Author Topic:   operators - Strings
eram
ranch hand 		posted March 19, 2000 02:52 PM             
Q) Which operators are overloaded for String Objects ?

I think
+
+= and
.(dot operator)

Any corrections , please.
But the BOOne exam answers only + , += How?

maha anna
bartender 		posted March 19, 2000 03:10 PM             
eram,
In Java the foll are separators.
( ) { } [ ] ; , .
So .(dot) is a separator. It is NOT an operator.
regds
maha anna

eram
ranch hand 		posted March 19, 2000 03:13 PM             
Maha,

So this means obj.equals(obj1) , here the '.' is a seperator and not an operator.
Thanks

maha anna
bartender 		posted March 19, 2000 03:22 PM             
yes. Simillarly in the pacakge and import statements also .(dot) acts as a separator. And when we refer a member of a class like you mentioned.
package myPack1.myPack2.myPack3;
import java.util.*;
regds
maha anna

[This message has been edited by maha anna (edited March 19, 2000).]

Jane Rozen
ranch hand 		posted March 19, 2000 03:27 PM             
Eram, "." IS considered Object member access OPERATOR.

maha anna
bartender 		posted March 19, 2000 03:35 PM             
Jane, From which source , you say so.
please check here in JLS specifically 3.11 section
regds
maha anna

Jim Yingst
sheriff 		posted March 19, 2000 03:59 PM             
Even if the . is considered an operator (and I think M.A. is correct that it isn't), I don't see how it is "overloaded" for Strings. Specific methods may be overloaded, but the . itself just what it always does for objects, right?

Jane Rozen
ranch hand 		posted March 19, 2000 04:24 PM             
Acc. to David Flanagan("Java in a Nutshell"):
quote:

There are five language constructs in Java that sometimes considered operators and sometimes considered simply part of the basic language syntax...
1. object member access(.)
2. array element access([])
3. method invocation (())
4. object creation(new)
5. type conversion or casting(())

I did not try to imply that "." is overloaded for String, just that it sometimes is considered to be a special type of operator.

Jim Yingst
sheriff 		posted March 19, 2000 04:46 PM             
Sorry Jane - I didn't mean to imply that you had said otherwise; I just was going back to eram's original question.

Off-topic alert: the following is not going to come up on an exam, so don't worry about it unless you're interested.

I like Flanagan's Java in a Nutshell (at least, the earlier parts that aren't completely superfluous compared to the online API) but in this case I'm inclined to think the JLS outranks him. I wonder who exactly is using this alternate terminology? C/C++ programmers who haven't read the JLS perhaps? Or perhaps there are documents from Sun somewhere that perpetuate this usage - after all, Sun's JDK is written in C, and the people who made it could well have used C terminology when describing it, unaware that the JLS had amended it in this case. Oh well...

eram
ranch hand 		posted March 19, 2000 05:08 PM             
Thanks to all.
I think I would rather go by JLS.

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