Output of C++ Program | Set 10

　　Predict the output of following C++ programs.

Question 1

 1 #include<iostream>
 2 #include<string.h>
 3 using namespace std;
 4
 5 class String
 6 {
 7     char *p;
 8     int len;
 9 public:
10     String(const char *a);
11 };
12
13 String::String(const char *a)
14 {
15     int length = strlen(a);
16     p = new char[length +1];
17     strcpy(p, a);
18     cout << "Constructor Called " << endl;
19 }
20
21 int main()
22 {
23     String s1("Geeks");
24     const char *name = "forGeeks";
25     s1 = name;
26     return 0;
27 }

　　Output:

　　Constructor called
　　Constructor called
　　The first line of output is printed by statement “String s1(“Geeks”);” and the second line is printed by statement “s1 = name;”. The reason for the second call is, a single parameter constructor also works as a conversion operator.

Question 2

 1 #include<iostream>
 2
 3 using namespace std;
 4
 5 class A
 6 {
 7 public:
 8     virtual void fun()
 9     {
10         cout << "A" << endl ;
11     }
12 };
13 class B: public A
14 {
15 public:
16     virtual void fun()
17     {
18         cout << "B" << endl;
19     }
20 };
21 class C: public B
22 {
23 public:
24     virtual void fun()
25     {
26         cout << "C" << endl;
27     }
28 };
29
30 int main()
31 {
32     A *a = new C;
33     A *b = new B;
34     a->fun();
35     b->fun();
36     return 0;
37 }

　　Output:

　　C
　　B
　　A base class pointer can point to objects of children classes. A base class pointer can also point to objects of grandchildren classes. Therefor, the line “A *a = new C;” is valid. The line “a->fun();” prints “C” because the object pointed is of class C and fun() is declared virtual in both A and B. The second line of output is printed by statement “b->fun();”.

　　

　　Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

　　转载请注明：http://www.cnblogs.com/iloveyouforever/

　　2013-11-27　　15:56:12