the fact that the "->" operator dereferences the value it returns is a quirky
rule of C++. "->" behaves a bit like a unary operator but not completely.
"a->setName()" would resolve to something nonsensical if "(a->)" simply
resolved to its return type.
so it is a C++ issue.
a->setName() as you have seen is the correct way to call functions on the
ref_ptr.
Andy
-----Original Message-----
From: [EMAIL PROTECTED] on behalf of ? ?
Sent: Thu 7/3/2008 10:18 PM
To: [email protected]
Subject: [osg-users] how to ref_ptr -> operator work?
osg::ref_ptr<osg::Node> a = new osg:Node;
a->setName(); //OK
but ref_ptr overload -> operator and return a pointer to osg:Node,
so may write (a->)->setName() is right? WHY?
Thanks !
---------------------------------
????,??????!
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