Ans: [Probably this would be the first question for a Java/c++ technical interview for freshers and sometimes for experienced as well. Try to give examples when you answer this question.]
Class defines a datatype, it's type definition of category of thing(s). But a class actually does not define the data, it just specifies the structure of data. To use them you need to create objects out of the class. Class can be considered as a blueprint of a building, you can not stay inside blueprint of building, you need to construct building(s) out of that plan. You can create any number of buildings from the blueprint, similarly you can create any number of objects from a class.
class Vehicle
{
public:
int numberOfTyres;
double engineCapacity;
void drive(){
// code to drive the car
}
};
Ans: Object is the instance of a class, which is concrete. From the above example, we can create instance of class Vehicle as given below
Vehicle vehicleObject;
We can have different objects of the class Vehicle, for example we can have Vehicle objects with 2 tyres, 4tyres etc. Similarly different engine capacities as well.
Ans: [Questions regarding access specifiers are common not just in c++ interview but for other object oriented language interviews as well.]
Access specifiers are used to define how the members (functions and variables) can be accessed outside the class. There are three access specifiers defined which are public, private, and protected
Ans:
Refer Questions 2 and 3 for the concepts about classes and objects
Encapsulation is the mechanism by which data and associated operations/methods are bound together and thus hide the data from outside world. It's also called data hiding. In c++, encapsulation achieved using the access specifiers (private, public and protected). Data members will be declared as private (thus protecting from direct access from outside) and public methods will be provided to access these data. Consider the below class
class Person
{
private:
int age;
public:
int getAge(){
return age;
}
int setAge(int value){
if(value > 0){
age = value;
}
}
};
In the class Person, access to the data field age is protected by declaring it as private and providing public access methods. What would have happened if there was no access methods and the field age was public? Anybody who has a Person object can set an invalid value (negative or very large value) for the age field. So by encapsulation we can preventing direct access from outside, and thus have complete control, protection and integrity of the data.
Data abstraction refers to hiding the internal implementations and show only the necessary details to the outside world. In C++ data abstraction is implemented using interfaces and abstract classes.
class Stack
{
public:
virtual void push(int)=0;
virtual int pop()=0;
};
class MyStack : public Stack
{
private:
int arrayToHoldData[]; //Holds the data from stack
public:
void push(int) {
// implement push operation using array
}
int pop()
{
// implement pop operation using array
}
};
In the above example, the outside world only need to know about the Stack class and its push, pop operations. Internally stack can be implemented using arrays or linked lists or queues or anything that you can think of. This means, as long as the push and pop method performs the operations work as expected, you have the freedom to change the internal implementation with out affecting other applications that use your Stack class.
Inheritance allows one class to inherit properties of another class. In other words, inheritance allows one class to be defined in terms of another class.
class SymmetricShape
{
public:
int getSize()
{
return size;
}
void setSize(int w)
{
size = w;
}
protected:
int size;
};
// Derived class
class Square: public SymmetricShape
{
public:
int getArea()
{
return (size * size);
}
};
In the above example, class Square inherits the properties and methods of class SymmetricShape. Inheritance is the one of the very important concepts in C++/OOP. It helps to modularise the code, improve reusability and reduces tight coupling between components of the system.
Ans: Most of the times compilers will do optimization to the code to speed up the program. For example in the below code,
compiler may think that value of 'a' is not getting changed from the program and replace it with 'while(true)', which will result in an infinite loop. In actual scenario the value of 'a' may be getting updated from outside of the program.
Volatile keyword is used to tell compiler that the variable declared using volatile may be used from outside the current scope so that compiler wont apply any optimization. This matters only in case of multi-threaded applications.
In the above example if variable 'a' was declared using volatile, compiler will not optimize it. In shot, value of the volatile variables will be read from the memory location directly.
Ans: In c++, variables can be initialized in two ways, the traditional C++ initialization using "=" operator and second using the constructor notation.
