Pramati Technical C Plus Plus Questions Part I

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Note: All the programs are tested under Turbo C + + 3.0, 4.5 and Microsoft VC + + 6.0 compilers.

It is assumed that.

Programs run under Windows environment.

The underlying machine is an x86 based system.

Program is compiled using Turbo C/C + + compiler.

The program output may depend on the information based on this assumptions (for example sizeof (int) = = 2 may be assumed).

  1. class Sample { public: Int * ptr; Sample (int i) { ptr = new int (i); } ~Sample () { delete ptr; }void PrintVal () { cout << “The value is” << * ptr; }}; void SomeFunc (Sample x) { cout << “Say i am in someFunc” << endl; }int main () { Sample s1 = 10; SomeFunc (s1); s1. PrintVal (); } Answer: Say i am in someFunc Null pointer assignment (Run-time error) Explanation: As the object is passed by value to SomeFunc the destructor of the object is called when the control returns from the function. So when PrintVal is called it meets up with ptr that has been freed. The solution is to pass the Sample object by reference to SomeFunc: Void SomeFunc (Sample &x) { cout << “Say i am in someFunc” << endl; } because when we pass objects by refernece that object is not destroyed. While returning from the function.

  2. Which is the parameter that is added to every non-static member function when it is called? Answer: This pointer

  3. class base { public: Int bval; base () { bval = 0; } }; class deri: Public base { public: Int dval; deri () { dval = 1; }}; void SomeFunc (base * arr, int size) { for int i = 0; i cout < bval; cout < } int main () { base BaseArr[5]; SomeFunc (BaseArr, 5); deri DeriArr[5]; SomeFunc (DeriArr, 5); } Answer: 00000 01010 Explanation: The function SomeFunc expects two arguments. The first one is a pointer to an array of base class objects and the second one is the sizeof the array. The first call of someFunc calls it with an array of bae objects, so it works correctly and prints the bval of all the objects. When Somefunc is called the second time the argument passed is the pointeer to an array of derived class objects and not the array of base class objects. But that is what the function expects to be sent. So the derived class pointer is promoted to base class pointer and the address is sent to the function. SomeFunc () knows nothing about this and just treats the pointer as an array of base class objects. So when arr + + is met, the size of base class object is taken into consideration and is incremented by sizeof (int) bytes for bval (the deri class objects have bval and dval as members and so is of size > = sizeof (int) + sizeof (int) ).

  4. class base public: Void baseFun () { cout<< “from base” < }; class deri: Public base {public: Void baseFun () { cout<< “from derived” < }}; void SomeFunc (base * baseObj) { baseObj → baseFun (); } int main () { base baseObject; SomeFunc (&baseObject); deri deriObject; SomeFunc (&deriObject); } Answer: From base from base Explanation: As we have seen in the previous case, SomeFunc expects a pointer to a base class. Since a pointer to a derived class object is passed, it treats the argument only as a base class pointer and the corresponding base function is called.

