friend AND inline method, what's the point ?

Question

I see in a header that I didn\'t write myself the following:

class MonitorObjectString: public MonitorObject {
   // some other declarations
   friend inlin         


        

Answer 1

Grammatically speaking...

The friend keyword is still needed to tell the compiler that this function is not a member of a class, EDIT: but instead a non-member function that can see the private members of the class.

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However, this could have been implemented more cleanly like this:

/* friend */ inline bool operator ==(const MonitorObjectString& rhs) const
{ return fVal == rhs.fVal; }

(Of course, I'm assuming fVal is of a suitable type that can be compared without affecting its const-ness.)