Output of C++ Program

　　Difficulty Level: Rookie

　　Predict the output of below C++ programs.

　　Question 1

 1 #include<iostream>

 2 using namespace std;

 3

 4 int x = 10;

 5 void fun()

 6 {

 7     int x = 2;

 8     {

 9         int x = 1;

10         cout << ::x << endl;

11     }

12 }

13

14 int main()

15 {

16     fun();

17     return 0;

18 }

　　Output: 10
　　If Scope Resolution Operator is placed before a variable name then the global variable is referenced. So if we remove the following line from the above program then it will fail in compilation.

1 int x = 10;

　　Question 2

 1 #include<iostream>

 2 using namespace std;

 3

 4 class Point

 5 {

 6 private:

 7     int x;

 8     int y;

 9 public:

10     Point(int i, int j);  // Constructor

11 };

12

13 Point::Point(int i = 0, int j = 0)

14 {

15     x = i;

16     y = j;

17     cout << "Constructor called";

18 }

19

20 int main()

21 {

22     Point t1, *t2;

23     return 0;

24 }

　　Output: Constructor called.
　　If we take a closer look at the statement “Point t1, *t2;:” then we can see that only one object is constructed here. t2 is just a pointer variable, not an object.

　　Question 3

 1 #include<iostream>

 2 using namespace std;

 3

 4 class Point

 5 {

 6 private:

 7     int x;

 8     int y;

 9 public:

10     Point(int i = 0, int j = 0);    // Normal Constructor

11     Point(const Point &t);            // Copy Constructor

12 };

13

14 Point::Point(int i, int j)

15 {

16     x = i;

17     y = j;

18     cout << "Normal Constructor called\n";

19 }

20

21 Point::Point(const Point &t)

22 {

23     y = t.y;

24     cout << "Copy Constructor called\n";

25 }

26

27 int main()

28 {

29     Point *t1, *t2;

30     t1 = new Point(10, 15);

31     t2 = new Point(*t1);

32     Point t3 = *t1;

33     Point t4;

34     t4 = t3;        //assignment operator

35     return 0;

36 }

　　Output:
　　Normal Constructor called
　　Copy Constructor called
　　Copy Constructor called
　　Normal Constructor called

　　See following comments for explanation:

1 Point *t1, *t2;             // No constructor call

2 t1 = new Point(10, 15);  // Normal constructor call

3 t2 = new Point(*t1);     // Copy constructor call

4 Point t3 = *t1;             // Copy Constructor call

5 Point t4;                 // Normal Constructor call

6 t4 = t3;                 // Assignment operator call

　　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　　15:16:34

巴特西

Output of C++ Program | Set 4

最新文章

热门文章