Understanding C++ Placement New and Constructors: Use and Types of Constructors, Study notes of Printing

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C++ Interview Questions Compiled by Dr. Fatih Kocan, Wael Kdouh, and Kathryn Patterson for the Data Structures in C++ course(CSE 3358) Spring 2008

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Q: Can a copy constructor accept an object of the same class as parameter, instead of reference of the object? A: No. It is specified in the definition of the copy constructor itself. It should generate an error if a programmer specifies a copy constructor with a first argument that is an object and not a reference.

Q: What is conversion constructor? A: constructor with a single argument makes that constructor as conversion ctor and it can be used for type conversion. for example: class Boo { public: Boo( int i ); }; Boo BooObject = 10 ; // assigning int 10 Boo object

Q: What is conversion operator?? A: class can have a public method for specific data type conversions. for example: class Boo { double value; public: Boo(int i ) operator double() { return value; } }; Boo BooObject; double i = BooObject; // assigning object to variable i of type double. now conversion operator gets called to assign the value.

Q: How can I handle a constructor that fails? A: throw an exception. Constructors don't have a return type, so it's not possible to use return codes. The best way to signal constructor failure is therefore to throw an exception.

Q: How can I handle a destructor that fails? A: Write a message to a log-_le. But do not throw an exception. The C++ rule is that you must never throw an exception from a destructor that is being called during the "stack unwinding" process of another exception. For example, if someone says throw Foo(), the stack will be unwound so all the stack frames between the throw Foo() and the } catch (Foo e) { will get popped. This is called stack unwinding. During stack unwinding, all the local objects in all

those stack frames are destructed. If one of those destructors throws an exception (say it throws a Bar object), the C++ runtime system is in a no-win situation: should it ignore the Bar and end up in the } catch (Foo e) { where it was originally headed? Should it ignore the Foo and look for a } catch (Bare) { handler? There is no good answer:either choice loses information. So the C++ language guarantees that it will call terminate() at this point, and terminate() kills the process. Bang you're dead.

Q: What is Virtual Destructor? A: Using virtual destructors, you can destroy objects without knowing their type - the correct destructor for the object is invoked using the virtual function mechanism. Note that destructors can also be declared as pure virtual functions for abstract classes. if someone will derive from your class, and if someone will say "new Derived", where "Derived" is derived from your class, and if someone will say delete p, where the actual object's type is "Derived" but the pointer p's type is your class.

Q: Can a copy constructor accept an object of the same class as parameter, instead of reference of the object? A: No. It is specified in the definition of the copy constructor itself. It should generate an error if a programmer specifies a copy constructor with a first argument that is an object and not a reference.

Q: What's the order that local objects are destructed? A: In reverse order of construction: First constructed, last destructed.

In the following example, b's destructor will be executed first, then a's destructor:

void userCode() { Fred a; Fred b; ... }

Q: What's the order that objects in an array are destructed? A: In reverse order of construction: First constructed, last destructed.

In the following example, the order for destructors will be a[9], a[8], ..., a[1], a[0]:

void userCode() { Fred a[10]; ... }

Q: Can I overload the destructor for my class?

File f; ...insert code that should execute when f is still open... } f's destructor will automagically be called here!

...insert code here that should execute after f is closed...}

Q: What if I can't wrap the local in an artificial block? A: Most of the time, you can limit the lifetime of a local by wrapping the local in an artificial block ({...}). But if for some reason you can't do that, add a member function that has a similar effect as the destructor. But do not call the destructor itself!

For example, in the case of class File, you might add a close() method. Typically the destructor will simply call this close() method. Note that the close() method will need to mark the File object so a subsequent call won't re-close an already-closed File. E.g., it might set the fileHandle_ data member to some nonsensical value such as -1, and it might check at the beginning to see if the fileHandle_ is already equal to -1:

class File { public: void close(); ~File(); ... private: int fileHandle_; // fileHandle_ >= 0 if/only-if it's open };

File::~File() { close(); }

void File::close() { if (fileHandle_ >= 0) { ...insert code to call the OS to close the file... fileHandle_ = -1; } } Note that the other File methods may also need to check if the fileHandle_ is -1 (i.e., check if the File is closed).