Traditional C++ initilization
variable i will get initialized to 10.
Using C++ constructor notation
Implicit conversions are performed when a type (say T) is used in a context where a compatible type (Say F) is expected so that the type T will be promoted to type F.
short a = 2000 + 20;
In the above example, variable a will get automatically promoted from short to int. This is called implicit conversion/coercion in c++.
Ans: Implicit conversions are performed when a type (say T) is used in a context where a compatible type (Say F) is expected so that the type T will be promoted to type F.
short a = 2000 + 20;
In the above example, variable a will get automatically promoted from short to int. This is called implicit conversion/coercion in c++.
Ans: C++ inline functions are special functions, for which the compiler replaces the function call with body/definition of function. Inline functions makes the program execute faster than the normal functions, since the overhead involved in saving current state to stack on the function call is avoided. By giving developer the control of making a function as inline, he can further optimize the code based on application logic. But actually, it's the compiler that decides whether to make a function inline or not regardless of it's declaration. Compiler may choose to make a non inline function inline and vice versa. Declaring a function as inline is in effect a request to the compiler to make it inline, which compiler may ignore. So, please note this point for the interview that, it is upto the compiler to make a function inline or not.
inline int min(int a, int b)
{
return (a < b)? a : b;
}
int main( )
{
cout << "min (20,10): " << min(20,10) << endl;
cout << "min (0,200): " << min(0,200) << endl;
cout << "min (100,1010): " << min(100,1010) << endl;
return 0;
}
If the complier decides to make the function min as inline, then the above code will internally look as if it was written like
int main( )
{
cout << "min (20,10): " << ((20 < 10)? 20 : 10) << endl;
cout << "min (0,200): " << ((0 < 200)? 0 : 200) << endl;
cout << "min (100,1010): " << ((100 < 1010)? 100 : 1010) << endl;
return 0;
}
Q10. What do you mean by translation unit in c++?
Ans: We organize our C++ programs into different source files (.cpp, .cxx etc). When you consider a source file, at the preprocessing stage, some extra content may get added to the source code ( for example, the contents of header files included) and some content may get removed ( for example, the part of the code in the #ifdef of #ifndef block which resolve to false/0 based on the symbols defined). This effective content is called a translation unit. In other words, a translation unit consists of
Ans: A symbol is said to be linked internally when it can be accessed only from with-in the scope of a single translation unit. By external linking a symbol can be accessed from other translation units as well. This linkage can be controlled by using static and extern keywords.
Ans: Storage class are used to specify the visibility/scope and life time of symbols(functions and variables). That means, storage classes specify where all a variable or function can be accessed and till what time those variables will be available during the execution of program.
Ans: Storage class are used to specify the visibility/scope and life time of symbols(functions and variables). That means, storage classes specify where all a variable or function can be accessed and till what time those variables will be available during the execution of program.
Following storage classes are available in C++
It's the default storage class for local variables. They can be accessed only from with in the declaration scope. auto variables are allocated at the beginning of enclosing block and deallocated at the end of enclosing block.
void changeValue(void)
{
auto int i = 1 ;
i++;
printf ( "%d ", i ) ;
}
int main()
{
changeValue();
changeValue();
changeValue();
changeValue();
return 0;
}
Output:-
In the above example, every time the method changeValue is invoked, memory is allocated for i and de allocated at the end of the method. So it's output will be same.
It's similar to auto variables. Difference is that register variables might be stored on the processor register instead of RAM, that means the maximum size of register variable should be the size of CPU register ( like 16bit, 32bit or 64bit). This is normally used for frequently accessed variables like counters, to improve performance. But note that, declaring a variable as register does not mean that they will be stored in the register. It depends on the hardware and implementation.
int main()
{
register int i;
int array[10] = {0,1,2,3,4,5,6,7,8,9};
for (i=0;i<10;i++)
{
printf("%d ", array[i]);
}
return 0;
}
Output:-
0 1 2 3 4 5 6 7 8 9
The variable i might be stored on the CPU register and due to which the access of i in the loop will be faster.
