pure virtual function

[jackel7777]

 #include <iostream>
  2 
  3 using namespace std;
  4 
  5 
  6 class base
  7 {
  8 protected:
  9         base(int i){val = i;cout<<"base"<<endl;}
 10         int val;
 11 public:
 12         virtual void show() = 0;
 13 
 14 };
 15 
 16 class derived1:virtual public base
 17 {
 18 public:
 19         derived1(int i):base(i){cout<<"derived1"<<endl;}
 20         void show(){cout<<hex<<val<<endl;}
 21 };
 22 
 23 class derived2:virtual public base
 24 {
 25 public:
 26         derived2(int i):base(i){cout<<"derived2"<<endl;}
 27         void show(){cout<<oct<<val<<endl;}
 28 
 29 };
 30 
 31 
 32 class derived3:public derived1,public derived2
 33 {
 34 public:
 35         derived3(int i):derived1(i),derived2(i),base(i){cout<<"derived3"<<endl;}
 36         void show(){cout<<val<<endl;}
 37 
 38 };
 39 
 40 
 41 int main()
 42 {
 43         derived1 d1(100);
 44         derived2 d2(120);
 45         cout<<"************************"<<endl;
 46         derived3 d3(12);
 47 
 48 
 49         d1.show();
 50         d2.show();
 51         d3.show();
 52 
 53 return 0;
 54 }

base
derived1
base
derived2
************************
base
derived1
derived2
derived3
64
170
14

why does d3.show() gives the octal result and not the decimal result?

[D80DSL3A]

Because show() is not virtual in derived2.

[firedraco]

IIRC, virtual is automatically there if the function is defined virtual in the base class...but maybe the virtual inheritance changes it.

[simeonz]

The reason is that the effect from the oct manipulator is permanent. You have to use the dec manipulator in derived3.

[despicable]

Yes simeonz is correct, the stream manipulator in C++ is permanent , so we need to set it explicitly if we have to use any other.

[jackel7777]

my reply to fun2code: show() is virtual in all derived classes as in lieu with inheritance.
To the rest: simeonz is correct abt the effect of the oct manipulator. Thanks.