Note also that any constructors that don't actually open a file should set fileHandle_ to -1.

Q: But can I explicitly call a destructor if I've allocated my object with new? A: Probably not.

Unless you used placement new, you should simply delete the object rather than explicitly calling the destructor. For example, suppose you allocated the object via a typical new expression:

Fred* p = new Fred(); Then the destructor Fred::~Fred() will automagically get called when you delete it via:

delete p; // Automagically calls p->~Fred() You should not explicitly call the destructor, since doing so won't release the memory that was allocated for the Fred object itself. Remember: delete p does two things: it calls the destructor and it deallocates the memory.

Q: What is "placement new" and why would I use it? A: There are many uses of placement new. The simplest use is to place an object at a particular location in memory. This is done by supplying the place as a pointer parameter to the new part of a new expression:

#include // Must #include this to use "placement new" #include "Fred.h" // Declaration of class Fred

void someCode() { char memory[sizeof(Fred)]; // Line # void* place = memory; // Line #

Fred* f = new(place) Fred(); // Line #3 (see "DANGER" below) // The pointers f and place will be equal

... } Line #1 creates an array of sizeof(Fred) bytes of memory, which is big enough to hold a Fred object. Line #2 creates a pointer place that points to the first byte of this memory (experienced C programmers will note that this step was unnecessary; it's there only to make the code more obvious). Line #3 essentially just calls the constructor Fred::Fred(). The this pointer in the Fred constructor will be equal to place. The returned pointer f will therefore be equal to place.

ADVICE: Don't use this "placement new" syntax unless you have to. Use it only when you really care that an object is placed at a particular location in memory. For example, when your hardware has a memory-mapped I/O timer device, and you want to place a Clock object at that memory location.

DANGER: You are taking sole responsibility that the pointer you pass to the "placement new"

};

Fred::~Fred() { // Compiler automagically calls z_.~Member() // Compiler automagically calls y_.~Member() // Compiler automagically calls x_.~Member() }

Q: When I write a derived class's destructor, do I need to explicitly call the destructor for my base class? A: No. You never need to explicitly call a destructor (except with placement new).

A derived class's destructor (whether or not you explicitly define one) automagically invokes the destructors for base class subobjects. Base classes are destructed after member objects. In the event of multiple inheritance, direct base classes are destructed in the reverse order of their appearance in the inheritance list.

class Member { public: ~Member(); ... };

class Base { public: virtual ~Base(); // A virtual destructor ... };

class Derived : public Base { public: ~Derived(); ... private: Member x_; };

Derived::~Derived() { // Compiler automagically calls x_.~Member() // Compiler automagically calls Base::~Base() } Note: Order dependencies with virtual inheritance are trickier. If you are relying on order dependencies in a virtual inheritance hierarchy, you'll need a lot more information than is in this

FAQ.

Q: Is there any difference between List x; and List x();? A: A big difference!

Suppose that List is the name of some class. Then function f() declares a local List object called x:

void f() { List x; // Local object named x (of class List) ... } But function g() declares a function called x() that returns a List:

void g() { List x(); // Function named x (that returns a List) ... }

Q: Can one constructor of a class call another constructor of the same class to initialize the this object? A: Nope.

Let's work an example. Suppose you want your constructor Foo::Foo(char) to call another constructor of the same class, say Foo::Foo(char,int), in order that Foo::Foo(char,int) would help initialize the this object. Unfortunately there's no way to do this in C++.

Some people do it anyway. Unfortunately it doesn't do what they want. For example, the line Foo(x, 0); does not call Foo::Foo(char,int) on the this object. Instead it calls Foo::Foo(char,int) to initialize a temporary, local object (not this), then it immediately destructs that temporary when control flows over the ;.

class Foo { public: Foo(char x); Foo(char x, int y); ... };

Foo::Foo(char x) {

Q: Is the default constructor for Fred always Fred::Fred()? A: No. A "default constructor" is a constructor that can be called with no arguments. One example of this is a constructor that takes no parameters:

class Fred { public: Fred(); // Default constructor: can be called with no args ... }; Another example of a "default constructor" is one that can take arguments, provided they are given default values:

class Fred { public: Fred(int i=3, int j=5); // Default constructor: can be called with no args ... };