A static variable will be kept in existence till the end of the program unlike creating and destroying each time they move into and out of the scope. This helps to maintain their value even if control goes out of the scope. When static is used with global variables, they will have internal linkage, that means it cannot be accessed by other source files. When static is used in case of a class member, it will be shared by all the objects of a class instead of creating separate copies for each object.
void changeValue(void)
{
static int i = 1 ;
i++;
printf ( "%d ", i ) ;
}
int main(){
changeValue();
changeValue();
changeValue();
changeValue();
return 0;
}
Output:-
Since static variable will be kept in existence till the end of program, variable i will retain it's value across the method invocations.
extern is used to tell compiler that the symbol is defined in another translation unit (or in a way, source files) and not in the current one. Which means the symbol is linked externally. extern symbols have static storage duration, that is accessible through out the life of program. Since no storage is allocated for extern variable as part of declaration, they cannot be initialized while declaring.
int x = 10;
int main( )
{
extern int y ;
printf("x: %d ", x );
printf("y: %d", y);
return 0;
}
int y = 70 ;
Output:-
x: 10 y: 70
extern variable is like global variable, it's scope is through out the program. It can be defined anywhere in the c++ program.
mutable storage class can be used only on non static non const data a member of a class. Mutable data member of a class can be modified even is it's part of an object which is declared as const.
class Test
{
public:
Test(): x(1), y(1) {};
mutable int x;
int y;
};
int main()
{
const Test object;
x = 123;
//object.y = 123;
/*
* The above line if uncommented, will create compilation error.
*/
cout<< "X:"<< object.x << ", Y:" << object.y;
return 0;
}
Output:-
X:123, Y:1
In the above example, we are able to change the value of member variable x though it's part of an object which is declared as const. This is because the variable x is declared as mutable. But if you try to modify the value of member variable y, compiler will throw error.
You can find the summary of c++ storage class specifiers below
| C++ Storage Specifier | Storage Location | Scope Of Variable | Life Time |
| auto | Memory (RAM) | Local | With in function |
| static | Memory (RAM) | Local | Life time is from when the flow reaches the first declaration to the termination of program. |
| register | CPU register | Local | With in function |
| extern | Memory (RAM) | Global | Till the end of main program |
Q14. What is 'Copy Constructor' and when it is called?
Ans: This is a frequent c++ interview question. Copy constructor is a special constructor of a class which is used to create copy of an object. Compiler will give a default copy constructor if you don't define one. This implicit constructor will copy all the members of source object to target object.
Implicit copy constructors are not recommended, because if the source object contains pointers they will be copied to target object, and it may cause heap corruption when both the objects with pointers referring to the same location does an update to the memory location. In this case its better to define a custom copy constructor and do a deep copy of the object.
class SampleClass{
public:
int* ptr;
SampleClass();
// Copy constructor declaration
SampleClass(SampleClass &obj);
};
SampleClass::SampleClass(){
ptr = new int();
*ptr = 5;
}
// Copy constructor definition
SampleClass::SampleClass(SampleClass &obj){
//create a new object for the pointer
ptr = new int();
// Now manually assign the value
*ptr = *(obj.ptr);
cout<<"Copy constructor...\n";
}
Ans:
This function is used to change the size of memory object pointed by address ptr to the size given by size. If ptr is a null pointer, then realloc will behave like malloc(). If the ptr is an invalid pointer, then defined behaviour may occur depending the implementation. Undefined behaviour may occur if the ptr has previously been deallocated by free(), or dealloc() or ptr do not match a pointer returned by an malloc(), calloc() or realloc().
This function is used to deallocate a block of memory that was allocated using malloc(), calloc() or realloc(). If ptr is null, this function does not doe anything.
Ans:
[This question can be expected in any interviews, not just c++ interviews. This is a usual question in most of the java interviews.]
