Pointer is not a data structure. It is a variable that contains the address of another variable and is not implemented in STL.

Scope resolution operator is represented by :: . Hence, this is the correct answer.

Templates are used to implement generic programming. They are of two types: function templates and class templates. Function templates are methods that can operate on any data type and class template can be used to write single code for multiple data types.

Scope is an enclosing context where values and expressions are associated. The scope resolution operator helps to identify and specifies the context to which an identifier refers. Hence, this is the correct answer.

The operator + is a binary operator used for addition of two elements. Hence, this is the correct answer.

POD means plain old data and is used to describe C style structure which contains only data members.This term as used  in C++ structures,can contain data members as well as member functions. Hence, this is the incorrect answer.

This is correct as all members share same memory location and the size of union is the size of the largest data element.

Friend functions are non-member functions which can access all the private and protected members of the class. An example of a friend function is class abc{ private: int f,s; public: friend int sum(abc x);}; Here, sum function can access even private members of class and is defined below -:int sum(abc x){return x.f+x.s;}.

Inline functions are very useful when we need to create efficient code as their code is expanded each time they are invoked. Inline functions are preceded by inline keyword. For example: inline int findlarge(int x, int y) {return x > y ? x : y;}. Hence, this is the correct answer.

Encapsulation is a mechanism that binds together code and data which it manipulates and keeps both of them safe from outside interference and misuse. Hence, this is the correct answer.

A far pointer is a pointer that can access all the segments of RAM and its size is 4 bytes.

A very simple concept has been used in this program, i.e. name of array basically contains base address and here we are initializing pointer to base address and then setting and incrementing pointer. Therefore, the correct answer is 10, 20, 30, 40, 50. 

It is not possible to create objects of abstract class as it contains pure virtual function which is not defined. This statement is false. Hence, it is the correct answer.

Clog is a built-in stream that refers to the buffered version of cerr. The  above given four options are built in streams. Hence, this is the correct option.

There is no iterator by the name linear in C++.

This option is correct as the only problem with the above function declaration is that a is intialized, but b is not. However, if a is intialized then b should also be initialized as explained in above option.

Bjarne Strourstrup has acknowledged that Simula 67 was the greatest influence on him in the development of C++. Hence, this is the correct option.

Let Initial value be: i It does not affect any thing in the program. Checking condition: ++i It is incrementing the value of i by one in each iteration.  We know in c, zero means false. When value of i is zero, loop will terminate. Increment value: i >= 99 It is not incrementing the value of variable i.

It is clear, loop will terminate when value of variable i is zero. Hence, printf function will print: 1 to 32767 and then, from -32768 to -1   (range of short int) Hence, this option is correct.

Here we have set basefield flag as hex which will display100 in hex and output is 64. Hence, the given option is correct.

The const_cast is used explicitly override const and/or volatile attribute of variable. Hence, this is the correct answer.

In C++, there is no finally keyword, which is defined in Java. Hence, this is the correct answer.

The basic thing that we need to remember in C is that zero means false. Now, the code inside for loop as condition is checked. Variable i is zero since we are using post-increment operator. Hence, loop is terminated and after that 1 is printed which is the final value of i. Therefore, this is the correct answer. 

When expressions are seperated by comma operator, they are evaluated from left and the rightmost expressions only used for assignment. Hence, 5 is assigned to a and 5 is printed on the screen. This is the correct answer.

Overload keyword is now obsolete and is no longer used or supported. It is not a part of Standard C++. It was preceded with function name in its definition to indicate that it was an overloaded function. Hence, this is the correct answer.

As shown in the code, i is incremented two times. In first cout, 1 will be displayed and in second cout -since a is incremented twice by then, 3 is displayed. Hence, 1 3 is the correct output.