6. What is the output of this program?
   

   #include 
       using namespace std;
       int main()
       {
           int num = 10;
           void *ptr = #
           int *ptr0 = static_cast(ptr);
           cout << *ptr0;
           return 0;
       }
   

1. 1. 10

   
   
   

   2. Compilation Error

   
   
   

   3. Runtime Error

   
   
   

   4. Garbage value

   
   
   

   5. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: A

   We just casted this from void to int, so it prints 10.

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7. What we can’t do on a void pointer?

1. 1. pointer functions

   
   
   

   2. pointer arithmetic

   
   
   

   3. Both A and B

   
   
   

   4. pointer static

   
   
   

   5. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: B

   Because the void pointer is used to cast the variables only, So pointer arithmetic can’t be done in a void pointer.

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8. What is the output of this program?
   

   #include 
       using namespace std;
       int main()
       {
           int p = 10, add;
           void *ptr = &p;
           double q = 2.24;
           ptr = &q;
           add = p + q;
           cout << add << '\n' << ptr;
           return 0;
       }
   

1. 1. 2.24

   
   
   

   2. 10

   
   
   

   3. 2.24
      10

   
   
   

   4. 0x7115378f4d48
      12

   
   
   

   5. 12
      0x7115378f4d48

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: E

   In this program, we are just adding the two values and printing it.

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9. The void pointer can point to which type of objects?

1. 1. float

   
   
   

   2. double

   
   
   

   3. int

   
   
   

   4. All of above

   
   
   

   5. None of these

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   Because it doesn’t know the type of object it is pointing to, So it can point to all objects.

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10. What is the output of this program?
    

    #include 
        using namespace std;
        int main()
        {
            int num;
            char ch;
            void *value;
            num = 5;
            ch = 'd';
            value = #
            cout << "The value points to the integer value " << value << endl;
            value = &ch;
            cout << "The value now points to the character " << value;
            return 0;
        }
    

1. 1. The value points to the integer value 0x7c9edb4e77ec

   
   
   

   2. The value now points to the character 0x7c9edb4e77eb

   
   
   

   3. Compilation Error

   
   
   

   4. The value points to the integer value 0x7c9edb4e77ec
      The value now points to the character 0x7c9edb4e77eb

   
   
   

   5. 5

   
   
   

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1. View Hint View Answer Discuss in Forum

   NA

   Correct Option: D

   Because the data points to the address value of the variables only, So it is printing the memory address of these two variable.

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