When you do a shallow copy, all the fields of the source object is copied to target object as it is. That means, if there is a dynamically created field in the source object, shallow copy will copy the same pointer to target object. So you will have two objects with fields that are pointing to same memory location which is not what you usually want.
In case of deep copy, instead of copying the pointer, the object itself is copied to target. In this case if you modify the target object, it will not affect the source. By default copy constructors and assignment operators do shallow copy. To make it as deep copy, you need to create a custom copy constructor and override assignment operator.
Ans: Persistent objects are the ones which we can be serialized and written to disk, or any other stream. So before stopping your application, you can serialize the object and on restart you can deserialize it. [ Drawing applications usually use serializations.]
Objects that can not be serialized are called non persistent objects. [ Usually database objects are not serialized because connection and session will not be existing when you restart the application. ]
Ans: Technically it is possible to generate the source code from binary. It is called reverse engineering. There are lot of reverse engineering tools available. But, in actual case most of them will not re generate the exact source code back because many information will be lost due to compiler optimization and other interpretations.
Ans: Virtual functions are member functions of class which is declared using keyword 'virtual'. When a base class type reference is initialized using object of sub class type and an overridden method which is declared as virtual is invoked using the base reference, the method in child class object will get invoked.
class Base
{
int a;
public:
Base()
{
a = 1;
}
virtual void method()
{
cout << a;
}
};
class Child: public Base
{
int b;
public:
Child()
{
b = 2;
}
virtual void method()
{
cout << b;}
};
int main()
{
Base *pBase;
Child oChild;
pBase = &oChild;
pBase->method();
return 0;
}
In the above example even though the method in invoked on Base class reference, method of the child will get invoked since its declared as virtual.
Ans: Pure virtual function is a function which doesn't have an implementation and the same needs to be implemented by the the next immediate non-abstract class. (A class will become an abstract class if there is at-least a single pure virtual function and thus pure virtual functions are used to create interfaces in c++).
Ans: A function is made as pure virtual function by the using a specific signature, " = 0" appended to the function declaration as given below,
class SymmetricShape
{
public:
// draw() is a pure virtual function.
virtual void draw() = 0;
};
Ans: Pure virtual functions are used when it doesn't make sense to provide definition of a virtual function in the base class or a proper definition does not exists in the context of base class. Consider the above example, class SymmetricShape is used as base class for shapes with symmetric structure(Circle, square, equilateral triangle etc). In this case, there exists no proper definition for function draw() in the base class SymmetricShape instead the child classes of SymmetricShape (Cirlce, Square etc) can implement this method and draw proper shape.
Ans: [This c++ interview question is in a way related to polymorphism.]
Virtual destructors are used for the same purpose as virtual functions. When you remove an object of subclass, which is referenced by a parent class pointer, only destructor of base class will get executed. But if the destructor is defined using virtual keyword, both the destructors [ of parent and sub class ] will get invoked.
Ans: [This c++ interview question is related to question about virtual functions ]
Binding is the process of linking actual address of functions or identifiers to their reference. This happens mainly two times.
For all the direct function references compiler will replace the reference with actual address of the method.
In case of virtual function calls using a Base reference, as in shown in the example of question no: 2, compiler does not know which method will get called at run time. In this case compiler will replace the reference with code to get the address of function at runtime.
Dynamic binding is another name for late binding.
Ans: In C++, reference variable allows you create an alias (second name) for an already existing variable. A reference variable can be used to access (read/write) the original data. That means, both the variable and reference variable are attached to same memory location. In effect, if you change the value of a variable using reference variable, both will get changed (because both are attached to same memory location).
Ans: Appending an ampersand (&) to the end of datatype makes a variable eligible to use as reference variable.
The first statement initializes a an integer variable a. Second statement creates an integer reference initialized to variable a
Take a look at the below example to see how reference variables work.
Above code creates following output.
