I'm getting a conversion issue with this program. When I ran this program using the Windows command prompt it compiles and runs without issues. Although, when I run it using the zyBooks compiler it keeps giving me these conversion errors in the following lines:

main.c: In function ‘main’: 
main.c:53:24: error: conversion to ‘float’ from ‘long int’ may alter its value [-Werror=conversion] averageCpi += cpi[i] * instructionCount[i] * Million;

main.c:57:29: error: conversion to ‘float’ from ‘long int’ may alter its value [-Werror=conversion] averageCpi = averageCpi / (totalInstructions * Million);

main.c:58:47: error: conversion to ‘float’ from ‘long int’ may alter its value [-Werror=conversion] mips = (frequency * Million) / (averageCpi * Million);

main.c:58:37: error: conversion to ‘float’ from ‘long int’ may alter its value [-Werror=conversion] mips = (frequency * Million) / (averageCpi * Million);

main.c:59:32: error: conversion to ‘float’ from ‘long int’ may alter its value [-Werror=conversion] executionTime = ((averageCpi * (totalInstructions * Million)) / (frequency * Million)) * 1000;

main.c:59:65: error: conversion to ‘float’ from ‘long int’ may alter its value [-Werror=conversion] executionTime = ((averageCpi * (totalInstructions * Million)) / (frequency * Million)) * 1000;

cc1: all warnings being treated as errors


Is there a more efficient way to convert from float to long int?

Here is the program: 

#include <stdio.h>

int main() {

    int instructionCount[100], totalInstructions = 0, cpi[100], noOfClass = 0, frequency = 0, ch = 0;
    float averageCpi = 0 , executionTime , mips;
    const long MILLION = 1000000;
    do{
       
        printf("\n Performance Assessment: ");
        printf("\n ----------- -----------");
        printf("\n 1) Enter Parameters");
        printf("\n 2) Print Results");
        printf("\n 3) Quit");
        printf("\n Enter Selection: ");
        scanf("%d",&ch);

        if(ch == 1){

           printf("\n Enter the number of instruction classes:  ");
           scanf("%d",&noOfClass);
           printf("\n Enter the frequency of the machine (MHz): ");
           scanf("%d",&frequency);

            for (int i = 0; i < noOfClass; ++i) {
                printf("\n Enter CPI of class %d :  ",(i+1));
                scanf("%d", &cpi[i]);
                printf("\n Enter instruction count of class %d (millions): ",(i+1));
                scanf("%d", &instructionCount[i]);
                totalInstructions += instructionCount[i];

            }
        } else if(ch == 2){

            printf("\nFREQUENCY (MHz): %d", frequency);
            printf("\nINSTRUCTION DISTRIBUTION");
            printf("\nCLASS \t CPI \t COUNT");

            for (int i = 0; i < noOfClass; ++i) {

                printf("\n %d \t %d \t %d",(i+1), cpi[i], instructionCount[i]);
                averageCpi += (cpi[i] * instructionCount[i] * MILLION);

            }

            averageCpi = averageCpi / (totalInstructions * MILLION);
            mips = (frequency * MILLION) / (averageCpi * MILLION);
            executionTime = ((averageCpi * (totalInstructions * MILLION)) / (frequency * MILLION)) * 1000;


            printf("\n PERFORMANCE VALUES");
            printf("\n AVERAGE CPI \t %.2f", averageCpi);
            printf("\n TIME (ms) \t %.2f", executionTime );
            printf("\n MIPS \t %.2f", mips);
            printf("\n");
        }

    }while(ch != 3);

    return 0;
}

Answer each of the following. Assume that single-precision floating-point numbers are stored in 8 bytes, and that the starting address of the array is at location 2030100 in memory. Each part of the exercise should use the results of previous parts where appropriate.

a) Define an array of type float called numbers with 5 elements, and initialize the elements to the values 0.11, 0.22, 0.33, …, 0.55. Assume the symbolic constant SIZE has been defined as 5.

b) Define a pointer, nPtr, that points to an object of typefloat.

c) Print the elements of array numbers using array subscript notation. Use a for statement and assume the integer control variable i has been defined. Print each number with 2 position of precision to the right of the decimal point.

d) Give two separate statements that assign the address of last element of array numbers to the pointer variable nPtr.

e) Printthe elements of array numbers using pointer/offset notation with the pointer nPtr.

f) Print the elements of array numbers using pointer/offset notation with the array name as the pointer.