  5. class base public: Virtual void baseFun () { cout<< “from base” < }; class deri: Public base { public: Void baseFun () { cout<< “from derived” < }}; void SomeFunc (base * baseObj) { baseObj → baseFun (); } int main () { base baseObject; SomeFunc (&baseObject); deri deriObject; SomeFunc (&deriObject); } Answer: From base from derived Explanation: Remember that baseFunc is a virtual function. That means that it supports run-time polymorphism. So the function corresponding to the derived class object is called. Answer: Compiler Error: ‘ra’ reference must be initialized Explanation: Pointers are different from references. One of the main differences is that the pointers can be both initialized and assigned, whereas references can only be initialized. So this code issues an error. Const int size = 5; void print (int * ptr) { cout< } void print (int ptr[size]) { cout< } void main () { int a[size] = { 1, 2, 3, 4, 5}; int * b = new int (size); print (a); print (b); } Answer: Compiler Error: Function ‘void print (int * )’ already has a body Explanation: Arrays cannot be passed to functions, only pointers (for arrays, base addresses) can be passed. So the arguments int * ptr and int prt[size] have no difference as function arguments. In other words, both the functoins have the same signature and so cannot be overloaded. */class some { public: ~some () { cout<< “some's destructor” < }}; void main () { some s; s. ~some (); }/* Answer: Some's destructor some's destructor Explanation: Destructors can be called explicitly. Here's. ~some () ‘explicitly calls the destructor of's’ When main () returns, destructor of s is called again, hence the result. */#include class fig2d { int dim1; int dim2; public: Fig2d () { dim1 = 5; dim2 = 6; } virtual void operator<< (ostream & rhs); }; void fig2d: Operator<< (ostream &rhs) { rhs <dim1<< “” <dim2<< “” }/* class fig3d: Public fig2d { int dim3; public: Fig3d () { dim3 = 7; } virtual void operator<< (ostream &rhs); }; void fig3d: Operator<< (ostream &rhs) { fig2d: Operator << (rhs); rhs <dim3; } */void main () { fig2d obj1;//fig3d obj2; obj1 << cout;//obj2 << cout; }/* Answer: 5 6 Explanation: In this program, the << operator is overloaded with ostream as argument. This enables the ‘cout’ to be present at the right-hand-side. Normally, ‘cout’ is implemented as global function, but it doesn't mean that ‘cout’ is not possible to be overloaded as member function. Overloading << as virtual member function becomes handy when the class in which it is overloaded is inherited, and this becomes available to be overrided. This is as opposed to global friend functions, where friend's are not inherited. */class opOverload { public: Bool operator = = (opOverload temp); }; bool opOverload: Operator = = (opOverload temp) { if ( * this = = temp) { cout<< “The both are same objects\n” return true; } else { cout<< “The both are different\n” return false; } } void main () { opOverload a1, a2; a1 = = a2; } Answer: Runtime Error: Stack Overflow Explanation: Just like normal functions, operator functions can be called recursively. This program just illustrates that point, by calling the operator = = function recursively, leading to an infinite loop. Class complex { double re; double im; public: Complex (): Re (1), im (0.5) { } bool operator = = (complex &rhs); operator int () { }}; bool complex: Operator = = (complex &rhs) { if ( (this → re = = rhs. Re) && (this → im = = rhs. Im) ) return true; else return false; } int main () { complex c1; cout<< c1; } Answer: Garbage value Explanation: The programmer wishes to print the complex object using output re-direction operator, which he has not defined for his lass. But the compiler instead of giving an error sees the conversion function and converts the user defined object to standard object and prints some garbage value. Class complex { double re; double im; public: Complex (): Re (0), im (0) { }} complex (double n) { re = n, im = n; }; complex (int m, int n) { re = m, im = n; } void print () { cout< }; void main () { complex c3; double i = 5; c3 = i; c3. Print (); } Answer: 5, 5 Explanation: Though no operator = function taking complex, double is defined, the double on the rhs is converted into a temporary object using the single argument constructor taking double and assigned to the lvalue. Answer: Compiler Error: ‘ra’ reference must be initialized Explanation: Pointers are different from references. One of the main differences is that the pointers can be both initialized and assigned, whereas references can only be initialized. So this code issues an error.

Try it Yourself

  1. Determine the output of the ‘C + +’ Codelet. Class base { public: Out () { cout<< “base” } }; class deri { public: Out () { cout<< “deri” } }; void main () { deri dp[3]; base * bp = (base * ) dp; for (int i = 0; i<3; i + + ) (bp + + ) → out (); }

  2. Justify the use of virtual constructors and destructors in C + +.

  3. Each C + + object possesses the 4 member fns (which can be declared by the programmer explicitly or by the implementation if they are not available), What are those 4 functions?

  4. What is wrong with this class declaration? class something { char * str; public: Something () { st = new char[10]; } ~something () { delete str; } }

  5. Inheritance is also known as____relationship. Containership as ________ relationship.

  6. When is it necessary to use member-wise initialization list (also known as header initialization list) in C + +?

  7. Which is the only operator in C + + which can be overloaded but NOT inherited.

  8. Is there anything wrong with this C + + class declaration? class temp { int value1; mutable int value2; public: Void fun (int val) const { ( (temp * ) this) → value1 = 10; value2 = 10; }}

Solve These

  1. What is a modifier? Answer: A modifier, also called a modifying function is a member function that changes the value of at least one data member. In other words, an operation that modifies the state of an object. Modifiers are also known as mutatorS'

  2. What is an accessor? Answer: An accessor is a class operation that does not modify the state of an object. The accessor functions need to be declared as const operations

  3. Differentiate between a template class and class template. Answer: Template class: A generic definition or a parameterized class not instantiated until the client provides the needed information. It's jargon for plain templates. Class template: A class template specifies how individual classes can be constructed much like the way a class specifies how individual objects can be constructed. It's jargon for plain classes.

  4. When does a name clash occur? Answer: A name clash occurs when a name is defined in more than one place. For example. two different class libraries could give two different classes the same name. If you try to use many class libraries at the same time, there is a fair chance that you will be unable to compile or link the program because of name clashes.

  5. Define namespace. Answer: It is a feature in c + + to minimize name collisions in the global name space. This namespace keyword assigns a distinct name to a library that allows other libraries to use the same identifier names without creating any name collisions. Furthermore, the compiler uses the namespace signature for differentiating the definitions.

  6. What is the use of using declaration. Answer: A using declaration makes it possible to use a name from a namespace without the scope operator.

  7. What is an Iterator class? Answer: A class that is used to traverse through the objects maintained by a container class. There are five categories of iterators: Input iterators, output iterators, forward iterators, bidirectional iterators, random access.

