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jeka57 [31]
3 years ago
5

Among object-oriented languages, one feature that varies considerably is whether the language allows multiple inheritance. C++ d

oes but Ada does not. Java takes a middle ground approach of allowing multiple inheritance of interfaces but not classes. Using a C++ example, illustrate some of the complexities that multiple inheritance introduces. How does C++ deal with them? Why does Java's middle ground approach offer some of the benefits of multiple inheritance while avoids its problems.
Computers and Technology
1 answer:
Amiraneli [1.4K]3 years ago
4 0

Multiple inheritance causes Diamond problem which happens when:

Class A is parent of class B and C

Now when class D will be inherited from both Class B and C it will have all the members of class A and B which if same will confuse the compiler to import which one?

C++ solves it by using virtual keyword with them and thus telling the compiler which one to inherit.

Java has introduced the interface concept rather then allowing multiple inheritance.

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You have a notebook computer and wish to connect to an IEEE 802.11n wireless network. The computer does not have a built-in wire
Ira Lisetskai [31]

Answer:

The wireless adapter to use is the Wireless USB 2.0 Extender

Explanation:

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3 0
3 years ago
Programming challenge description: Given a string comprising just of the characters (,),{,},[,] determine if it is well-formed o
nadezda [96]

Answer:

Employing Java language,

Explanation:

//for using stack data structure

import java.util.Stack;

//class definition

public class BracketMatch {

   //function to find the string is well formed or not

   public static boolean isWellFormed(String str) {

       //variable to store the character value

       char ch;

       //check if the first character itself is the end of the brackets

       if (str.charAt(0) == ')'||str.charAt(0) =='}'||str.charAt(0) ==']')

           return false;

       //creating a stack of character type

       Stack<Character> stack = new Stack<Character>();

       //iterating through the characters in string

       for(int i=0; i < str.length(); i++) {

           //storing the current character from the string in the variable

           ch = str.charAt(i);

           if(ch == '('||ch=='{'||ch=='[')

               //if the character is starting of the brackets push it into stack

               stack.push(ch);

           else if(ch == ')')

               //if it is closing bracket and the stack is empty then the string is not well formatted

               if(stack.empty())

                   return false;

               else if(stack.peek() == '(')

                   stack.pop();

               else

                   return false;

           else if(ch == '}')

               if(stack.empty())

                   return false;

               else if(stack.peek() == '{')

                   stack.pop();

               else

                   return false;

       }

       //returning the stack content, if the string will be well formed the stack would be empty

       //after all the operations

       return stack.empty();

   }

   //main method

   public static void main(String[] args) {

       //calling function to find if the string is welll formatted or not

       System.out.println(isWellFormed("{}"));

       //calling function to find if the string is welll formatted or not

       System.out.println(isWellFormed("([)]"));

   }

}

5 0
3 years ago
If AL contains binary 1000 1111, which one will show the information of the binary bits in AL after performing operation SHR AL,
padilas [110]

Answer:

a) AL will contains 0011 1100

Explanation:

In assembly language, shifting bits in registers is a common and important practice. One of the shifting operations is the SHR AL, x where the x specifies that the bits be shifted to the right by x places.

SHR AL, 2 therefore means that the bits contained in the AL should be shifted to the right by two (2) places.

For example, if the AL contains binary 1000 1111, the SHR AL, 2 operation will cause the following to happen

Original bit =>                                        |   1  |  0 |  0  |  0  |  1 |  1 |  1 |  1 |

Shift once to the right =>                |  1 |   0 |  0  |  0  |  1 |  1 |  1 |  1 | (0) |

Shift once to the right =>          |  1 |  0 |  0  |  0  |  1 |  1 |  1 |  1 | (0) | (0) |

Notice;

(i) that there are two shifts - one at a time.

(ii) that the bits in bold face are the bits in the AL after the shift. Those that in regular face are those in the carry flag.

(iii) that the new bits added to the AL after a shift are the ones in bracket. They are always set to 0.

5 0
3 years ago
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