Output of C++ Program | Set 4

Difficulty Level: Rookie

Predict the output of below C++ programs.

Question 1

#include<iostream> 
using namespace std; 
  
int x = 10; 
void fun() 
{ 
    int x = 2; 
    { 
        int x = 1; 
        cout << ::x << endl;  
    } 
} 
  
int main() 
{ 
    fun(); 
    return 0; 
}

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.

  int x = 10;

Question 2

#include<iostream> 
using namespace std; 
class Point { 
private: 
    int x; 
    int y; 
public: 
    Point(int i, int j);  // Constructor 
}; 
  
Point::Point(int i = 0, int j = 0)  { 
    x = i; 
    y = j; 
    cout << "Constructor called"; 
} 
  
int main() 
{ 
   Point t1, *t2; 
   return 0; 
} 

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

#include<iostream> 
using namespace std; 
  
class Point { 
private: 
    int x; 
    int y; 
public: 
    Point(int i = 0, int j = 0);    // Normal Constructor 
    Point(const Point &t); // Copy Constructor 
}; 
  
Point::Point(int i, int j)  { 
    x = i; 
    y = j; 
    cout << "Normal Constructor called\n"; 
} 
  
Point::Point(const Point &t) { 
   y = t.y; 
   cout << "Copy Constructor called\n"; 
} 
  
int main() 
{ 
   Point *t1, *t2; 
   t1 = new Point(10, 15); 
   t2 = new Point(*t1); 
   Point t3 = *t1; 
   Point t4; 
   t4 = t3; 
   return 0; 
} 

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

See following comments for explanation:

Point *t1, *t2;   // No constructor call 
t1 = new Point(10, 15);  // Normal constructor call 
t2 = new Point(*t1);   // Copy constructor call  
Point t3 = *t1;  // Copy Constructor call 
Point t4;   // Normal Constructor call 
t4 = t3;   // Assignment operator call 

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