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Super class method

 
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How is the output as "SuperSuperBase" ? why second invocation calls Super method again?
[ July 24, 2008: Message edited by: amarkirt saroay ]
 
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Because it's and overloading, not an overriding...
The parameters of the methods are of diferent types. So, since its an overloading, its decided on compile time, not run time :-)

See these 2 classes:
 
amarkirt saroay
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Thanks ! got it.
 
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