XStream within the REPL

I don't know about changing name of class in REPL but you could alias
it in xstream :

xstream.alias("F", classOf[F])

/davidB

On Fri, Jan 8, 2010 at 05:32, turicum wrote:
>
> Dear Scala-ers,
>
> I need a suggestion regarding deserializing classes using XStream within the
> REPL
>
> Let's suppose that, in the REPL, I define
>
> class F { var a = 10 }
>
> and then I try the following:
>
> val xml = "  33  "
> xstream.fromXML(xml).asInstanceOf[F]
>
> an exception will be thrown because XStream does not know anything of class
> F. In fact, an instance of class F, when it's defined within the REPL, is
> serialized to the following XML:
>
>   10  
>
> Is there a way to change the name the REPL assigns to classes defined
> on-the-go, so to avoid this problem?
>
> Thanks!
> Alex
>
> --
> View this message in context: http://old.nabble.com/XStream-within-the-REPL-tp27071261p27071261.html
> Sent from the Scala mailing list archive at Nabble.com.
>
>

It works! Thanks!
Alex

David Bernard-3 wrote:
>
> I don't know about changing name of class in REPL but you could alias
> it in xstream :
>
> xstream.alias("F", classOf[F])
>
> /davidB
>
> On Fri, Jan 8, 2010 at 05:32, turicum wrote:
>>
>> Dear Scala-ers,
>>
>> I need a suggestion regarding deserializing classes using XStream within
>> the
>> REPL
>>
>> Let's suppose that, in the REPL, I define
>>
>> class F { var a = 10 }
>>
>> and then I try the following:
>>
>> val xml = "  33  "
>> xstream.fromXML(xml).asInstanceOf[F]
>>
>> an exception will be thrown because XStream does not know anything of
>> class
>> F. In fact, an instance of class F, when it's defined within the REPL, is
>> serialized to the following XML:
>>
>>   10  
>>
>> Is there a way to change the name the REPL assigns to classes defined
>> on-the-go, so to avoid this problem?
>>
>> Thanks!
>> Alex
>>
>> --
>> View this message in context:
>> http://old.nabble.com/XStream-within-the-REPL-tp27071261p27071261.html
>> Sent from the Scala mailing list archive at Nabble.com.
>>
>>
>
>

new problem now: it does not work in an embedded interpreter:

import scala.tools.nsc._
val interp = new Interpreter(new Settings)
val code = ""class F { var a = 10 };"
interp.compileString(code)
interp.bind("xml", "String", """ 33 """)
interp.interpret("""
val xstream = new XStream
println(xstream.toXML(new F)) // this works: 10 (but
different from what I get in the REPL)
xstream.alias("F", classOf[F]) // this is fine
val g = xstream.fromXML(xml)) // throws XStream's
CannotResolveClassException: F : F
""")

Adding

xstream.setClassLoader((new F).getClass.getClassLoader)

solves the problem,
also, in the REPL, but not in the embedded interpreter

xstream.setClassLoader(classOf[F].getClass.getClassLoader)

works too.

Is there a more elegant solution?

Thanks!
Alex

David Bernard-3 wrote:
>
> I don't know about changing name of class in REPL but you could alias
> it in xstream :
>
> xstream.alias("F", classOf[F])
>
> /davidB
>
> On Fri, Jan 8, 2010 at 05:32, turicum wrote:
>>
>> Dear Scala-ers,
>>
>> I need a suggestion regarding deserializing classes using XStream within
>> the
>> REPL
>>
>> Let's suppose that, in the REPL, I define
>>
>> class F { var a = 10 }
>>
>> and then I try the following:
>>
>> val xml = "  33  "
>> xstream.fromXML(xml).asInstanceOf[F]
>>
>> an exception will be thrown because XStream does not know anything of
>> class
>> F. In fact, an instance of class F, when it's defined within the REPL, is
>> serialized to the following XML:
>>
>>   10  
>>
>> Is there a way to change the name the REPL assigns to classes defined
>> on-the-go, so to avoid this problem?
>>
>> Thanks!
>> Alex
>>
>> --
>> View this message in context:
>> http://old.nabble.com/XStream-within-the-REPL-tp27071261p27071261.html
>> Sent from the Scala mailing list archive at Nabble.com.
>>
>>
>
>

I don't known about your classloader issue. but it's better to set
alias (and other xstream initialisation) before any call to toXml(...)
or fromXML(...).