Value of a : 10
Value of a reference (b) : 10
Value of a : 20
Value of a reference (b) : 20
Ans: [This question is usually asked in a twisted way during c++ interviews. Sometimes the interviewer might use examples and ask you to find the error.]
| Pointers | Reference Variables |
| Pointers can be assigned to NULL | References cannot be assigned NULL. It should always be associated with actual memory, not NULL. |
| Pointers can be (re)pointed to any object, at any time, any number of times during the execution. | Reference variables should be initialized with an object when they are created and they cannot be reinitialized to refer to another object |
| Pointer has own memory address and location on stack | Reference variables has location on stack, but shares the same memory location with the object it refer to. |
Ans: cout << 25u - 50;
The answer is not -25. Rather, the answer (which will surprise many) is 4294967271, assuming 32 bit integers. Why?
In C++, if the types of two operands differ from one another, then the operand with the “lower type” will be promoted to the type of the “higher type” operand, using the following type hierarchy (listed here from highest type to lowest type): long double, double, float, unsigned long int, long int, unsigned int, int (lowest).
So when the two operands are, as in our example, 25u (unsigned int) and 50 (int), the 50 is promoted to also being an unsigned integer (i.e., 50u).
Moreover, the result of the operation will be of the type of the operands. Therefore, the result of 25u - 50u will itself be an unsigned integer as well. So the result of -25 converts to 4294967271 when promoted to being an unsigned integer.
C++ supports multiple inheritance. What is the “diamond problem” that can occur with multiple inheritance? Give an example.
Let’s consider a simple example. A university has people who are affiliated with it. Some are students, some are faculty members, some are administrators, and so on. So a simple inheritance scheme might have different types of people in different roles, all of whom inherit from one common “Person” class. The Person class could define an abstract getRole() method which would then be overridden by its subclasses to return the correct role type.
But now what happens if we want to model a the role of a Teaching Assistant (TA)? Typically, a TA is both a grad student and a faculty member. This yields the classic diamond problem of multiple inheritance and the resulting ambiguity regarding the TA’s getRole() method:
Which getRole() implementation should the TA inherit? That of the Faculty Member or that of the Grad Student? The simple answer might be to have the TA class override the getRole() method and return newly-defined role called “TA”. But that answer is also imperfect as it would hide the fact that a TA is, in fact, both a faculty member and a grad student.
Ans: my_struct_t *bar;
/* ... do stuff, including setting bar to point to a defined my_struct_t object ... */
memset(bar, 0, sizeof(bar));
The last argument to memset should be sizeof(*bar), not sizeof(bar). sizeof(bar) calculates the size of bar (i.e., the pointer itself) rather than the size of the structure pointed to by bar.
The code can therefore be corrected by using sizeof(*bar) as the last argument in the call to memset.
A sharp candidate might point out that using *bar will cause a dereferencing error if bar has not been assigned. Therefore an even safer solution would be to use sizeof(my_struct_t). However, an even sharper candidate must know that in this case using *bar is absolutely safe within the call to sizeof, even if bar has not been initialized yet, since sizeof is a compile time construct.
Ans: int i = 5;
int j = i++;
After the above code executes, i will equal 6, but j will equal 5.
Understanding the reason for this is fundamental to understanding how the unary increment (++) and decrement (--) operators work in C++.
When these operators precede a variable, the value of the variable is modified first and then the modified value is used. For example, if we modified the above code snippet to instead say int j = ++i;, i would be incremented to 6 and then j would be set to that modified value, so both would end up being equal to 6.
However, when these operators follow a variable, the unmodified value of the variable is used and then it is incremented or decremented. That’s why, in the statement int j = i++; in the above code snippet, j is first set to the unmodified value of i (i.e., 5) and then i is incremented to 6.
Ans: size_t sz = buf->size();
while ( --sz >= 0 )
{
/* do something */
}
The problem in the above code is that --sz >= 0 will always be true so you’ll never exit the while loop (so you’ll probably end up corrupting memory or causing some sort of memory violation or having some other program failure, depending on what you’re doing inside the loop).
