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method parameter binding

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  method parameter binding Philipp Kraus <philipp.kraus.flashpixx@gmail.com> - 2015-12-10 07:44 -0800
    Re: method parameter binding Eric Douglas <e.d.programmer@gmail.com> - 2015-12-10 07:53 -0800
      Re: method parameter binding Jan Burse <janburse@fastmail.fm> - 2015-12-10 17:14 +0100
        Re: method parameter binding Eric Douglas <e.d.programmer@gmail.com> - 2015-12-10 08:20 -0800
          Re: method parameter binding Jan Burse <janburse@fastmail.fm> - 2015-12-10 17:23 +0100
            Re: method parameter binding Jan Burse <janburse@fastmail.fm> - 2015-12-10 17:23 +0100
        Re: method parameter binding Jan Burse <janburse@fastmail.fm> - 2015-12-10 17:21 +0100
          Re: method parameter binding Jan Burse <janburse@fastmail.fm> - 2015-12-10 17:27 +0100
            Re: method parameter binding Jan Burse <janburse@fastmail.fm> - 2015-12-10 17:28 +0100
              Re: method parameter binding Eric Sosman <esosman@comcast-dot-net.invalid> - 2015-12-10 11:35 -0500
                Re: method parameter binding Martin Gregorie <martin@address-in-sig.invalid> - 2015-12-10 18:00 +0000
            Re: method parameter binding Eric Douglas <e.d.programmer@gmail.com> - 2015-12-10 08:41 -0800
              Re: method parameter binding j4n bur53 <janburse@fastmail.fm> - 2015-12-10 19:29 +0100
    Re: method parameter binding Eric Sosman <esosman@comcast-dot-net.invalid> - 2015-12-10 11:33 -0500
    Re: method parameter binding Jan Burse <janburse@fastmail.fm> - 2015-12-10 22:34 +0100
      Re: method parameter binding Eric Sosman <esosman@comcast-dot-net.invalid> - 2015-12-10 17:09 -0500
        Re: method parameter binding j4n bur53 <janburse@fastmail.fm> - 2015-12-11 09:27 +0100

#33327 — method parameter binding

Hello,

I have got a class with two methods like:

Number get( Double x, Double y )
{
    return new Double(x.doubleValue() + y.doubleValue());
}

Number get( Long x, Long y )
{
    return new Long(x.longValue() + y.longValue());
}

Number get( Number x, Number y )
{
   throw new IllegalStateException("test");
}


If I run

this.get( new Long/Double(3), new Long/Double(4) );

it calls everytime the method with the exception definition. IMHO the compiler should bind the method to the exact parameter type and if the compiler cannot match the generic type bind should be used. 
In C++  I want expect this binding structure, so can I do an equal behaviour with Java?

I have changed the Number to a generic type: <T extends Number> Number get( T x, T y )
and get also the exception.

Can you explain me, why the last/bottom method is used?

Thanks

Phil

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#33328

On Thursday, December 10, 2015 at 10:44:55 AM UTC-5, Philipp Kraus wrote:
> Hello,
> 
> I have got a class with two methods like:
> 
> Number get( Double x, Double y )
> {
>     return new Double(x.doubleValue() + y.doubleValue());
> }
> 
> Number get( Long x, Long y )
> {
>     return new Long(x.longValue() + y.longValue());
> }
> 
> Number get( Number x, Number y )
> {
>    throw new IllegalStateException("test");
> }
> 
> 
> If I run
> 
> this.get( new Long/Double(3), new Long/Double(4) );
> 
> it calls everytime the method with the exception definition. IMHO the compiler should bind the method to the exact parameter type and if the compiler cannot match the generic type bind should be used. 
> In C++  I want expect this binding structure, so can I do an equal behaviour with Java?
> 
> I have changed the Number to a generic type: <T extends Number> Number get( T x, T y )
> and get also the exception.
> 
> Can you explain me, why the last/bottom method is used?
> 
> Thanks
> 
> Phil

I pasted your method overloads exactly in Eclipse, then added static in front so I could call them from main(), then put this line in my main method:
System.out.println(get(new Double(5), new Double(4)));
My output was:
9.0

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#33329

Hi,

Eric Douglas schrieb:
> I pasted your method overloads exactly in Eclipse, then added static in front so I could call them from main(), then put this line in my main method:
> System.out.println(get(new Double(5), new Double(4)));
> My output was:
> 9.0

Static method resolution is slightly different than
dynamic method resolution.

