Thread: How to protect a parameter beeing modified

Hi,

I'm a total beginner with Java - coming from C++
In the code snippet below, I want to protect the parameter 'V" in method display
from beeing modified. How can this be achieved (the attribute 'final" doesn't seem to work).
Many thanks for a hint,
Helmut.

class Test {
public static void main(String[] args) {
int[] A = new int[3];
for (int i=0; i < A.length;i++) A[i]= 7*i;
display(A);
System.out.println(A[0]);
}

static void display( final int[] V) {
V[0]=17; // WHY NO ERROR here <========================== ?
for (int i=0; i < V.length;i++)
System.out.println(i+" "+V[i]);
}
}

Originally Posted by John Joe

What error you expect here ?

I thought that declaring the parameter V as final would make any assignments to any element of the array V illegal.
If this does not work how can I protect a parameter of a method from being modified?
In C++ I would declare this parameter as
const Array& V
which would make any assignments to any element of V an error at compile time.
Helmut

Originally Posted by John Joe

Have a looked on this : https://stackoverflow.com/questions/...utable-in-java
final has no effect on that

Many thanks for explaining this.

FLAMES ON
But let me say that this is fundamental flaw of Java.
The same is with the lack of "constant" or "final" methods.
/FLAMES OFF