g) Print the elements of array numbers by subscripting pointer nPtr. h) Refer to element 2 of array numbers using array subscript notation, pointer/offset notation with the array name as the pointer, pointer subscript notation with nPtr and pointer/offset notation with nPtr. i) Assuming that nPtr pointsto the end of array numbers (i.e., the memory location after the last element of the array), what addressisreferenced by nPtr - 5? What value is stored at thatlocation? j) Assuming that nPtr points to numbers[5], what address is referenced by nPtr – = 2? What’s the value stored at that location?

def isPower(x, y):
    """
        >>> isPower(4, 64)
        3
        >>> isPower(5, 81)
        -1
        >>> isPower(7, 16807)
        5
    """
    # --- YOU CODE STARTS HERE
    def isPower(x, y):
        num = x
        count = 1
        while x < y:
            x *= num
            count += 1
        if x == y:
            return count
        return -1


def translate(translation, txt):
    """
        >>> myDict = {'up': 'down', 'down': 'up', 'left': 'right', 'right': 'left', '1': 'one'} 
        >>> text = 'Up down left Right forward 1 time' 
        >>> translate(myDict, text) 
        'down up right left forward one time'
    """
    # --- YOU CODE STARTS HERE
    def translate(translationDict, txt):
        words = txt.lower().split()
        for i in range(len(words)):
            if words[i] in translationDict:
                words[i] = translationDict[words[i]]
        return ' '.join(words)




def onlyTwo(x, y, z):
    """
        >>> onlyTwo(1, 2, 3)
        13
        >>> onlyTwo(3, 3, 2)
        18
        >>> onlyTwo(5, 5, 5)
        50
    """
    # --- YOU CODE STARTS HERE
    def onlyTwo(x, y, z):
        m1 = max(x, y, z)
        m2 = (x + y + z) - min(x, y, z) - m1
        return m1 * m1 + m2 * m2



def largeFactor(num):
    """
        >>> largeFactor(15) 
        5
        >>> largeFactor(24)
        12
        >>> largeFactor(7)
        1
    """
    # --- YOU CODE STARTS HERE
    def largeFactor(num):
        n = num - 1
        while n >= 1:
            if num % n == 0:
                return n
            n -= 1




def hailstone(num):
    """
        >>> hailstone(5)
        [5, 16, 8, 4, 2, 1]
        >>> hailstone(6)
        [6, 3, 10, 5, 16, 8, 4, 2, 1]
    """
    # --- YOU CODE STARTS HERE
    def hailstone(num):
        result = []
        while num > 1:
            result.append(num)
            if num % 2 == 0:
                num = num // 2
            else:
                num = 3 * num + 1
        result.append(num)
        return result

if any wrong with my code

What is the minimum length in hop by hop extension header in ipv6 ?

In this assignment, you are expected to rewrite the program from assignment7 using structs. That is, instead of using multiple arrays (one for each field, you need to create a single array of the datatype you should create as a struct).

So, write a C++ program (use struct and dynamic memory allocation) that reads N customer records from a text file (customers.txt) such as each record has 4 fields (pieces of information) as shown below:

Account Number (integer)

Customer full name (string)

Customer email (string)

Account Balance (double)

The program is expected to print the records in the format shown below, sorted in decreasing order based on the account balance:

Account Number : 1201077

Name                     : Jonathan I. Maletic

Email                     : [email protected]

Balance                   : 10,000.17

-------------------------------------

Note: The records stored in the following format (4 lines for each account). You need to create a text file that contains sample records as explained in class:

First Line is the account number

Second Line is full name

Third line is email

Forth line is the available balance

Important Note: You must create a struct called (Customer) with enough members (to hold record information) and use it to declare a dynamic array to store all the records (information) read from the file. Then sort the array and then print it. Use the example shared in class (see the slides) as a model for your program.

• Make sure your programs adhere to proper programming style (e.g., good identifiers, comments, etc.)

This is C++, please insert screenshot of output please.

Part1: Palindrome detector

Write a program that will test if some string is a palindrome

1. Ask the user to enter a string
2. Pass the string to a Boolean function that uses recursion to determine if something is a palindrome or not. The function should return true if the string is a palindrome and false if the string is not a palindrome.
3. Main displays the result ( if the string is a palindrome or not

Problem 1 (Ransom Note Problem) in python

A kidnapper kidnaps you and writes a ransom note. He does not write it by hand to avoid having his hand writing being recognized, so he uses a magazine to create a ransom note. We need to find out, given the ransom string and magazine string, is it possible to create a given ransom note. The kidnapper can use individual characters of words.