The reasons that --sz >= 0 will always be true is that the type of sz is size_t. size_t is really just an alias to one of the fundamental unsigned integer types. Therefore, since sz is unsigned, it can never be less than zero (so the condition can never be true).
One example of an alternative implementation that would avoid this problem would be to instead use a for loop as follows:
for (size_t i = 0; i < sz; i++)
{
/* do something */
}
Option 1:
vector vec;
/* ... .. ... */
for (auto itr = vec.begin(); itr != vec.end(); itr++) {
itr->print();
}
Option 2:
vector vec;
/* ... .. ... */
for (auto itr = vec.begin(); itr != vec.end(); ++itr) {
itr->print();
}
Ans: Although both options will accomplish precisely the same thing, the second option is better from a performance standpoint. This is because the post-increment operator (i.e., itr++) is more expensive than pre-increment operator (i.e., ++itr). The underlying implementation of the post-increment operator makes a copy of the element before incrementing it and then returns the copy.
Ans: This question tests understanding of C++ templates. An experienced developer will know that this is already a part of the C++11 std library (std::is_base_of) or part of the boost library for C++ (boost::is_base_of). Even an interviewee with only passing knowledge should write something similar to this, mostly likely involving a helper class:
template<typename D, typename B>
class IsDerivedFromHelper
{
class No { };
class Yes { No no[3]; };
static Yes Test( B* );
static No Test( ... );
public:
enum { Is = sizeof(Test(static_cast<D*>(0))) == sizeof(Yes) };
};
template <class C, class P>
bool IsDerivedFrom() {
return IsDerivedFromHelper<C, P>::Is;
}
Ans: template <typename T, typename U>
struct is_same
{
static const bool value = false;
};
template <typename T>
struct is_same<T, T>
{
static const bool value = true;
};
template <class A, class B>
bool IsSameClass() {
return is_same<A, B>::value;
Ans: Although you can call an inline function from within itself, the compiler will not generate inline code since the compiler cannot determine the depth of recursion at compile time.
Ans: #include <iostream>
class A {
public:
A() {}
~A() {
throw 42;
}
};
int main(int argc, const char * argv[]) {
try {
A a;
throw 32;
} catch(int a) {
std::cout << a;
}
}
This program will terminate abnormally. throw 32 will start unwinding the stack and destroy class A. The class A destructor will throw another exception during the exception handling, which will cause program to crash. This question is testing if developer has experience working with exceptions.
class Something {
public:
Something() {
topSecretValue = 42;
}
public:
bool somePublicBool;
int somePublicInt;
std::string somePublicString;
private:
int topSecretValue;
};
Implement a method to get topSecretValue for any given Something* object. The method should be cross-platform compatible and not depend on sizeof (int, bool, string).
Ans: Create another class which has all the members of Something in the same order, but has additional public method which returns the value. Your replica Something class should look like:
class SomethingReplica {
public:
int getTopSecretValue() { return topSecretValue; }
public:
bool somePublicBool;
int somePublicInt;
std::string somePublicString;
private:
int topSecretValue;
};
Then, to get the value:
int main(int argc, const char * argv[]) {
Something a;
SomethingReplica* b = reinterpret_cast<SomethingReplica*>(&a);
std::cout << b->getTopSecretValue();
}
It’s important to avoid code like this in a final product, but it’s nevertheless a good technique when dealing with legacy code, as it can be used to extract intermediate calculation values from a library class. (Note: If it turns out that the alignment of the external library is mismatched to your code, you can resolve this using #pragma pack.)
