You need to write a program that reads in the mass of an object (in kg) and output the weight (in N) on the Earth, on the Moon, and on Venus. An object's mass can be measured in kilograms. The weight is measured in newtons. So an object of a specific mass (in kilograms) would have one weight (in newtons) on the earth and a different weight on the moon. Your program will read in the mass (in kilograms) and convert it to newtons for the Earth, the Moon, and Venus. So, on the Earth we can convert kilograms to newtons with the following expression: weight = mass * 9.81 where 9.81 is the acceleration due to gravity on earth (in meters per second squared or m/s^2). On the Moon this formula would be: weight = mass * 1.62 where 1.62 is the acceleration due to gravity on the moon (m/s^2) Finally, for Venus it would be: weight = mass * 8.87

Answer:

184

Explanation:

Given the codes

The initial value of cost is 82.

The if condition in Line 3 will pass and therefore cost = 82 + 10 = 92

The if condition in Line 8 will pass and therefore cost = 92 * 2 = 184

The if condition in Line 13 will fail l and therefore the if block will be skipped.

At last, the we get cost = 184.

Answer:

See the pictures attached

Explanation:

Final answer:

The question involves writing a program in C++ and Python that counts multiples of 3 and 5 between two user-entered integers. Examples of both a C++ and Python program are provided, demonstrating how to prompt for input and calculate the counts.

Explanation:

Counting Multiples in C++ and Python:

A program that counts multiples of 3 and 5 between two integers can be written in multiple programming languages. Below are examples in both C++ and Python that accomplish the task. Both programs prompt the user for two integers and then iterate through the range, counting the numbers that are multiples of 3 and 5, respectively.

C++ Program

#include
using namespace std;
int main() {
   int start, end, count3 = 0, count5 = 0;
   cout << "Enter two integers: ";
   cin >> start >> end;
   for (int i = start; i <= end; ++i) {
       if (i % 3 == 0) count3++;
       if (i % 5 == 0) count5++;
   }
   cout << "Multiples of 3: " << count3 << endl;
   cout << "Multiples of 5: " << count5 << endl;
   return 0;
}

Python Program

start = int(input("Enter the first integer: "))
end = int(input("Enter the second integer: "))
count3 = len([i for i in range(start, end+1) if i % 3 == 0])
count5 = len([i for i in range(start, end+1) if i % 5 == 0])
print(f"Multiples of 3: {count3}")
print(f"Multiples of 5: {count5}")

Final answer:

To solve this problem, you can write a program that prompts the user to enter two integers and then displays the number of multiples of 3 and 5 between those two integers.

Explanation:

To solve this problem, you can use a loop to iterate through each number between the two integers provided by the user. Within the loop, you can check if a number is a multiple of 3 or 5 by using the modulus operator (%). If the remainder of dividing the number by 3 is 0, then it is a multiple of 3. If the remainder of dividing the number by 5 is 0, then it is a multiple of 5. You can keep track of the count of multiples using two separate counters and display the results at the end.

Here's an example of how the program might look in Python:

Answer:

The actual methods used to protect against data loss are Technical controls, but the program that identifies which data to protect is a Procedural control.

Explanation:

In technical control, different software has been installed and protocols has been deployed to avoid any data loss in network security. As all the devices and protocols that we need while installing a network needs technical support. This is the reason, these things are called technical control of the system.

Another control in network security is procedural control. In this type of control it has been identified that, which data should be protected or which data should not be accessed to unauthorized users. This is called procedural control.

Answer:

Explanation: see attachment below

Explanation:

The "re" module in python provides various functions to perform complex string operations easily.

For example, you can check whether a specific word exists in a given string or to replace it with some other word.  

Some of the functions in the re module are:

re.findall

re.search

re.sub()

We are particularly interested in the re.sub() function, we can use this function to replace the occurrences of space, coma or dot with semicolon.  

Syntax:

re.sub(pattern, replace, target_string, n).  

Where pattern = space, coma or dot  

replace = semicolon  

target_string = any string containing spaces, coma and dots

n = number of times you want the replacement to be done  

Code:

import re  

target_string = "Hello world,code,code."  

replace = ":"

pattern = "[ ,.]"

print(target_string)

print(re.sub(pattern, replace, target_string, 2) )

Output:

Hello world,code,code.

Hello:world:code,code.