Q: Which constructor gets called when I create an array of Fred objects? A: Fred's default constructor (except as discussed below).

class Fred { public: Fred(); ... };

int main() { Fred a[10]; calls the default constructor 10 times Fred* p = new Fred[10]; calls the default constructor 10 times ... } If your class doesn't have a default constructor, you'll get a compile-time error when you attempt to create an array using the above simple syntax:

class Fred { public: Fred(int i, int j); assume there is no default constructor ... };

int main() { Fred a[10]; ERROR: Fred doesn't have a default constructor Fred* p = new Fred[10]; ERROR: Fred doesn't have a default constructor ... } However, even if your class already has a default constructor, you should try to use std::vector rather than an array (arrays are evil). std::vector lets you decide to use any constructor, not just the default constructor:

#include

int main() { std::vector a(10, Fred(5,7)); the 10 Fred objects in std::vector a will be initialized with Fred(5,7) ... } Even though you ought to use a std::vector rather than an array, there are times when an array might be the right thing to do, and for those, you might need the "explicit initialization of arrays" syntax. Here's how:

class Fred { public: Fred(int i, int j); assume there is no default constructor ... };

int main() { Fred a[10] = { Fred(5,7), Fred(5,7), Fred(5,7), Fred(5,7), Fred(5,7), // The 10 Fred objects are Fred(5,7), Fred(5,7), Fred(5,7), Fred(5,7), Fred(5,7) // initialized using Fred(5,7) }; ... } Of course you don't have to do Fred(5,7) for every entry you can put in any numbers you want, even parameters or other variables.

Finally, you can use placement-new to manually initialize the elements of the array. Warning: it's ugly: the raw array can't be of type Fred, so you'll need a bunch of pointer-casts to do things like compute array index operations. Warning: it's compiler- and hardware-dependent: you'll need to make sure the storage is aligned with an alignment that is at least as strict as is required for objects of class Fred. Warning: it's tedious to make it exception-safe: you'll need to manually

different orders. Or it might happen when two data members are self-referential. Or when a data- member needs a reference to the this object, and you want to avoid a compiler warning about using the this keyword prior to the { that begins the constructor's body (when your particular compiler happens to issue that particular warning). Or when you need to do an if/throw test on a variable (parameter, global, etc.) prior to using that variable to initialize one of your this members. This list is not exhaustive; please don't write me asking me to add another "Or when...". The point is simply this: use common sense.

Q: Should you use the this pointer in the constructor? A: Some people feel you should not use the this pointer in a constructor because the object is not fully formed yet. However you can use this in the constructor (in the {body} and even in the initialization list) if you are careful.

Here is something that always works: the {body} of a constructor (or a function called from the constructor) can reliably access the data members declared in a base class and/or the data members declared in the constructor's own class. This is because all those data members are guaranteed to have been fully constructed by the time the constructor's {body} starts executing.

Here is something that never works: the {body} of a constructor (or a function called from the constructor) cannot get down to a derived class by calling a virtual member function that is overridden in the derived class. If your goal was to get to the overridden function in the derived class, you won't get what you want. Note that you won't get to the override in the derived class independent of how you call the virtual member function: explicitly using the this pointer (e.g., this->method()), implicitly using the this pointer (e.g., method()), or even calling some other function that calls the virtual member function on your this object. The bottom line is this: even if the caller is constructing an object of a derived class, during the constructor of the base class, your object is not yet of that derived class. You have been warned.

Here is something that sometimes works: if you pass any of the data members in this object to another data member's initializer, you must make sure that the other data member has already been initialized. The good news is that you can determine whether the other data member has (or has not) been initialized using some straightforward language rules that are independent of the particular compiler you're using. The bad news it that you have to know those language rules (e.g., base class sub-objects are initialized first (look up the order if you have multiple and/or virtual inheritance!), then data members defined in the class are initialized in the order in which they appear in the class declaration). If you don't know these rules, then don't pass any data member from the this object (regardless of whether or not you explicitly use the this keyword) to any other data member's initializer! And if you do know the rules, please be careful.

