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int n = (2 + 3);
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It may take you a while to become comfortable with this notion of return by const value In the meantime, a good rule of thumb is to always return class types by const value unless you have an explicit reason not to do so For most simple programs this will have no effect on your program other than to ag some subtle errors Note that although it is legal, it is pointless to return basic types, such as int, by const value The const has no effect in the case of basic types When a function or operator returns a value of one of the basic types, such as int, double, or char, it returns the value, such as 5, 55, or A It does not return a variable or anything like a variable1 Unlike a variable, the value cannot be changed you cannot change 5 Values of a basic type cannot be changed whether there is a const before the returned type or not On the other hand, values of a class type that is, objects can be changed, since they have member variables, and so the const modi er has an effect on the object returned
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Unless the value returned is returned by reference, but return by reference is a topic covered later in this chapter Here we assume the value is not returned by reference
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Operator Overloading, Friends, and References
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RETURNING MEMBER VARIABLES
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OF A
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CLASS TYPE
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When returning a member variable of a class type, in almost all cases it is important to return the member value by const value To see why, suppose you do not, as in the example outlined in what follows:
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class Employee { public: Money getSalary( ) { return salary; } private: Money salary; };
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In this example, salary is a private member variable that should not be changeable except by using some accessor function of the class Employee However, this privateness is easily circumvented as follows:
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Employee joe; (joegetSalary( ))input( );
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The lucky employee named joe can now enter any salary she wishes! On the other hand, suppose getSalary returns its value by const value, as follows:
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class Employee { public: const Money getSalary( ) { return salary; } private: Money salary; };
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In this case, the following will give a compiler error message
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(joegetSalary( ))input( );
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(The declaration for getSalary should ideally be
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const Money getSalary( ) const { return salary; }
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but we did not want to confuse the issue with another kind of const)
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Self-Test Exercises
4 Suppose you omit the const at the beginning of the declaration and definition of the overloaded plus operator for the class Money, so that the value is not returned by const value Is the following legal
Money m1(1099), m2(2357), m3(1234); (m1 + m2) = m3;
Is it legal if the de nition of the class Money is as shown in Display 81, so that the plus operator returns its value by const value
s OVERLOADING UNARY OPERATORS
In addition to the binary operators, such as + in x + y, C++ has unary operators, such as the operator - when it is used to mean negation A unary operator is an operator that takes only one operand (one argument) In the statement below, the unary operator - is used to set the value of a variable x equal to the negative of the value of the variable y:
x = -y;
unary operator
The increment and decrement operators, ++ and --, are other examples of unary operators You can overload unary operators as well as binary operators For example, we have overloaded the minus operator - for the type Money (Display 81) so that it has both a unary and a binary operator version of the subtraction/negation operator - For example, suppose your program contains this class de nition and the following code:
Money amount1(10), amount2(6), amount3;
Then the following sets the value of amount3 to amount1 minus amount2:
amount3 = amount1 - amount2;
The following will, then, output $400 to the screen:
amount3output( );
On the other hand, the following will set amount3 equal to the negative of amount1:
amount3 = -amount1;
The following will, then, output -$1000 to the screen:
amount3output( );