It replaced the first white space and coma with semicolon, if we increase the n to 4 then the output is

Hello:world:code:code:

As expected, it replaced all of the space, coma, and dots with semicolons.

Final answer:

The solution is to write a Python program that replaces up to two occurrences of spaces, commas, or dots with colons using the 're.sub()' function from the 're' module.

Explanation:

The question involves writing a Python program using the module re to perform a substitution on a string. The task is to replace at most 2 occurrences of a space, comma, or dot with a colon. Here is an example of how to do this using the re.sub() function:

import re
def replace_characters(text):
   # The regex pattern matches a space, comma, or dot
   pattern = '[ ,.]'
   # This will replace at most 2 occurrences of the pattern with a colon
   result = re.sub(pattern, ':', text, 2)
   return result

# Example usage
input_text = "This is a sentence. With spaces, and commas."
output_text = replace_characters(input_text)
print(output_text)

This program will output "This:is:a sentence. With spaces, and commas." after replacing the first two occurrences of the matched characters with colons.

Answer:

Explanation:

def the_perfect(n):

  try: #exception handling if n is a negative number

       n > 0

   except: #return -1 if the error is reached

       return -1

  else:

        total = 0

        for i in range(1, n): #for loop from 1 to the target number

           if n % i == 0:

               total += i

   return total == n  #should return true if number is perfect number

print(perfect_number(8))

Answer:

500kbps

Explanation:

Bandwidth is the actual amount of data that a network is capable of transferring theoretically.

Throughput is the actual amount of data passing through a connection. It is the rate (usually measured in bps- bits per sec or pps - packets per second) at which packets or bits are successfully delivered over a network channel

If R₁=500 kbps, R₂=2 Mbps, and R₃ = 1 Mbps.

Since there is no other traffic in the link from Host A to Host B, the Throughput is the minimum of the links between Host A and Host B. Therefore:

Throughput= Mininum(R₁,R₂,R₃)

=500kbps

Answer:

Access functions explained below

Explanation:

Access functions for Three Dimensional Arrays:

Let M, N and P refer to the size of the 1st 2nd and 3rd dimensions respectively.

Element_Size is the memory size the array element.

Generic access functions for Row Major:

If character data is used then Element_Size is 1.

For i=0 to P do

For i=0 to N do

  For i=0 to M do

     Address_of _array[i, j, k] = address_of_array[0,0,0] + (i * P + j * N + k) * Element_Size

    Store data into the Address_of _array[i, j, k]

    Or

  Display data from Address_of _array[i, j, k]

end

end

end        

Generic access function for Column Major:

if For i=0 to M do

For i=0 to N do

  For i=0 to P do

     Address_of _array[i, j, k] = address_of_array[0,0,0] + (i * M + j * N + k) * Element_Size

    Store data into the Address_of _array[i, j, k]

    Or

  Display data from Address_of _array[i, j, k]

end

end

end      

You can get complete answer in attached document.please have a look.

Explanation:

(2) Without rolloff, as given fc =1kHz Let us assume it is a first order RC filter ats transfer function is H(s)-RC RC withou

Then the transfer function becom

and step response is flat zero for without rolloff..

Answer:

top down approach

Explanation:

Top down approach is process of breaking down system into sub-system which helps in formulating the overall system. At the end every sub system can be refined with more details.

While bottom up approach is process of gathering systems to design large complex system.

Top down is better as basic information is present and resources can be utilized effectively without impacting maintennance and operations.

Final answer:

Choosing between a top-down approach and a bottom-up approach for developing a school registration system depends on the project goals and context. A top-down approach is hierarchically structured and policy-driven, while a bottom-up approach focuses on detailed, user-centric components. A hybrid approach may offer the best of both methodologies.

Explanation:

When developing a logical model of a registration system at a school or college, choosing between a top-down approach and a bottom-up approach depends on the specific context and goals of the project. A top-down approach generally starts with defining the high-level processes and structures of the system based on the policies and requirements of the institution. It is methodical, following a hierarchical path from the general to the specific, which can be beneficial when there are clear objectives and standards to adhere to.

On the other hand, a bottom-up approach builds the system from the lowest level of detailed requirements, often focusing on user needs and individual system components before integrating them into the full model. This can be advantageous when needing to understand complex details from the outset or when user feedback is critical in shaping the system's functionality.

