R help - Nov 2012 - Question on callNextMethod

I don't understand why I get the following results. I define two classes
'Base' and 'Derived', the latter of which 'contains' the
first. I then
define a generic method 'test' and overload it for each of these
classes. I
call 'callNextMethod()' in the overload for Derived. From the output, it
appears that the overload for Base gets called twice. Why is this? Test
code follows:

setClass('Base')
setClass('Derived', contains='Base')
setGeneric('test', function(x) standardGeneric('test'))
setMethod('test', signature(x='Base'), function(x)
print('base called'))
setMethod('test', signature(x='Derived'), function(x)
{print('derived
called'); callNextMethod()})

d = new('Derived')
test(d)


Produces the output:

[1] "derived called"
[1] "base called"
[1] "base called"


and I was expecting:

[1] "derived called"
[1] "base called"


Thanx in advance,
Simon Knapp

	[[alternative HTML version deleted]]

Simon Knapp <sleepingwell at gmail.com> writes:
> I don't understand why I get the following results. I define two
classes
> 'Base' and 'Derived', the latter of which
'contains' the first. I then
> define a generic method 'test' and overload it for each of these
classes. I
> call 'callNextMethod()' in the overload for Derived. From the
output, it
> appears that the overload for Base gets called twice. Why is this? Test
Autoprinting has tricked you.
> a <-test(d) # What you expected
[1] "derived\ncalled"
[1] "base called"
> a # what was autoprinted
[1] "base called"
>
HTH,

Chuck

> code follows:
>
> setClass('Base')
> setClass('Derived', contains='Base')
> setGeneric('test', function(x) standardGeneric('test'))
> setMethod('test', signature(x='Base'), function(x)
print('base called'))
> setMethod('test', signature(x='Derived'), function(x)
{print('derived
> called'); callNextMethod()})
>
> d = new('Derived')
> test(d)
>
>
> Produces the output:
>
> [1] "derived called"
> [1] "base called"
> [1] "base called"
>
>
> and I was expecting:
>
> [1] "derived called"
> [1] "base called"
>
>
> Thanx in advance,
> Simon Knapp
>
> 	[[alternative HTML version deleted]]
>
-- 
Charles C. Berry                            Dept of Family/Preventive Medicine
cberry at ucsd edu			    UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901

On 11/06/2012 07:03 AM, Simon Knapp wrote:
> I don't understand why I get the following results. I define two
classes
> 'Base' and 'Derived', the latter of which
'contains' the first. I then
> define a generic method 'test' and overload it for each of these
classes. I
> call 'callNextMethod()' in the overload for Derived. From the
output, it
> appears that the overload for Base gets called twice. Why is this? Test
> code follows:
>
> setClass('Base')
> setClass('Derived', contains='Base')
> setGeneric('test', function(x) standardGeneric('test'))
> setMethod('test', signature(x='Base'), function(x)
print('base called'))
> setMethod('test', signature(x='Derived'), function(x)
{print('derived
> called'); callNextMethod()})
>
> d = new('Derived')
> test(d)
>
>
> Produces the output:
>
> [1] "derived called"
> [1] "base called"
> [1] "base called"
Fun; I think you're seeing the print command, and also the (normally
invisible)
return value from print

 > d = new('Derived')
 > res = test(d)
[1] "derived called"
[1] "base called"
 > res
[1] "base called"

similar to

 > print("x")
[1] "x"
 > (print("x"))
[1] "x"
[1] "x"

Martin
>
>
> and I was expecting:
>
> [1] "derived called"
> [1] "base called"
>
>
> Thanx in advance,
> Simon Knapp
>
> 	[[alternative HTML version deleted]]
>
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>

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