- Which of the following statement is accurate characteristic overloading in C++?
- Function overloading is performed by having two or greater functions with equal names and with different or specific return types.
- Function overloading is carried out by having two or extra functions with identical names but unique argument lists.
- Function overloading involves two or more functions with special names but the same argument lists.
- C++ no longer assists function overloading.
Explanation:
The correct answer is option (b). Overloading is accomplished with the aid of defining multiple functions that have the same name but differ in both varieties of arguments or different sorts of arguments.
- Which of the following statements approximately define overloading functions and default arguments in C++?
- Function overloading and default arguments can not be used together.
- Function overloading and default arguments can be used together.
- Default arguments override function overloading.
- Function overloading must be disabled to use default arguments.
Explanation:
The accurate choice is alternative (b). When working with C++, it is possible to combine function overloading and default arguments.
- What is the correct syntax for implementing operator overloading of the '+' operator within a class?
class A {
public:
A operator+(const A&);
};
- A operator+(A a);
- A operator+(const A&);
- A operator+;
- A operator+(A&);
Explanation:
The accurate choice is alternative (b). The proper method of operator overloading involves declaring it as a member function with a single parameter of class type.
- What is the return type for overloading the stream insertion operator (<<) for a class?
class MyClass {
friend std::ostream& operator<<(std::ostream&, const MyClass&);
};
- void
- MyClass
- std::ostream&
- std::istream&
Explanation:
The accurate solution is alternative (c). The stream insertion operator << generally provides a reference back with the std::ostream object, enabling the concatenation of several << actions.
- Which of these statements accurately overloads the equality operator == for a class?
class A {
public:
bool operator==(const A&) const;
};
- bool operator==(A&);
- bool operator==(const A&);
- bool operator==(const A&) const;
- bool operator==(A&) const;
Explanation:
The accurate choice is alternative (c). The equality operator == is commonly redefined as a constant member function that accepts a non-const reference to the other operand.
