERFACE
{
int var=99;
//test
void printx()
{
System.out.println("var = " + var);
}
}
class hello
{
public static void main(String[] args)
{
new A_CLASS().printx();
}
}
and it prints out var = 99
Is var get overridden? I am totally confused. Thank you for any suggestions!
Thank you very much everyone! I am pretty new to this interface thing. "Shadow" is the key word to understand this. I look up the related materials and understand it now.
It is not overridden, but shadowed, with additional confusion because the constant in the interface is also static.
Try this:
A_INTERFACE o = new A_CLASS();
System.out.println(o.var);
You should get a compile-time warning about accessing a static field in a non-static way.
And now this
A_CLASS o = new A_CLASS();
System.out.println(o.var);
System.out.println(A_INTERFACE.var); // bad name, btw since it is const
You did not override the variable, you shadowed it with a brand-new instance variable declared in a more specific scope. This is the variable printed in your printx method.
Default signature for any variable in an interface is
public static final ...
So you cannot override it anyhow.
The variable you declared in that interface is not visible to the class that implemented it.
If you declare a variable in an static and final, i.e. a constant, THEN it is visible to implementors.
来源:https://stackoverflow.com/questions/8814153/overriding-interfaces-variable