In the implementation phase of a policy process, one might consider elements from both strategies. For example, important federal policies might require adherence to regulations (top-down), but the practical and on-the-ground impact of policies might be best understood by considering local circumstances (bottom-up). Thus, developing a registration system could potentially benefit from a hybrid approach, combining the structured, policy-driven aspects of the top-down method with the detailed, user-driven focus of the bottom-up method.

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

Answer:

It will send ARP request broad cast because of security and pubilicity of data to the public.

Answer:

Invoice.java

import java.util.*;

public class Invoice {

   static final String endPrices = "END_PRICES";

   static final String endInvoice = "END_INVOICE";

   static final String quit = "QUIT";

   public static void main(String... args) {

       Scanner sc = new Scanner(System.in);

       //HashMap to store fruit name as key and price as value

       Map<String, Float> fruits = new HashMap<>();

       //Loop until input is not "END_PRICES"

       while (true){

           String input = sc.next();

           //Come out of loop if input is "END_PRICES"

           if(input.equals(endPrices))

               break;

           float price = Float.parseFloat(sc.next());

           //add fruit to hash map

           fruits.put(input, price);

       }

       //ArrayList to store total cost of each invoice

       List<Float> totalList = new ArrayList<>();

       Float total = 0f;

       //loop until input becomes "QUIT"

       while (true){

           String input = sc.next();

           //Break out of the loop if input is "QUIT"

           if(input.equals(quit)){

               break;

           }

           //Add total price of the invoice to array list and set total to "0f" to store total of next invoice

           if(input.equals(endInvoice)){

               totalList.add(total);

               total = 0f;

               continue;

           }

           int quantity = sc.nextInt();

           total += (fruits.get(input)*quantity);

       }

       //Iterate through all objects in the total Array List and print them

       Iterator itr = totalList.iterator();

       while (itr.hasNext()){

           System.out.printf("Total: %.2f \n", itr.next());

       }

   }

}

Answer:

The memory and the processing unit communicate through the Memory Address Register (MAR) and the Memory Data Register (MDR).

Explanation:

Final answer:

The memory and processing unit in a computer communicates through the system bus, utilizing RAM for fast access to data. This interaction is optimized with techniques like caching and, when necessary, swapping to manage memory usage effectively.

Explanation:

The interface between the memory and the processing unit in a computer system facilitates communication and data exchange crucial for computational tasks. These two components communicate through a data pathway known as the system bus. This bus includes data lines for transferring data, address lines to specify where the data should be sent, and control lines to manage the timing and direction of data flow.

Random Access Memory (RAM) plays a pivotal role in this communication. When a process is initiated, data from the long-term storage (like a hard drive) is loaded into RAM, because RAM is much faster, though more expensive. The processing unit (CPU) then reads instructions and data directly from RAM, performs computations, and may write the results back to RAM or send them to output devices or long-term storage.

Another method to facilitate communication involves caching. CPU caches are small, faster memory locations that store copies of the data from frequently used main memory locations. These caches help reduce the time it takes for the CPU to access data from RAM, speeding up the computation process.

To manage and optimize memory use, operating systems may employ techniques like swapping, where data is moved back and forth between RAM and a hard drive, particularly when RAM is full. However, this can significantly slow down system performance due to the difference in speed between RAM and hard drives.

Answer:

The three important concepts in the Federal Networking Council's definition of the Internet are, namely;

Explanation:

The important concepts in the Federal Networking Council's definition of the Internet include:

It should be noted that packet switching means the transfer of data that are grouped into different packets and then sent through a digital network to different destinations.

The TCP/IP communications protocol is important as it enables a computer to connect with another computer through the internet.

Lastly, client/server computing describes how a server is able to provide services by taking requests from client's computers and then sharing the resources with the computers.

In conclusion, packet switching, TCP/IP communications protocol, and client/server computing are important to the functioning of the internet.

Learn more about the internet on:

https://brainly.com/question/2780939

Answer:

value=int(input("Enter the value from where the user wants to starts: "))#Take the value from the user.

i=1#intialize the value of a variable by 1.

while(value<1000) and (i<9):#while loop which prints the value.

   print(value,end=" ")# print the value.

   value=value*2#calculate the value to print.

   i=i+1#calculate the value for the count.

Output:

If the user enter 5, then the output is : "5 10 20 40 80 160 320 640".

If the user enter 5, then the output is : "3 6 12 24 48 96 192 384".

Explanation:

Answer:

This program is written in C++.  The explanation of the code is given below in the explanation section.

