   Overloading Operators  Question...What about normal operator implementation
  In learning C++ I have come across Overloading operators.  At first I was thinking that it was strictly because you "can" manipulate the meaning for an operator so when it is used with a class you are actually calling a member function to perform some operation.However, I have two question since learning about overloading operators:1) what happens when you just use the operatore in your program expecting it to behave as the default representation?  What happens when you want to use that operatore that you have overloaded with class objects and want to use it in as the default representation and not be calling a member function?2) Is there any "must use" circumstances that you have to define a overload operator for a class or is it strictly for objective reasons?Thanks in advance,   
Quote:1) what happens when you just use the operatore in your program expecting it to behave as the default representation? What happens when you want to use that operatore that you have overloaded with class objects and want to use it in as the default representation and not be calling a member function?Well most operators you overload don't have a default definition with classes, so the overloaded version is the only version.  In the cases where you do have a default definition (e.g. =) I don't think you can use it if you've overloaded it.Quote:2) Is there any "must use" circumstances that you have to define a overload operator for a class or is it strictly for objective reasons?In the STL there's various constraints on the types you use with various containers and functions, and in some cases you do have to have an overloaded operator defined.  E.g. the std::sort function requires that the elements you're sorting have a < operator defined (otherwise it doesn't know what order to sort them in).   ;  You misunderstand operator overloading. The goal of operator overloading is to define the operator's behavior (ie the way it operates), not to change its meaning. = is not +, and + is not /. Moreover, as soon as you quit the realm of artithmetic types, most arithmetic operators become undefined. You can't add two structures? You can't multiply them. If you want to do so, you have two possibilities: writing a function of overloading the corresponding operator.Eitherway, it is your choice to overload the operator - it is never a requirement, unless you use a library that needs it. Most of the time, a function or a member function is just as good as any operator overload. HTH,;Quote:Original post by MonderQuote:2) Is there any "must use" circumstances that you have to define a overload operator for a class or is it strictly for objective reasons?In the STL there's various constraints on the types you use with various containers and functions, and in some cases you do have to have an overloaded operator defined.  E.g. the std::sort function requires that the elements you're sorting have a < operator defined (otherwise it doesn't know what order to sort them in).Not *required*; you can also tell it by providing a comparison function. However, if you want to use an object for "the comparison function" (note that this is a *good* idea generally), that object in turn needs to have an overloaded operator defined: the 'operator()' (which allows you to "call" the object as if it were a function). This is particular useful when there are several, unrelated, meaningful ways to sort the objects. (If they're related, you can sometimes take shortcuts: e.g. if you want descending order instead of ascending, you can use 'std::greater<your object type>' for the comparison function.)   