Meaning of ((CaseInsensitiveString) o).s in Effective java example

Question

I was going through 'equal' method concept in effective java, and there I found below code:

@Override
public boolean equals(Object o) {
    if (o instanceof CaseInsensitiveString)
        return s.equalsIgnoreCase(((CaseInsensitiveString) o).s);
    if (o instanceof String)  // One-way interoperability!            
        return s.equalsIgnoreCase((String) o);
    return false;
}

Here I am not able to get particular line that is ((CaseInsensitiveString) o).s . Now what I understand from this piece of code is object 'o' is been typecast to CaseInsensitiveString Class. Now what does ).s mean.

Don't format large pieces of code with backticks. Indent by four spaces instead. I fixed that, but you should try to fix the other formatting problems as well. — Carcigenicate, Oct 26 '17 at 13:26
The `s` is just some type specific field of that particular class — EpicPandaForce, Oct 26 '17 at 13:28

score 5 · Answer 1 · answered Oct 26 '17 at 13:28

5

You can't read the ).s in isolation:

return s.equalsIgnoreCase(((CaseInsensitiveString) o).s);

Is like:

CaseInsensitiveString c = (CaseInsensitiveString) o;
return s.equalsIgnoreCase(c.s);

It's just accessing a field from the instance of CastInsensitiveString.

answered Oct 26 '17 at 13:28

Andy Turner

137,514
11
162
243

2

And OP's confusion about what `.s` means is a great example of why you never have public members with names like `s`. If it were a more meaningful name, it might have been more obvious. – Mike Harris Oct 26 '17 at 13:36

score 2 · Accepted Answer · answered Oct 26 '17 at 13:28

2

The class CaseInsensitiveString has a member s of type String.

To access that member, you need to

cast the Object o to CaseInsensitiveString using ((CaseInsensitiveString) o)
then access the field s using ((CaseInsensitiveString) o).s.

answered Oct 26 '17 at 13:28

f1sh

11,489
3
25
51

Meaning of ((CaseInsensitiveString) o).s in Effective java example

2 Answers2