Explanation:

#include <iostream>

using namespace std;

//factorial calculation

int fact(int n) {

  if (n == 0 || n == 1)

     return 1;

  else

     return n * fact(n - 1);

}

int main() {

  int n, r, comb; /* n represent number of people (population), r represent the number of selection from population and combination represent number of ways or combination */

  cout<<"Enter n : ";// prompt user to enter population i.e. total number of people

  cin>>n;//store into n

  cout<<"\nEnter r : ";//prompt user to enter number of selection from population

  cin>>r;

  comb = fact(n) / (fact(r) * fact(n-r));// calcuate combination- number of differnt ways to form a committe

  cout << "\nCombination : " << comb; //show the result

  return 0;

}

A battery has capacity (instance variable) which can be drained. Capacity is a double value measured in milliampere hours (mAh) -A typical AA battery has a capacity of 2000 to 3000 mAh

Explanation:

public class Battery

{

     private double fullCharge;

     private double batteryCapacity;

public Battery(doublecapacity)

{

     battery Capacity = capacity;

     full Charge = capacity;

}

public void drain(double amount)

{

    battery Capacity = battery  Capacity - amount;

}

public void charge()

{

     battery Capacity = fullCharge;

}

public double get RemainingCapacity()

{

    return battery Capacity;

}

}  

Answer:

TCP convention is totally connection oriented, it searches for the information that was going in the system definitely and will guarantee an affirmation on the parcel sent, likewise if the uplayer convention is down, it won't permit to move the information, fundamentally it will guarantee there is no any bundle misfortune during the exchange.  

~ However UDP is connectionless convention, there is no any affirmation business in UDP convention, it regularly utilized where not many datagram misfortune doesn't a make a difference. It was nearly a light gauge convention.  

So underneath are sway we will confront, if TCP is expelled.  

1. There will association foundation required in information move, so the application will work minimal quicker.  

2. without appropriate information control and affirmations, there is gigantic possibility for information misfortune.  

3. Presently the application needs to depend totally on UDP, since UDP doesn't has blockage control, the application will get clogged.  

4. Likewise the communication with top layer convention will be a wreck, it won't look if the layers are down or not. so if any upper layer is down, consequently the information will stuck the system, can't capable arrive at the goal.

Explanation:

Switching to TCP Reno from Tahoe requires adapting to faster congestion recovery and retransmission strategies. Applications should adjust for improved throughput and handling of packet loss signals.

Switching the TCP algorithm from Tahoe to Reno in the operating system affects how your application interacts with the network, particularly in terms of congestion control and handling of packet loss. Here are the key changes you would need to consider and potentially address in your application:

1. Congestion Control Behavior: Reno TCP introduces Fast Recovery and Fast Retransmit mechanisms, which allow faster recovery from packet loss compared to Tahoe. Your application should be aware of these behaviors, especially in scenarios where it needs to manage large data transfers or real-time communications. It may need to adjust its own congestion control strategies or settings to optimize performance with Reno TCP.

2. Handling of Packet Loss: Reno TCP reacts differently to packet loss compared to Tahoe. Instead of immediately halving the congestion window (as Tahoe does), Reno enters Fast Recovery, retransmits the lost packet, and then adjusts its congestion window dynamically. Your application should be prepared for potentially different patterns of packet loss and recovery, which could affect application-level retransmission strategies or timeouts.

3. Performance Monitoring and Tuning: With the change to Reno TCP, monitoring and tuning network performance metrics becomes crucial. Your application may need to adapt its monitoring mechanisms to track parameters specific to Reno's congestion control algorithms, such as congestion window size, packet retransmission rates, and round-trip time estimates.

4. Compatibility and Testing: Ensure compatibility with both Tahoe and Reno TCP variants during transitional periods. Your application should be able to gracefully handle connections with both types of TCP algorithms to support users who have not yet upgraded their operating systems.

5.Documentation and User Communication: Update documentation and inform users about any changes in network behavior or performance optimizations that may result from switching to Reno TCP. Provide guidance on configuring your application to leverage Reno's improvements effectively.

By considering these factors and making necessary adjustments in your application, you can ensure optimal performance and compatibility with the new Reno TCP algorithm introduced in the operating system update.

Answer: true

Explanation:

Double-clicking which is the act of pressing a computer mouse button twice quickly without moving the mouse allows two different actions to be associated with the same mouse button. It was developed by Bill Atkinson of Apple Computer, Double-clicking executes the function associated with that objec and also opens the application on which you double clicked.

