Re: confusion of array clone() method
Patricia Shanahan wrote:
hyena wrote:
hi,
I just came across this situation that the cloned array linked to
the original array and changes made to the clone also applied to the
original one. I do not think I understand fully the description of
deep clone/shallow clone concept and got a bit confused here.
code:
private int[][] decode(int[][] tr_seq) {
// translate the sequces into real travel times
int[][] tt = (int[][]) tr_seq.clone();
for(int i =0; i < tt.length; i++){
for(int j =0; j < tt[0].length; j++){
tt[i][j] = 100;
}
}
return tt;
}
tr_seq was changed after this method was called. I was assuming a
clone is apart from the original object and has its own reference
address. isn't it?
from java document ,
"Creates and returns a copy of this object. The precise meaning of
"copy" may depend on the class of the object. The general intent is
that, for any object x, the expression:
x.clone() != xwill be true".
this statement does not agree on this example.
could someone shed some light on this? Thanks.
The key issue is what constitutes an array. Java does not really have
multidimensional arrays. tr_seq is an array whose elements are
references to arrays.
clone makes a new array whose elements point to the same arrays as the
original.
For any i from 0 to tt.length-1, tt[i] and tr_seq[i] are identical. They
are either both null, or are pointers to the same array. Changes to an
element of an array they both reference apply regardless of whether you
find that array through tr_seq or tt.
Rereading this, I saw an ambiguity. tt[i] and tr_seq[i] are identical
immediately after the clone call. An assignment e.g. tt[i] = null; could
make them different.
Patricia
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