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The previous expression returns the int value of finalDollars
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1 What is the difference between a (binary) operator and a function 2 Suppose you wish to overload the operator < so that it applies to the type Money defined in Display 81 What do you need to add to the definition of Money given in Display 81 3 Is it possible using operator overloading to change the behavior of + on integers Why or why not
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s RETURNING BY const VALUE
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Notice the returned types in the declarations for overloaded operators for the class Money in Display 81 For example, the following is the declaration for the overloaded plus operator as it appears in Display 81:
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const Money operator +(const Money& amount1, const Money& amount2);
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This subsection explains the const at the start of the line But before we discuss that rst const, let s make sure we understand all the other details about returning a value So, let s rst consider the case where that const does not appear in either the declaration or
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de nition of the overloaded plus operator Let s suppose that the declaration reads as follows:
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Money operator +(const Money& amount1, const Money& amount2);
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and let s see what we can do with the value returned When an object is returned, for example, (m1 + m2), where m1 and m2 are of type Money, the object can be used to invoke a member function, which may or may not change the value of the member variables in the object (m1 + m2) For example,
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(m1 + m2)output( );
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is perfectly legal In this case, it does not change the object (m1 + m2) However, if we omitted the const before the type returned for the plus operator, then the following would be legal and would change the values of the member variables of the object (m1 + m2):
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(m1 + m2)input( );
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So, objects can be changed, even when they are not associated with any variable One way to make sense of this is to note that objects have member variables and thus have some kinds of variables that can be changed Now let s assume that everything is as shown in Display 81; that is, there is a const before the returned type of each operator that returns an object of type Money For example, below is the declaration for the overloaded plus operator as it appears in Display 81:
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const Money operator +(const Money& amount1, const Money& amount2);
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The rst const on the line is a new use of the const modi er This is called returning a value as const or returning by const value or returning by constant value What the const modi er means in this case is that the returned object cannot be changed For example, consider the following code:
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Money m1(1099), m2(2357); (m1 + m2)output( );
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The invocation of output is perfectly legal because it does not change the object (m1 + m2) However, with that const before the returned type, the following will produce a compiler error message:
(m1 + m2)input( );
Why would you want to return by const value It provides a kind of automatic error checking When you construct (m1 + m2), you do not want to inadvertently change it At rst this protection from changing an object may seem like too much protection, since you can have
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Money m3; m3 = (m1 + m2);
and you very well may want to change m3 No problem the following is perfectly legal:
m3 = (m1 + m2); m3input( );
The values of m3 and (m1 + m2) are two different objects The assignment operator does not make m3 the same as the object (m1 + m2) Instead, it copies the values of the member variables of (m1 + m2) into the member variables of m3 With objects of a class, the default assignment operator does not make the two objects the same object, it only copies values of member variables from one object to another object This distinction is subtle but important It may help you understand the details if you recall that a variable of a class type and an object of a class type are not the same thing An object is a value of a class type and may be stored in a variable of a class type, but the variable and the object are not the same thing In the code
m3 = (m1 + m2);
the variable m3 and its value (m1 + m2) are different things, just as n and 5 are different things in
int n = 5;