ba9807fd   
   From: no.spam.here@its.invalid   
      
   On 3/15/2012 12:07 AM, Michael wrote:   
   > This has probably been asked before but...   
   >   
   > Is there a particular reason why when you define an operator ==()   
   > for a type:   
   >   
   > struct SomeType   
   > {   
   >    bool operator== (const SomeType&  rhs) const { return ...; }   
   > };   
   >   
   > ... that the compiler doesn't also go ahead and implicitly define   
   > operator !=() for you?   
   >   
   > This omission has become a pet peeve of mine lately...   
   >   
      
   Probably because operator== doesn't have to return a bool, or even   
   mean "equality operator".  You can use it for some other purpose,   
   much in the same way that operator<< doesn't do a shift for ostreams.   
      
   To be honest, that would probably be bad form, and horrendous to read,   
   but it's perfectly legal to do so.   
      
      
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