Answer:

C++ code is explained

Explanation:

#include <iostream>

#include <string>

#include <sstream>

using namespace std;

int firstDigit(int n){

// Converting int to string

ostringstream convert;

convert << n;

string s = convert.str();

char first_char = s[0];

int first_int = first_char - '0';

return first_int;

}

int lastDigit(int n){

// Converting int to string

ostringstream convert;

convert << n;

string s = convert.str();

char last_char = s[s.length() - 1];

int last_int = last_char - '0';

return last_int;

}

int digits(int n){

// Converting int to string

ostringstream convert;

convert << n;

string s = convert.str();

int length = s.length();

return length;

}

int main() {

int number;

cout << "Enter integer: " << endl;

cin >> number;

cout << "The first digit of the number is " << firstDigit(number) << endl;

cout << "The last digit of the number is " << lastDigit(number) << endl;

cout << "The number of digits of the number is " << digits(number) << endl;

}

Answer:

The correct answer to the following question will be "Iterative procedure".

Explanation:

Iterative production or procedure seems to be a technique of distinguishing product progression from expansive implementation to littler bits. This including code in regurgitated processes is designed, developed and attempted.

Regularly iterative architecture is used in conjunction with gradual improvement where a certain more sketched-out process of code progress is composed of littler components building on each other.

This will improve observer performance, as each move is systematic and cheap to run.

The shorter cycles of Iterative production affect the ability of the analyst to manage system’s requirements in a way that improve the observer performance as each move is systematic and cheap to run.

An Iterative production means a technique of distinguishing product progression from expansive implementation to littler bits.

An Iterative production which includes code in regurgitated processes is designed, developed and attempted.

In conclusion, the shorter cycles of Iterative production affect the ability of the analyst to manage system’s requirements in a way that improve the observer performance as each move is systematic and cheap to run.

Read more about Iterative production

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Answer:

See the pictures attached

Explanation:

Answer:

The explanation of the question is described in the section below.

Explanation:

(a)

Go Back N :

A gives in a maximum of 9 pieces. Typically they will be sent back sections 1, 2, 3, 4, 5 and then re-sent segments 2, 3, 4 and 5.

B sends out 8 ACK's. They are 4 ACKS including 1 series and 4 ACKS with 2, 3, 4 and 5 series amounts.

Selective Repeat :

A sends in such max of 6 bits. Subsequently, segments 1, 2, 3, 4, 5 and earlier re-sent Segments 2 will be sent.

B assigns five ACKs. We are 4our ACKS with numbers 1, 3, 4, 5. However there is one sequence quantity 2 ACK.

TCP :

A assigns in a total of 6 bits. Originally, segments 1, 2, 3, 4, 5 and future re-sent Segments 2 have always been sent.

B sends five ACKs. There are 4 ACKS with either the number 2 series. There has been one ACK with a sequence of numbers 6. Remember that TCP always needs to send an ACK with a sequence number you anticipate.

(b)

This is although TCP utilizes convenient retransmission without searching for the time out.

So, it's the right answer.

Answer:

Explanation:

To check if n is divisible by d, we can use modulus operator such as %. If n % d equal to 0, this means n is divisible by d. Modulus operator is to calculate the remainder obtained from a division. If a number is divisible by another number, the remainder will be 0.

By making use of this concept we can create if and else statement to check if n % d is equal to 0, return true (divisible) else return false (not divisible).

Answer:

Here is the function in Python:

def isDivisibleBy(n,d):

   if(n%d)==0:

       return True;

   else:

       return False;    

Explanation:

Lets see how this works. You can call this function in main program to check if this function isDivisibleBy() works properly.

if(isDivisibleBy(8,2)==True):

   print("It is divisible")

else:

   print(" It is not divisible")

Now the bool function isDivisibleBy() takes two arguments, n and d. bool function returns either true or false. If the number in n is divisible by d the function returns true otherwise it returns false. The number n is completely divisible by d when the remainder of this division is 0. So modulus operator is used to check if n is completely divisible by d.

In main program if condition is used which calls the isDivisibleBy() function and checks if the function returns true. The function is passed two values 8 and 2 where 8 is n and 2 is d. If this condition evaluates to true then the message displayed in output is: It is divisible. When the if condition evaluates to false the messages displayed in output is: It is not divisible.