On Fri, Jan 8, 2010 at 07:46, turicum wrote:
>
> new problem now: it does not work in an embedded interpreter:
>
> import scala.tools.nsc._
> val interp = new Interpreter(new Settings)
> val code = ""class F { var a = 10 };"
> interp.compileString(code)
> interp.bind("xml", "String", """  33  """)
> interp.interpret("""
>    val xstream = new XStream
>    println(xstream.toXML(new F)) // this works:  10   (but
> different from what I get in the REPL)
>    xstream.alias("F", classOf[F]) // this is fine
>    val g = xstream.fromXML(xml)) // throws XStream's
> CannotResolveClassException: F : F
>    """)
>
> Adding
>
> xstream.setClassLoader((new F).getClass.getClassLoader)
>
> solves the problem,
> also, in the REPL, but not in the embedded interpreter
>
> xstream.setClassLoader(classOf[F].getClass.getClassLoader)
>
> works too.
>
> Is there a more elegant solution?
>
> Thanks!
> Alex
>
>
> David Bernard-3 wrote:
>>
>> I don't know about changing name of class in REPL but you could alias
>> it in xstream :
>>
>> xstream.alias("F", classOf[F])
>>
>> /davidB
>>
>> On Fri, Jan 8, 2010 at 05:32, turicum wrote:
>>>
>>> Dear Scala-ers,
>>>
>>> I need a suggestion regarding deserializing classes using XStream within
>>> the
>>> REPL
>>>
>>> Let's suppose that, in the REPL, I define
>>>
>>> class F { var a = 10 }
>>>
>>> and then I try the following:
>>>
>>> val xml = "  33  "
>>> xstream.fromXML(xml).asInstanceOf[F]
>>>
>>> an exception will be thrown because XStream does not know anything of
>>> class
>>> F. In fact, an instance of class F, when it's defined within the REPL, is
>>> serialized to the following XML:
>>>
>>>   10  
>>>
>>> Is there a way to change the name the REPL assigns to classes defined
>>> on-the-go, so to avoid this problem?
>>>
>>> Thanks!
>>> Alex
>>>
>>> --
>>> View this message in context:
>>> http://old.nabble.com/XStream-within-the-REPL-tp27071261p27071261.html
>>> Sent from the Scala mailing list archive at Nabble.com.
>>>
>>>
>>
>>
>
> --
> View this message in context: http://old.nabble.com/XStream-within-the-REPL-tp27071261p27071974.html
> Sent from the Scala mailing list archive at Nabble.com.
>
>

using
xstream.setClassLoader(this.getClass.getClassLoader)
does the job!
Alex

turicum wrote:
>
> new problem now: it does not work in an embedded interpreter:
>
> import scala.tools.nsc._
> val interp = new Interpreter(new Settings)
> val code = ""class F { var a = 10 };"
> interp.compileString(code)
> interp.bind("xml", "String", """ 33 """)
> interp.interpret("""
> val xstream = new XStream
> println(xstream.toXML(new F)) // this works: 10 (but
> different from what I get in the REPL)
> xstream.alias("F", classOf[F]) // this is fine
> val g = xstream.fromXML(xml)) // throws XStream's
> CannotResolveClassException: F : F
> """)
>
> Adding
>
> xstream.setClassLoader((new F).getClass.getClassLoader)
>
> solves the problem,
> also, in the REPL, but not in the embedded interpreter
>
> xstream.setClassLoader(classOf[F].getClass.getClassLoader)
>
> works too.
>
> Is there a more elegant solution?
>
> Thanks!
> Alex
>
>
> David Bernard-3 wrote:
>>
>> I don't know about changing name of class in REPL but you could alias
>> it in xstream :
>>
>> xstream.alias("F", classOf[F])
>>
>> /davidB
>>
>> On Fri, Jan 8, 2010 at 05:32, turicum wrote:
>>>
>>> Dear Scala-ers,
>>>
>>> I need a suggestion regarding deserializing classes using XStream within
>>> the
>>> REPL
>>>
>>> Let's suppose that, in the REPL, I define
>>>
>>> class F { var a = 10 }
>>>
>>> and then I try the following:
>>>
>>> val xml = "  33  "
>>> xstream.fromXML(xml).asInstanceOf[F]
>>>
>>> an exception will be thrown because XStream does not know anything of
>>> class
>>> F. In fact, an instance of class F, when it's defined within the REPL,
>>> is
>>> serialized to the following XML:
>>>
>>>   10  
>>>
>>> Is there a way to change the name the REPL assigns to classes defined
>>> on-the-go, so to avoid this problem?
>>>
>>> Thanks!
>>> Alex
>>>
>>> --
>>> View this message in context:
>>> http://old.nabble.com/XStream-within-the-REPL-tp27071261p27071261.html
>>> Sent from the Scala mailing list archive at Nabble.com.
>>>
>>>
>>
>>
>
>