In dynamic method resolution a sub class could still
have overridden the method and thrown the exception.

Need more code...

Bye

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#33330

On Thursday, December 10, 2015 at 11:15:10 AM UTC-5, Jan Burse wrote:
> Static method resolution is slightly different than
> dynamic method resolution.
> 
> In dynamic method resolution a sub class could still
> have overridden the method and thrown the exception.
> 
> Need more code...
> 
> Bye

I'm not sure I get what you're saying.  Certainly if it's an override it should choose that.

class One {
Number get( Double x, Double y ) 
{ 
    return new Double(x.doubleValue() + y.doubleValue()); 
} 

Number get( Long x, Long y ) 
{ 
    return new Long(x.longValue() + y.longValue()); 
} 
}
Class Two extends One {
Number get( Number x, Number y ) 
{ 
   throw new IllegalStateException("test"); 
} 
}

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#33332

Eric Douglas schrieb:
> Class Two extends One {
> Number get( Number x, Number y )
> {
>     throw new IllegalStateException("test");
> }
> }

If you override get( Double x, Double y )
or get( Long x, Long y ) you can also throw
an exception there.

For static methods, you can override like
that, I mean you would write Two.get and
not One.get, but this is not overriding,
for static methods if you write One.get
you get One.get.

For dynamic methods if you write x.get and
x is of type One, you might also get
Two.get invoked.

Bye

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#33333

Jan Burse schrieb:
> For static methods, you can override like
can --> cannot

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#33331

Jan Burse schrieb:
> Need more code...

Works fine, as expected:

Main Method:

     public static void main(String[] args) {
         Test test = new Test();
         System.err.println("test.get(1L, 2L)="+test.get(1L, 2L));
         System.err.println("test.get(1.0, 2.0)="+test.get(1.0, 2.0));
         System.err.println("test.get(1, 2)="+test.get(1, 2));
     }

Results:

     test.get(1L, 2L)=3
     test.get(1.0, 2.0)=3.0
     Exception in thread "main" java.lang.IllegalStateException: test
	at protect.Test.get(Test.java:20)
	at protect.Test.main(Test.java:27)

In the last case the signature is Integer,Integer and
it only finds the throwing get method, since Number is
super class of Integer.

Bye

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#33334

Jan Burse schrieb:
>          System.err.println("test.get(1L, 2L)="+test.get(1L, 2L));
>          System.err.println("test.get(1.0, 2.0)="+test.get(1.0, 2.0));
>          System.err.println("test.get(1, 2)="+test.get(1, 2));

I put on purpose System.err.println and not System.out.println,
since my IDE, IntelliJ, since some recent release scrambles
err and out, so that it doesn't show in the order the
println were emitted.

Very annoying, sometimes I don't find which exceptions
matches to which step in the program.

Bye

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#33335

Jan Burse schrieb:
> Very annoying, sometimes I don't find which exceptions
> matches to which step in the program.

Maybe some f*** up Logger that sits somewhere inbetween.

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#33337

On 12/10/2015 11:28 AM, Jan Burse wrote:
> Jan Burse schrieb:
>> Very annoying, sometimes I don't find which exceptions
>> matches to which step in the program.
>
> Maybe some f*** up Logger that sits somewhere inbetween.

     (Topic drift ...)

     Could it be that System.out and System.err are two different
and independent PrintStreams, each with its own buffering and
making its own decisions about when to flush buffers?

     Naaah, too simple ...