The program along with its output is attached in the screen shot.

Final answer:

The answer explains the concept of clipping in geometry and how it relates to implementing clippers in a pipeline. It also discusses the proofs and theorems in geometry regarding segments, polygons, and constructions using straight-edge and transferer of segments.

Explanation:

Clipping against the top of the clipping rectangle can be done independently of the other sides due to the nature of line segments. This concept is essential in implementing a clipper as a pipeline of four simple clippers.

Proofs and theorems in geometry demonstrate the relationships and properties of segments, polygons, and shapes in a structured manner.

Understanding the construction using straight-edge and transferer of segments allows for solving a variety of geometrical problems by drawing straight lines and laying off segments.

Answer:

Tom Servo

Bob (prints inside Cyborg constructor)

Bob (prints inside Cyberman constructor)

Explanation:

The above type of inheritance is multilevel inheritance where Robot is super class and Cyborg and Cyberman are sub classes.

So in above statement

Cyberman one = new Cyberman("Bob"); it calls the Cyberman(String name) in which super(name) calls Cyborg(String name) but before executing the body inside Cyborg constructor it first calls super class default constructor (i.e., Robot()) so the output of the above statement is

Tom Servo

Bob (prints inside Cyborg constructor)

Bob (prints inside Cyberman constructor)

Answer:

Code:

    #include <iostream>

    #include <fstream>

    #include <cctype>

    #include <cmath>

    #include <vector>

    #include <float.h>

    using namespace std;

    typedef vector<vector<double> > C2DArray;

    void Matrix_Mult(bool* error_flag, const int mRows, const int nCols, const int pCols, const C2DArray A_Matrix, const C2DArray B_Matrix, C2DArray& C_Matrix) {      

     int i = A_Matrix.size(), j = A_Matrix[0].size(), k = B_Matrix[0].size();

     double dummy;

     if ((j != B_Matrix.size()) || (i != mRows) || (j != nCols) || (k != pCols))  *error_flag = true;

     if ((C_Matrix.size() != i) || (k != C_Matrix[0].size()))  *error_flag = true;

     if (!(*error_flag)){

     for (k = 0; k < mRows; k++) { // Do each row of A with each column of B

      for (i = 0; i < pCols; i++) {

       dummy = 0.0;

       for (j = 0; j < nCols; j++) {

        dummy += A_Matrix[k][j]*B_Matrix[j][i];

          } // End for j

          C_Matrix[k][i] = dummy;

      } // End for i

     } // End for k

     }

        return;

    } // End Matrix_Mult

    int main()

    {char rflag = 0; //Readiness flag  

    cout << "                 Mat_Mul1 (7 June 2013)\n";

    cout << "=========================================================================== \n";

    cout << "This program multiplies two matrices, [A] and [B]: [A][B] = [C].\n";

    cout << "[A] is M X N, [B] is N X P, and [C] is M X P\n";

    cout << "The dimensions of the [A] and [B] matrices, M, N, and P, should \n";

    cout << "have been saved beforehand in a file named mulmatdat.\n";

    cout << "mulmatdat should be in the same folder as the Mat_Mul1 executable.\n";

    cout << "The next values of mulmatdat should be the entries for matrix [A],\n";

    cout << "with data for row 1 first, then row 2, then row 3, etc.\n";

    cout << "The entries for the [B] matrix should follow, in the same order.\n";

    cout << "\nThe data is assumed to be of type double. Variables used within this program\n";

    cout << "are type double.\n";

    cout << "\nThe output is written to the file mulmatout.txt.\n";

    cout << "\nIs everything ready (are you ready to continue?)? If yes, Enter y. \n";

    cout << "Otherwise Enter any other key. \n";

    cin >> rflag;

    if (toupper(rflag) == 'Y') {

    C2DArray A, B, C; // A, B, and C Matrices

    int i, j, mDim, nDim, pDim; // Array index variables and matrix dimensions

    bool erFlag = false;  // Error flag

    ifstream in("mulmatdat.txt", ios::in);

    if (!in) {

            cout << "\nCannot open the input file.\n";

         cout << "\nEnter any key to continue. \n";

         cin >> rflag;

            return 0;

    }

    in >> mDim >> nDim >> pDim; //Input the Matrices' dimensions from the file

    if ((mDim < 1) || (nDim < 1) || (pDim < 1)){

            cout << "\nInvalid dimension entered. Program terminated. \n";

         cout << "\nEnter any key to continue. \n";

         cin >> rflag;

            in.close(); //Close the input file before terminating

            return 0;

