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Chapter 11: Problem 35

In the following line of code, the class access specification for the base class is _______. class Pet : \(\mathbb{R}\) ish

Short Answer

Expert verified
Answer: In the given line of code, the class access specification for the base class 'Fish' is 'private'.

Step by step solution

01

Identify Base Class and Derived Class

In the given line of code, 'class Pet : Fish', we have two classes. The class 'Pet' is the derived class (or subclass), and the class 'Fish' is the base class (or superclass). Step 2:
02

Clarify Class Access Specification

In C++, there are three types of class access specifications available: 1. Public: Members declared as public are available for any function to access. 2. Private: Members declared as private are accessible only within the class they are declared. 3. Protected: Members declared as protected can be accessed by both the base class and the derived class. Step 3:
03

Identify Implicit Access Specification

In the given line of code, there is no explicit access specification mentioned. When no access specification is provided in C++, the default access specifiers come into play. In C++, the default specifiers are: - For normal classes, the default access specifier is 'private'. - For struct, the default access specifier is 'public'. However, the default access specifier only affects the member variables and member functions, not inheritance. Step 4:
04

Determine the Access Specification for Inheritance

Since there is no access specification mentioned in the given code, the class access specification for the base class used in this context is 'private' by default. This is because, by default, inheritance without specifying an access specifier is considered private inheritance. Conclusion:
05

Class Access Specification for the Base Class

In the given line of code, the class access specification for the base class 'Fish' is 'private'.

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Most popular questions from this chapter

Each of the following class declarations has errors, Locate as many as you can. A) clase Box 1 private: double width: double length: double height: public: Box(double w, 1. h) \(\ldots\) Other member functions follow... \\[ 1: \\] B) class Circle \(\mathbf{1}\) private: double diameter: Int centerX: Int centerY: public: Circle(double d, int \(x,\) int \(y\) ) I diameter \(=d:\) centerX \(=x ;\) center \(Y=y: \quad\) ) // Overloaded = operator void Circle"(Circle \(8 \mathrm{r} 1 \mathrm{ght})\) 1 diameter \(=r 1 g h t\). diameter \\[ \begin{array}{l} \text { centerX }=\operatorname{righ} t \text { . centerX; } \\ \text { center } Y=\operatorname{righ} t \text { . center } Y \text { ; } \end{array} \\] \(\ldots\) Other member functions follow \(\ldots\) 1: C) class Point 1 private: Int xcoord: Int yCoord: public: Point (int x, int y) 1 xcoord \(=x: y\) Coord \(=y: 1\) // Overloaded + operator void operatort (const \&Point Right) XCoord te right. xCoord: yCoord \(+=\) right. ycoord: \\[ 1 \\] \(\ldots\) Other member functions follow \(\ldots\) \\[ 1: \\] D) class Box 1 private: double width: double length: double height: public: Box (double \(w, 1, h)\) I width \(=w ;\) length \(=1:\) helght \(=h: 1\) // Overloaded prefix \(+\) operator void operatort \(+()\) 1 \(\quad\) twidth \(;\) thength: 1 // Overloaded postfix \(++\) operator void operatort \(+()\) ( \(\quad\) widtht \(;\) lengtht \(;\) ) Other member functions follow... \\[ 1: \\] E) class Yard 1 private: double length: public: Yard (double 1 ) I length \(=1: 1\) // double conversion function void operator double () ( return length: ) \(\ldots\) Other member functions follow... \\[ 1 \\]

Protected members of a base class are like _________ members, except they may be accessed by derived classes.

Assume a class named Collection exists. Write the header for a member function that overloads the \(|]\) operator for that class.

Explain why a class's copy constructor is called when an object of that class is passed by value into a function.

Complete the following table by filling in private, protected, public, or inaccessible in the right-hand column: $$\begin{array}{ll} \hline \begin{array}{l} \text { In a public base class, this base class } \\ \text { MEMBER access specification... } \end{array} & \begin{array}{l} \text { .. becomes this access specification } \\ \text { in the derived dass. } \end{array} \\ \hline \text { private } & \\ \text { protected } & \\ \text { public } & \\ \hline \end{array}$$
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