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/* d3 inherits both d1 and d2 However, now there is only one copy of base in d3 */ class d3 : public d1, public d2 { public: int m; }; int main() { d3 d; di dj dk dm = = = = 10; // no longer ambiguous 20; 30; 40;
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cout << di << " "; // no longer ambiguous cout << dj << " " << dk << " "; cout << dm; C++ return 0; }
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As this examples shows, the keyword virtual precedes the rest of the inherited class s specification Now that both d1 and d2 have inherited base as virtual, any multiple inheritance involving them will cause only one copy of base to be present Therefore, in d3, there is only one copy of base, so di = 10 is perfectly valid and unambiguous One further point to keep in mind: Even though both d1 and d2 specify base as virtual, base is still present in any objects of either type For example, the following sequence is perfectly valid:
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// define a class of type d1 d1 myclass; myclassi = 100;
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Virtual base classes and normal ones differ only when an object inherits the base more than once If virtual base classes are used, only one base class is present in the object Otherwise, multiple copies will be found
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const Member Functions and mutable
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Class member functions may be declared as const, which causes this to be treated as a const pointer Thus, a const function cannot modify the object that invokes it Also, a const object may not invoke a nonconst member function However, a const member function can be called by either const or nonconst objects To specify a member function as const, use the form shown in the following example
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class X { int some_var; public: int f1() const; // const member function };
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As you can see, the const follows the function s parameter declaration The purpose of declaring a member function as const is to prevent it from modifying the object that invokes it For example, consider the following program
/* Demonstrate const member functions This program won't compile */ #include <iostream> using namespace std; class Demo { int i; public: int geti() const { return i; // ok }
void seti(int x) const { i = x; // error! } }; int main() { Demo ob; obseti(1900);
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cout << obgeti(); return 0; }
This program will not compile because seti( ) is declared as const This means that it is not allowed to modify the invoking object Since it attempts to change i, the program is in error In contrast, since geti( ) does not attempt to modify i, it is perfectly acceptable Sometimes there will be one or more members of a class that you want a const function to be able to modify even though you don t want the function to be able to modify any of its other members You can accomplish this through the use of mutable It overrides constness That is, a mutable member can be modified by a const member function For example,
// Demonstrate mutable #include <iostream> using namespace std; class Demo { mutable int i; int j; public: int geti() const { return i; // ok } void seti(int x) const { i = x; // now, OK } /* The following function won't compile void setj(int x) const { j = x; // Still Wrong! } */ }; int main() { Demo ob;
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obseti(1900); cout << obgeti(); return 0; }
Here, i is specified as mutable, so it may be changed by the seti( ) function However, j is not mutable and setj( ) is unable to modify its value
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