    }

    ofstream out("mulmatout.txt", ios::out);

    if (!out) {

            cout << "\nCannot open the output file. Program terminated.\n";

         cout << "\nEnter any key to continue. \n";

         cin >> rflag;

            in.close(); //Close the input file before terminating

            return 0;

    }

    // Beginning of try block, if vector re-sizing unsuccessful

    try {

           // Resize the arrays to the appropriate sizes

            A.resize(mDim); // M rows

         B.resize(nDim); // N rows

         C.resize(mDim); // M rows

            for (i = 0; i < mDim; i++){

               A[i].resize(nDim); // M columns

            C[i].resize(pDim); // P columns

            } // End for i

         for (i = 0; i < nDim; i++){

               B[i].resize(pDim); // P columns

            } // End for i

    } // End of try block

    catch (bad_alloc& xa) { // Catch block, for exceptions

            in.close();

            out.close();

            cerr << "\nIn catch block, so an exception occurred: " << xa.what() << "\n";

            cout << "\nEnter any key to continue. \n";

            cin >> rflag;

            return 0;

    } // End of catch block

    for (i = 0; i < mDim; i++){ //Input the A Matrix from the file

            for (j = 0; j < nDim; j++){

              in >> A[i][j];

            }//End for j

    }//End for i

    for (i = 0; i < nDim; i++){ //Input the B Matrix from the file

            for (j = 0; j < pDim; j++){

             in >> B[i][j];

            }//End for j

    }//End for i

    in.close(); //Close the input file

    Matrix_Mult(&erFlag, mDim, nDim, pDim, A, B, C);

    if (erFlag){

      cout << "Inconsistent data sent to Matrix_Mult routine. Matrix multiplication NOT performed. Check data before running again.\n";

    }

    else {

     out.precision(DBL_DIG);

     out << "\nThe A matrix follows:\n";

     out << "\n";

     for (i = 0; i < mDim; i++){

      for (j = 0; j < nDim; j++){

     out << A[i][j] << " ";

      } // End for j

      out << "\n";

     }//End for i

     out << "\nThe B matrix follows:\n";

     out << "\n";

     for (i = 0; i < nDim; i++){

      for (j = 0; j < pDim; j++){

     out << B[i][j] << " ";

      } // End for j

      out << "\n";

    }//End for i

     out << "\nThe product matrix, C, follows:\n";

     out << "\n";

     for (i = 0; i < mDim; i++){

      for (j = 0; j < pDim; j++){

       out << C[i][j] << " ";

      } // End for j

      out << "\n";

     }//End for i

     cout << "\nDone! The solution is in the text file mulmatout.txt \n";

    } // End else !erFlag

    out.close();

    }

    else cout << "\nNot ready. Try again when ready with information. \n";

    cout << "\nEnter any key to continue. \n";

    cin >> rflag;

    return 0;

    }

Answer:

Hi Zoom! Good question. This is particularly useful for learning sorting functions in arrays and user input handling as well. Please find the implementation below.

Explanation:

I have implemented the code in Python 2+. You can save the below code into a file called ArrayDemo.py and then run it from the command line with the command "python ArrayDemo.py". I have declared the demo_array in the code to an array of 10 unsorted integer values to test this. You can change the code if needed to have this array input by user as well and it should work fine.

Note: if you're using Python 3+ then you'll need to change the "raw_input" in the code to just "input" to make it work.

ArrayDemo.py

demo_array = [2,1,3,4,5,7,6,8,10,9]

option = 0

while (option != 4):

 option = raw_input("Enter 1 to view list in ascending order, 2 to view list in descending order, 3 to choose specific poisition to view, 4 to quit the application: ")

 try:

   option = int(option)

   if option == 1:

     sort_by = 'asc'

     print(sorted(demo_array, reverse=False))

   elif option == 2:

     sort_by = 'desc'

     print(sorted(demo_array, reverse=True))

   elif option == 3:

     position = raw_input("Enter specific position to view (0-9): ")

     try:

       position = int(position)

       if position < 0 or position > 9:

         print("Position does not exist")

       else:

         print(demo_array[position])

     except ValueError:

       print("Position does not exist!")

   elif option == 4:

     break

   else:

     break

 except ValueError:

   print("Invalid input!")