  8. An iterator is an entity that gives access to the contents of a container object without violating encapsulation constraints. Access to the contents is granted on a one-at-a-time basis in order. The order can be storage order (as in lists and queues) or some arbitrary order (as in array indices) or according to some ordering relation (as in an ordered binary tree). The iterator is a construct, which provides an interface that, when called, yields either the next element in the container, or some value denoting the fact that there are no more elements to examine. Iterators hide the details of access to and update of the elements of a container class. The simplest and safest iterators are those that permit read-only access to the contents of a container class. The following code fragment shows how an iterator might appear in code: Cont_iter: = new cont_iterator (); x: = cont_iter. Next (); while x/= none do … s (x); … x: = cont_iter. Next (); end; In this example, cont_iter is the name of the iterator. It is created on the first line by instantiation of cont_iterator class, an iterator class defined to iterate over some container class, cont. Succesive elements from the container are carried to x. The loop terminates when x is bound to some empty value (Here, none). In the middle of the loop, there is s (x) an operation on x, the current element from the container. The next element of the container is obtained at the bottom of the loop.

  9. List out some of the OODBMS available. Answer: GEMSTONE/OPAL of Gemstone systems. ONTOS of Ontos. Objectivity of Objectivity inc. Versant of Versant object technology. Object store of Object Design. ARDENT of ARDENT software. POET of POET software.

  10. List out some of the object-oriented methodologies. Answer: Object Oriented Development (OOD) (Booch 1991, 1994). Object Oriented Analysis and Design (OOA/D) (Coad and Yourdon 1991). Object Modelling Techniques (OMT) (Rumbaugh 1991). Object Oriented Software Engineering (Objectory) (Jacobson 1992). Object Oriented Analysis (OOA) (Shlaer and Mellor 1992). The Fusion Method (Coleman 1991).

  11. What is an incomplete type? Answer: Incomplete types refers to pointers in which there is non availability of the implementation of the referenced location or it points to some location whose value is not available for modification. Example: Int * i = 0 × 400//i points to address 400 * i = 0;//set the value of memory location pointed by i. Incomplete types are otherwise called uninitialized pointers.

  12. What is a dangling pointer? Answer: A dangling pointer arises when you use the address of an object after its lifetime is over. This may occur in situations like returning addresses of the automatic variables from a function or using the address of the memory block after it is freed.

  13. Differentiate between the message and method. Answer: Message Method Objects communicate by sending messages Provides response to a message. To each other. A message is sent to invoke a method. It is an implementation of an operation.

  14. What is an adaptor class or Wrapper class? Answer: A class that has no functionality of its own. Its member functions hide the use of a third party software component or an object with the non-compatible interface or a non-object-oriented implementation.

  15. What is a Null object? Answer: It is an object of some class whose purpose is to indicate that a real object of that class does not exist. One common use for a null object is a return value from a member function that is supposed to return an object with some specified properties but cannot find such an object.

  16. What is class invariant? Answer: A class invariant is a condition that defines all valid states for an object. It is a logical condition to ensure the correct working of a class. Class invariants must hold when an object is created, and they must be preserved under all operations of the class. In particular all class invariants are both preconditions and post-conditions for all operations or member functions of the class.

  17. What do you mean by Stack unwinding? Answer: It is a process during exception handling when the destructor is called for all local objects between the place where the exception was thrown and where it is caught.

  18. Define precondition and post-condition to a member function. Answer: Precondition: A precondition is a condition that must be true on entry to a member function. A class is used correctly if preconditions are never false. An operation is not responsible for doing anything sensible if its precondition fails to hold. For example, the interface invariants of stack class say nothing about pushing yet another element on a stack that is already full. We say that isful () is a precondition of the push operation. Post-condition: A post-condition is a condition that must be true on exit from a member function if the precondition was valid on entry to that function. A class is implemented correctly if post-conditions are never false. For example, after pushing an element on the stack, we know that isempty () must necessarily hold. This is a post-condition of the push operation.

  19. What are the conditions that have to be met for a condition to be an invariant of the class? Answer: The condition should hold at the end of every constructor. The condition should hold at the end of every mutator (non-const) operation.

  20. What are proxy objects?

Answer: Objects that stand for other objects are called proxy objects or surrogates.

Example: Emplate

class Array2D { public: Class Array1D { public: T& operator[] (int index); const T& operator[] (int index) const; … }; Array1D operator[] (int index); const Array1D operator[] (int index) const; … }.

The following then becomes legal:

Array2Ddata (10, 20); ____

cout

Here data[3] yields an Array1D object and the operator [] invocation on that object yields the float in position (3, 6) of the original two dimensional array. Clients of the Array2D class need not be aware of the presence of the Array1D class. Objects of this latter class stand for one-dimensional array objects that, conceptually, do not exist for clients of Array2D. Such clients program as if they were using real, live, two-dimensional arrays. Each Array1D object stands for a one-dimensional array that is absent from a conceptual model used by the clients of Array2D. In the above example, Array1D is a proxy class. Its instances stand for one-dimensional arrays that, conceptually, do not exist.

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