Here is how your program should work to simulate the ransom problem:

• your program should prompt the user to enter a long String to represent the magazine and another short String to represent the required ransom note.
• your program needs to check if the magazine string contains all required characters in equal or greater number present in the ransom note.
• your program should print true if it is possible to create the given ransom note from the given magazine, and print false otherwise.
• Break up your code into a set of well-defined functions. Each function should include a comment block that briefly describes it, its parameters and any return values. 


Example: If the magazine string is “programming problems are weird”

If the ransom note is: “no see” your program should print true as all the characters in “no see” exist in the magazine string.

If the ransom note is “no show” your program should print false as not all the characters in “no show” exist in the magazine string as you can see the character ‘h’ does not exist.

1. What are some drawbacks to crowd sourced answers?

2. Do people generally utilize the diversity of sources on the Internet effectively?

3. How reliant are we and how reliant should we be on getting our news from social media?

Can anyone just check my code and tell me why the program doesn't show the end date & stops at the date before the end date? Tell me how i can fix it.

Question:

Write a program compare.cpp that asks the user to input two dates (the beginning and the end of the interval). The program should check each day in the interval and report which basin had higher elevation on that day by printing “East” or “West”, or print “Equal” if both basins are at the same level.

My code:

#include

#include

#include

#include

using namespace std;

int main()

{

string date;

string starting_date;

string ending_date;

double eastSt;

double eastEl;

double westSt;

double westEl;

int dateRange=0;

cout<< "Enter the starting date" << endl;

cin>> starting_date;

cout<< "Enter the ending date" << endl;

cin>> ending_date;

ifstream fin("Current_Reservoir_Levels.tsv");

if (fin.fail())

{

cerr << "File cannot be opened for reading." << endl;

exit(1);

}

string junk;

getline(fin, junk);

while(fin >> date >> eastSt >> eastEl >> westSt >> westEl)

{

fin.ignore(INT_MAX, '\n');

if (date == starting_date)

{

dateRange = 1;

}

if (date == ending_date)

{

dateRange= 0;

}

if (dateRange == 1)

{

if(eastEl > westEl)

{

cout<< date << " "<< "East " <

}

else if (eastEl < westEl)

{

cout<< date << " " << "West" <

}

else if (eastEl == westEl )

{

cout<< date << " " << "Equal" <

}

}

}

return 0;

}

Expected output

01/17/2018 West
  01/18/2018 West
  01/19/2018 West
  01/20/2018 West
  01/21/2018 West
  01/22/2018 West
  01/23/2018 West

Received output:

01/17/2018 West
  01/18/2018 West
  01/19/2018 West
  01/20/2018 West
  01/21/2018 West
  01/22/2018 West

*************** F.H**************

Hello, I keep getting the error C4430: missing type specifier - int assumed. Note: C++ does not support default-int for my code. What am I doing wrong? I only have #include <iostream>

int read_num(void);
main()
{
   int num1, num2, max;

   //find maximum of two 2 numbers
   num1 = read_num();
   num2 = read_num();
   if (num1 > num2)
   {
       max = num1;
   }
   else
   {
       max = num2;
   }
   printf("the maximum number is: %d\n", max);
  
   return;
}

int read_num(void)
{
   int num;

   printf("Enter a number: ");
   scanf_s("%d", &num);

   return (num);
}

1. Describe the basic components of network security and the differences between wired and wireless network security best practices.

Instruction

This task is about using the Java Collections Framework to accomplish some basic textprocessing tasks. These questions involve choosing the right abstraction (Collection, Set, List, Queue, Deque, SortedSet, Map, or SortedMap) to efficiently accomplish the task at hand. The best way to do these is to read the question and then think about what type of Collection is best to use to solve it. There are only a few lines of code you need to write to solve each of them. Unless specified otherwise, sorted order refers to the natural sorted order on Strings, as defined by String.compareTo(s). Part 0 in the assignment is an example specification and solution.