-- 
esosman@comcast-dot-net.invalid
"Don't be afraid of work. Make work afraid of you." -- TLM

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#33341

On Thu, 10 Dec 2015 11:35:48 -0500, Eric Sosman wrote:

>      Could it be that System.out and System.err are two different
> and independent PrintStreams, each with its own buffering and making its
> own decisions about when to flush buffers?
> 
>      Naaah, too simple ...
>
That's certainly the case in the UNIX/Linux world, where stdout, which 
System.out maps to, is quite extensively buffered while stderr, which 
System.err uses, is unbuffered. 

Normally it all works as expected if you only use stdout or stderr, but 
use both, e.g. for debugging, and the actual output from mixed stdout and 
stderr print statements does not necessarily appear in the same order as 
the statements were executed. As Jan says, this can be confusing, 
especially if you're capturing the output in a single stream using 
something like:

   myprog 2>&1 |less 

because you know the output is much bigger than one screen's worth and 
you want to search it or page through it using the 'less' utility.
 
Similarly, in C you can output a prompt with fprintf(stderr, "....") and 
follow it immediately with a read to get user input, but if you used print
("....") or fprintf(stdout, "....") you need to call fflush(stdout) to 
make sure the prompt has been displayed before attempting to read the 
user input.

I don't recall whether Windows does the same.
 

-- 
martin@   | Martin Gregorie
gregorie. | Essex, UK
org       |

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#33338

On Thursday, December 10, 2015 at 11:27:35 AM UTC-5, Jan Burse wrote:
> Jan Burse schrieb:
> >          System.err.println("test.get(1L, 2L)="+test.get(1L, 2L));
> >          System.err.println("test.get(1.0, 2.0)="+test.get(1.0, 2.0));
> >          System.err.println("test.get(1, 2)="+test.get(1, 2));
> 
> I put on purpose System.err.println and not System.out.println,
> since my IDE, IntelliJ, since some recent release scrambles
> err and out, so that it doesn't show in the order the
> println were emitted.
> 
> Very annoying, sometimes I don't find which exceptions
> matches to which step in the program.
> 
> Bye

I've only ever used Eclipse for Java code and never had such problem.  Do you use IntelliJ for any particular reason other than personal preference?

Everywhere in my custom API that I need to print a standard out message for logging I just call my own custom log class, so I can just edit that class if I want to hide or redirect all messages, and I can add common text to every message.  From there I call Thread.currentThread().getStackTrace() so I can show which class, which method, which line number the message came from.

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#33342

Eric Douglas schrieb:
> I've only ever used Eclipse for Java code and never
 > had such problem. Do you use IntelliJ for any particular
 > reason other than personal preference?

I saw the problem recently on IntelliJ 14.1.3, if
you execute a main() inside the IDE. But it could also
persist to other IDEs, Consoles, etc..

It seems there are or have been even more problems
than only interleaving, when using err and out
at the same time:

Printing to stderr and stdout sometimes gets lost in IDEA 12.1.4
https://youtrack.jetbrains.com/issue/IDEA-110325

Bye

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#33336

On 12/10/2015 10:44 AM, Philipp Kraus wrote:
> Hello,
>
> I have got a class with two methods like:
>
> Number get( Double x, Double y )
> {
>      return new Double(x.doubleValue() + y.doubleValue());
> }
>
> Number get( Long x, Long y )
> {
>      return new Long(x.longValue() + y.longValue());
> }
>
> Number get( Number x, Number y )
> {
>     throw new IllegalStateException("test");
> }
>
>
> If I run
>
> this.get( new Long/Double(3), new Long/Double(4) );
>
> it calls everytime the method with the exception definition. IMHO the compiler should bind the method to the exact parameter type and if the compiler cannot match the generic type bind should be used.
> In C++  I want expect this binding structure, so can I do an equal behaviour with Java?
>
> I have changed the Number to a generic type: <T extends Number> Number get( T x, T y )
> and get also the exception.
>
> Can you explain me, why the last/bottom method is used?