Q: What is the "Named Constructor Idiom"? A: A technique that provides more intuitive and/or safer construction operations for users of your class.

The problem is that constructors always have the same name as the class. Therefore the only way to differentiate between the various constructors of a class is by the parameter list. But if there are lots of constructors, the differences between them become somewhat subtle and error prone.

With the Named Constructor Idiom, you declare all the class's constructors in the private or protected sections, and you provide public static methods that return an object. These static methods are the so-called "Named Constructors." In general there is one such static method for each different way to construct an object.

For example, suppose we are building a Point class that represents a position on the X-Y plane. Turns out there are two common ways to specify a 2-space coordinate: rectangular coordinates (X+Y), polar coordinates (Radius+Angle). (Don't worry if you can't remember these; the point isn't the particulars of coordinate systems; the point is that there are several ways to create a Point object.) Unfortunately the parameters for these two coordinate systems are the same: two floats. This would create an ambiguity error in the overloaded constructors:

class Point { public: Point(float x, float y); // Rectangular coordinates Point(float r, float a); // Polar coordinates (radius and angle) // ERROR: Overload is Ambiguous: Point::Point(float,float) };

int main() { Point p = Point(5.7, 1.2); // Ambiguous: Which coordinate system? ... } One way to solve this ambiguity is to use the Named Constructor Idiom:

#include // To get sin() and cos()

class Point { public: static Point rectangular(float x, float y); // Rectangular coord's static Point polar(float radius, float angle); // Polar coordinates // These static methods are the so-called "named constructors" ... private: Point(float x, float y); // Rectangular coordinates float x_, y_; };

inline Point::Point(float x, float y) : x_(x), y_(y) { }

inline Point Point::rectangular(float x, float y)

Q: What is virtual function? A: When derived class overrides the base class method by redefining the same function, then if client wants to access redefined the method from derived class through a pointer from base class object, then you must define this function in base class as virtual function. class parent { void Show() { cout << "i'm parent" << endl; } };

class child: public parent { void Show() { cout << "i'm child" << endl; } }; parent * parent_object_ptr = new child; parent_object_ptr->show() // calls parent->show() now we goto virtual world... class parent { virtual void Show() { cout << "i'm parent" << endl; } }; class child: public parent { void Show() { cout << "i'm child" << endl; } }; parent * parent_object_ptr = new child; parent_object_ptr->show() // calls child->show()

Q: What is a "pure virtual" member function? A: The abstract class whose pure virtual method has to be implemented by all the classes which derive on these. Otherwise it would result in a compilation error. This construct should be used when one wants to ensure that all the derived classes implement the method defined as pure virtual in base class.

Q: How virtual functions are implemented C++? A: Virtual functions are implemented using a table of function pointers, called the vtable. There is one entry in the table per virtual function in the class. This table is created by the constructor of the class. When a derived class is constructed, its base class is constructed _rst which creates the vtable. If the derived class overrides any of the base classes virtual functions, those entries in the vtable are overwritten by the derived class constructor. This is why you should never call virtual functions from a constructor: because the vtable entries for the object may not have been set up by the derived class constructor yet, so you might end up calling base class implementations of those virtual functions

Q: What is pure virtual function? or what is abstract class? A: When you de_ne only function prototype in a base class without implementation and do the complete implementation in derived class. This base class is called abstract class and client won't able to instantiate an object using this base class. You can make a pure virtual function or abstract class this way.. class Boo { void foo() = 0; } Boo MyBoo; // compilation error

Q: What is Pure Virtual Function? Why and when it is used? A: The abstract class whose pure virtual method has to be implemented by all the classes which derive on these. Otherwise it would result in a compilation error. This construct should be used when one wants to ensure that all the derived classes implement the method defined as pure virtual in base class.

Q: How Virtual functions call up is maintained? A: Through Look up tables added by the compile to every class image. This also leads to performance penalty.

Q: What is a virtual destructor? A: The simple answer is that a virtual destructor is one that is declared with the virtual attribute. The behavior of a virtual destructor is what is important. If you destroy an object through a caller or reference to a base class, and the base-class destructor is not virtual, the derived-class destructors are not executed, and the destruction might not be complete.