Folks, the confusion, IMHO, is pretty simple: the difference between
returning a parameter and the parameter passed in to the method.
For example, had the code been written this way:
public class Param
{
public static void main(String[] args)
{
String temp = null;
manipParameter(temp);
System.out.println("temp = " + temp);
}
private static void manipParameter(String param)
{
param = "This should be modified";
}
}
the resulting output would be:
temp = null
because Java supports pass-by-value semantics, not pass-by-reference, even
for Object-derived types. (This is where Java's insistence that it has no
pointers really trips people up.) This means that the POINTER to String,
held by temp, is copied into the String reference named "param", and
subsequent modification of "param" has no effect on the original reference
"temp".
For all you C++-heads out there, the difference is one of
void manipParameter(String* pString); // THIS is what Java does, effectively
vs.
void manipParameter(String& pString); // THIS is NOT what Java does
Hence, Shiraz is correct in that the *parameter*, temp, (not the lvalue of
the return value's assignment) will not be modified.
However, because the original code was written to not only use the (wrongly)
assumed pass-by-reference semantics, but also to copy the return value into
temp:
>> >> temp = process_me(temp);
the code worked as expected.
In short, everybody's right, bur for different reasons. :)
Ted Neward
Patterns/C++/Java/CORBA/EJB/COM-DCOM spoken here
http://www.javageeks.com/~tneward
"I don't even speak for myself; my wife won't let me." --Me
-----Original Message-----
From: Thor Heinrichs-Wolpert <[EMAIL PROTECTED]>
To: [EMAIL PROTECTED] <[EMAIL PROTECTED]>
Date: Tuesday, August 10, 1999 7:04 PM
Subject: Re: will this code work?
>ummm ... you're wrong here.
>Yes it will work, as Chris said it would.
>
>It follows the basic right hand operation rules, and does indeed printout
>"my name".
>
>I do think it is convoluted and is not clear as to what is being attempted.
>The function is returning "my name", it is not over-writing the variable
>passed in.
>
>Passing in "temp" to the function does nothing, but then, it isn't being
>used to do anything inside of the function.
>
>Thor HW
>----- Original Message -----
>From: Shiraz Wasim Zaidi <[EMAIL PROTECTED]>
>To: <[EMAIL PROTECTED]>
>Sent: Thursday, August 12, 1999 7:06 PM
>Subject: Re: will this code work?
>
>
>> Hi!,
>>
>> No, It wont work....
>>
>> The result will be null . Method parameters variables are created
>> when a method is invoked and its value is initialized with the method
>> argument.
>> i.e copy of actual argument is passed not the actual argument.
>>
>> Shiraz
>>
>> -----Original Message-----
>> From: Chris Pratt <[EMAIL PROTECTED]>
>> To: [EMAIL PROTECTED] <[EMAIL PROTECTED]>
>> Date: Thursday, August 12, 1999 7:32 PM
>> Subject: Re: will this code work?
>>
>>
>> >Yes it will. But that is an awefully convoluted way of doing
>> >out.println("my name"); At this point I would suggest you read through
>the
>> >Java Tutorial on Sun's site, it's free and it will teach you these
>> >rudimentary things.
>> > (*Chris*)
>> >
>> >----- Original Message -----
>> >From: Duke Martin <[EMAIL PROTECTED]>
>> >To: <[EMAIL PROTECTED]>
>> >Sent: Thursday, August 12, 1999 4:57 PM
>> >Subject: will this code work?
>> >
>> >
>> >> String temp = null;
>> >> temp = process_me(temp);
>> >>
>> >> out.println(temp);
>> >>
>> >> public String process_me(String temp)
>> >> {
>> >> temp = "my name";
>> >> return temp;
>> >> }
>> >>
>> >> I would like to this the output of this servlet to print "my name" to
>the
>> >> screen. Will this work?
>> >>
>> >> thanks
>> >>
>> >>
>>
>>__________________________________________________________________________
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>> >
>>
>>__________________________________________________________________________
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>>
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