Part0

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Set;

public class Part0 {
  
   /**
   * Read lines one at a time from r. After reading all lines, output
   * all lines to w, outputting duplicate lines only once. Note: the order
   * of the output is unspecified and may have nothing to do with the order
   * that lines appear in r.
   * @param r the reader to read from
   * @param w the writer to write to
   * @throws IOException
   */
   public static void doIt(BufferedReader r, PrintWriter w) throws IOException {
Set<String> s = new HashSet<>();

for (String line = r.readLine(); line != null; line = r.readLine()) {
s.add(line);
}

for (String text : s) {
w.println(text);
}
   }

   /**
   * The driver. Open a BufferedReader and a PrintWriter, either from System.in
   * and System.out or from filenames specified on the command line, then call doIt.
   * @param args
   */
   public static void main(String[] args) {
       try {
           BufferedReader r;
           PrintWriter w;
           if (args.length == 0) {
               r = new BufferedReader(new InputStreamReader(System.in));
               w = new PrintWriter(System.out);
           } else if (args.length == 1) {
               r = new BufferedReader(new FileReader(args[0]));
               w = new PrintWriter(System.out);              
           } else {
               r = new BufferedReader(new FileReader(args[0]));
               w = new PrintWriter(new FileWriter(args[1]));
           }
           long start = System.nanoTime();
           doIt(r, w);
           w.flush();
           long stop = System.nanoTime();
           System.out.println("Execution time: " + 10e-9 * (stop-start));
       } catch (IOException e) {
           System.err.println(e);
           System.exit(-1);
       }
   }
}

Question 6

[5 marks] Read the input one line at a time and output the current line if and only if it is not a suffix of some previous line. For example, if the some line is "0xdeadbeef" and some subsequent line is "beef", then the subsequent line should not be output.

Template code

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;

public class Part6 {
  
   /**
   * Your code goes here - see Part0 for an example
   * @param r the reader to read from
   * @param w the writer to write to
   * @throws IOException
   */
   public static void doIt(BufferedReader r, PrintWriter w) throws IOException {
       // Your code goes here - see Part0 for an example
   }

   /**
   * The driver. Open a BufferedReader and a PrintWriter, either from System.in
   * and System.out or from filenames specified on the command line, then call doIt.
   * @param args
   */
   public static void main(String[] args) {
       try {
           BufferedReader r;
           PrintWriter w;
           if (args.length == 0) {
               r = new BufferedReader(new InputStreamReader(System.in));
               w = new PrintWriter(System.out);
           } else if (args.length == 1) {
               r = new BufferedReader(new FileReader(args[0]));
               w = new PrintWriter(System.out);              
           } else {
               r = new BufferedReader(new FileReader(args[0]));
               w = new PrintWriter(new FileWriter(args[1]));
           }
           long start = System.nanoTime();
           doIt(r, w);
           w.flush();
           long stop = System.nanoTime();
           System.out.println("Execution time: " + 10e-9 * (stop-start));
       } catch (IOException e) {
           System.err.println(e);
           System.exit(-1);
       }
   }
}

1. Briefly explain the principle of least privilege and how it should be applied. 2. Explain what describes a truly secure password.

Summary

In this lab the completed program should print the numbers 0 through 10, along with their values multiplied by 2 and by 10. You should accomplish this using a for loop instead of a counter-controlled while loop.

Instructions

1. Write a for loop that uses the loop control variable to take on the values 0 through 10.
2. In the body of the loop, multiply the value of the loop control variable by 2 and by 10.
3. Execute the program. Is the output the same?

4. // NewMultiply.java - This program prints the numbers 0 through 10 along
   // with these values multiplied by 2 and by 10.
   // Input: None.
   // Output: Prints the numbers 0 through 10 along with their values multiplied by 2 and by 10.

   
   public class NewMultiply
   {
   public static void main(String args[])
   {
     
   String head1 = "Number: ";
   String head2 = "Multiplied by 2: ";
   String head3 = "Multiplied by 10: ";   
   int numberCounter; // Numbers 0 through 10.
   int byTen; // Stores the number multiplied by 10.
   int byTwo; // Stores the number multiplied by 2.
   final int NUM_LOOPS = 10; // Constant used to control loop.

   // This is the work done in the housekeeping() method
   System.out.println("0 through 10 multiplied by 2 and by 10" + "\n");

   // This is the work done in the detailLoop() method
   // Write for loop
   
   // This is the work done in the endOfJob() method
   System.exit(0);
   } // End of main() method.

   } // End of NewMultiply class.

1. Briefly but completely explain the two most significant differences you discern from Windows and Linux file management.

2. Explain why Windows compatibility may cause issues for users.