     No, I cannot.  Also, I cannot reproduce your problem.  With:

package esosman.misc;
public class Kraus {

    public static void main(String[] unused) {
       Kraus kraus = new Kraus();
       System.err.println("Doubles: " + kraus.get(new Double(3), new 
Double(4)));
       System.err.println("Longs: " + kraus.get(new Long(3), new Long(4)));
       System.err.println("Integers: " + kraus.get(new Integer(3), new 
Integer(4)));
    }

    Number get(Double x, Double y) {
       return new Double(x.doubleValue() + y.doubleValue());
    }

    Number get(Long x, Long y) {
       return new Long(x.longValue() + y.longValue());
    }

    Number get(Number x, Number y) {
       throw new IllegalStateException("test");
    }
}

... I get the output:

Doubles: 7.0
Longs: 7
Exception in thread "main" java.lang.IllegalStateException: test
	at esosman.misc.Kraus.get(Kraus.java:21)
	at esosman.misc.Kraus.main(Kraus.java:9)

     Are you doing something slightly different from what you've shown?

-- 
esosman@comcast-dot-net.invalid
"Don't be afraid of work. Make work afraid of you." -- TLM

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#33344

Hi,

BTW: Situation changes totally when the argument types
are primitive types, since then the specificity rules are
different and widening applies, I guess.

See here:

Take this example program:

public class Test {

     double get( double x, double y ) {
         return x+y;
     }

     long get( long x, long y ) {
         return x+y;
     }

     Number get( Number x, Number y ) {
         throw new IllegalStateException("test");
     }

     public static void main(String[] args) {
         Test test = new Test();
         System.err.println("test.get(1L, 2L)="+test.get(1L, 2L));
         System.err.println("test.get(1.0, 2.0)="+test.get(1.0, 2.0));
         System.err.println("test.get(1, 2)="+test.get(1, 2));
     }

}

We get the following results:

test.get(1L, 2L)=3
test.get(1.0, 2.0)=3.0
test.get(1, 2)=3

So the two integers 1 and 2 are widened to long and
the long, long method is invoked.

Bye


Philipp Kraus schrieb:
> Hello,
>
> I have got a class with two methods like:
>
> Number get( Double x, Double y )
> {
>      return new Double(x.doubleValue() + y.doubleValue());
> }
>
> Number get( Long x, Long y )
> {
>      return new Long(x.longValue() + y.longValue());
> }
>
> Number get( Number x, Number y )
> {
>     throw new IllegalStateException("test");
> }
>
>
> If I run
>
> this.get( new Long/Double(3), new Long/Double(4) );
>
> it calls everytime the method with the exception definition. IMHO the compiler should bind the method to the exact parameter type and if the compiler cannot match the generic type bind should be used.
> In C++  I want expect this binding structure, so can I do an equal behaviour with Java?
>
> I have changed the Number to a generic type: <T extends Number> Number get( T x, T y )
> and get also the exception.
>
> Can you explain me, why the last/bottom method is used?
>
> Thanks
>
> Phil
>

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#33345

On 12/10/2015 4:34 PM, Jan Burse wrote:
> Hi,
>
> BTW: Situation changes totally when the argument types
> are primitive types, since then the specificity rules are
> different and widening applies, I guess.
> [...]

     Unless I've missed something, nobody has yet been able to reproduce
the problem you originally reported:

     - Eric Douglas tried it got the unsurprising result, but his use
       of `static' may have confused things (I think it's a red herring,
       but even so ...),

     - Jan Burse tried it and got the unsurprising result,

     - I tried it and got the unsurprising result.

     Only your reported result was surprising -- and it would be well
to understand why that is so before you go looking for dark corners in
the specificity rules.

-- 
esosman@comcast-dot-net.invalid
"Don't be afraid of work. Make work afraid of you." -- TLM

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#33347

Eric Sosman schrieb:
>
>      Only your reported result was surprising -- and it would be well
> to understand why that is so before you go looking for dark corners in
> the specificity rules.

Yeah, the dark force of specificity. Your friend and enemy.

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