For example: if A = {2, 1, 5, 9}, then B would be {45, 90, 18, 10}
Ans: This problem seems easy at first glance so a careless developer might write something like this:
void F(int* A, int* B, int N) {
int m = 1;
for (int i = 0; i < N; ++i) {
m *= A[i];
}
for (int i = 0; i < N; ++i) {
B[i] = m / A[i];
}
}
This will work for the given example, but when you add a 0 into input array A, the program will crash because of division by zero. The correct answer should take that edge case into account and look like this:
void F(int* A, int* B, int N) {
int m = 1;
int numZero = 0;
int zeroIndex = -1;
for (int i = 0; i < N; ++i) {
B[i] = 0;
if (A[i] == 0) {
++numZero;
zeroIndex = i;
} else {
m *= A[i];
}
}
if (numZero == 0) {
for (int i = 0; i < N; ++i) {
B[i] = m / A[i];
}
return;
}
if (numZero >= 2) {
return;
}
B[zeroIndex] = m;
}
The presented solution above has a Big O complexity of O(n). While there are simpler solutions available (ones that would ignore the need to take 0 into account), that simplicity has a price of complexity, generally running significantly slower.
Ans: While it’s ideal to avoid virtual inheritance altogether (you should know how your class is going to be used) having a solid understanding of how virtual inheritance works is still important:
So when you have a class (class A) which inherits from 2 parents (B and C), both of which share a parent (class D), as demonstrated below:
#include <iostream>
class D {
public:
void foo() {
std::cout << "Foooooo" << std::endl;
}
};
class C: public D {
};
class B: public D {
};
class A: public B, public C {
};
int main(int argc, const char * argv[]) {
A a;
a.foo();
}
If you don’t use virtual inheritance in this case, you will get two copies of D in class A: one from B and one from C. To fix this you need to change the declarations of classes C and B to be virtual, as follows:
class C: virtual public D {
};
class B: virtual public D {
};
Ans: The only difference between a class and struct are the access modifiers. Struct members are public by default; class members are private. It is good practice to use classes when you need an object that has methods and structs when you have a simple data object.
Ans: #include <iostream>
int main(int argc, const char * argv[]) {
int a[] = {1, 2, 3, 4, 5, 6};
std::cout << (1 + 3)[a] - a[0] + (a + 1)[2];
}
The above will output 8, since:
(1+3)[a] is the same as a[1+3] == 5
a[0] == 1
(a + 1)[2] is the same as a[3] == 4
This question is testing pointer arithmetic knowledge, and the magic behind square brackets with pointers.
While some might argue that this isn’t a valuable question as it appears to only test the capability of reading C constructs, it’s still important for a candidate to be able to work through it mentally; it’s not an answer they’re expected to know off the top of their head, but one where they talk about what conclusion they reach and how.
Ans: #include
class Base {
virtual void method() {std::cout << "from Base" << std::endl;}
public:
virtual ~Base() {method();}
void baseMethod() {method();}
};
class A : public Base
{
void method() {std::cout << "from A" << std::endl;}
public:
~A() {method();}
};
int main(void) {
Base* base = new A;
base->baseMethod();
delete base;
return 0;
}
The above will output:
from A
from A
from Base
The important thing to note here is the order of destruction of classes and how Base’s method reverts back to its own implementation once A has been destroyed.
Ans: The volatile keyword informs the compiler that a variable will be used by multiple threads. Variables that are declared as volatile will not be cached by the compiler to ensure the most up-to-date value is held.
The mutable keyword can be used for class member variables. Mutable variables are allowed to change from within const member functions of the class.
Ans: unsigned char half_limit = 150;
for (unsigned char i = 0; i < 2 * half_limit; ++i)
{
// do something;
}
If you said 300, you would have been correct if i had been declared as an int. However, since i was declared as an unsigned char, the corrct answer is that this code will result in an infinite loop.
Here’s why:
The expression 2 * half_limit will get promoted to an int (based on C++ conversion rules) and will have a value of 300. However, since i is an unsigned char, it is rerepsented by an 8-bit value which, after reaching 255, will overflow (so it will go back to 0) and the loop will therefore go on forever.
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