The parameter right was declared const. That means the object is const
inside the operator function. But, const objects can only call const
member functions because const objects must have a guarantee that they
won't be changed. Since the operator<() function wasn't defined as
const.
a const object can't call the function. The reason is that every class
method has an invisible parameter: the this pointer. The this pointer
can implicity or explicity be used inside the function to change the
object, e.g.:
setPrivateVar(10);
or
this->setPrivateVar(10);
When you declare a member function const, the compiler will ensure that
no changes can be made to the calling object.
posted on 2010-05-22 13:53
桂湖山 阅读(409)
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