Using either your own C-string functions of Lab 7.1 or the ones from the standrd C++ cstring library, create a String class, which — while having minimal functionality — illustrates the use of and need for the canonical form.

Overview

Here is the .h file for the class (note the class name — String with a capital S; trying to force the use of the classname string was to much of an issue:

class String {
        friend std::ostream &operator <<(std::ostream &os, const String &s);
        
        friend String operator +(const String &s1, const String &s2);
public:
        String(const char *cs="");
        String(const String &s);
        ~String();
        String &operator =(const String &rhs);
        char &operator [](int index);
        String &operator +=(const String &s);
        
        int length() const;
        
private:
        char *cs;
};

I've also supplied a String_Exception class and an app for testing your class (it will be the test driver once I get it all into Codelab).

Implementation Notes

• This exercise has potential name clashes with the standard C C-string library, named cstring, but originally named string.h — recall the naing conventions for the C++ wrapper libraries of standard C libraries. To avoid this, the files have been name mystring, e.g. mystring_app.cpp, mystring.h, etc. The only one that really affects you is the latter since you must #include it in your implementation file (mystring.cpp).
• The String)const char *cs="") constructor allows one to create a String from C-string (and "..." literals, which are of type const char * – i.e., C-strings).
• Operations on the cs buffer are performed using the C-string functions you wrote in lab 4.2.
  • Memory allocation involves making sure the cs data member (i.e., the pointer to the C-string buffer) is pointing to a sufficiently sized buffer.
    • For this implementation, we will use exact-sized buffer; i.e., enough elements in the char array to hold the characters of the C-string + the null terminator
    • This is relevant for the two constructors, the assignment operator and the += and + operators.
      • Using the String(const char *) constructor as an example:
        • when this constructor is called, the length of the argument is obtained using strlen and a buffer of the corresponding size is allocated (this can be done within the member initialization list)
        • the contents of the argument C-string is then copied to this new buffer using strcpy (this needs to be done in the body of the constructor; there is no way to work it into the member intialization list)

          String::String(const char *cs) : cs(new char[strlen(cs)+1) {    // the +1 is for the null terminator
                                                                          

      • Similar logic applies to the copy constructor, the assignment operator, and the += operator (you should be coding the + operator using the += operator as shown in class): in those three cases the source buffer (i.e., the C-string to be copied, assigned, or concatenated) will be the cs data member of another String object
        
        
        
        
        mystring.h

        #ifndef MYSTRING_H
        #define MYSTRING_H
        
        #include <iostream>
        
        
        class String {
                friend std::ostream &operator <<(std::ostream &os, const String &s);
        //      friend bool operator ==(const String &s1, const String &s2);
                friend String operator +(const String &s1, const String &s2);
        public:
                String(const char *cs="");
                String(const String &s);
                ~String();
                String &operator =(const String &rhs);
                char &operator [](int index);
                String &operator +=(const String &s);
        //      int find(char c) const;
                int length() const;
        //      void clear();
        private:
                char *cs;
        };
        
        #endif

mystring_app.cpp

#include <iostream>
#include <sys/sysinfo.h>
#include <cstdlib>

#include "mystring.h"

using namespace std;

int main() {
        String s = "Hello";

        cout << "s: " << s << " (" << s.length() << ")" << endl;

        cout << "s + \" world\": " << s + " world" << endl;

        cout << "s[1]: " << s[1] << endl;

        String s1 = s;          // making sure copy constructor makes deep copy
        String s2;
        s2 = s;                 // making sure assignment operator makes deep copy
        s[0] = 'j';
        cout << endl;
        cout << "s: " << s << " (" << s.length() << ")" << endl;
        cout << "s1: " << s1 << " (" << s1.length() << ")" << endl;
        cout << "s2: " << s2 << " (" << s2.length() << ")" << endl;
        cout << endl;

        for (int i = 0; i < 5; i++) {
                s += s;
                cout << "s: " << s << " (" << s.length() << ")" << endl;
        }
        cout << endl;

        for (int i = 0; i < 5; i++) 
                s += s;
        cout << "s: " << s << " (" << s.length() << ")" << endl;

        return 0;
}



mystring_exception.cpp

#ifndef MYSTRING_EXCEPTION
#define MYSTRING_EXCEPTION

#include <string>         // Note this is the C++ string class!

class String_Exception {
public:
    String_Exception(std::string what) : what(what) {}
        std::string getWhat() {return what;}
private:
    std::string what;
};

#endif

Effect on pipelining

We consider the transmission of a message between 2 host A and B via a router. We dispose of the following information :

- the distance between each host and the router is 1500m.

- the speed of propagation = 360.000 km/s(speed of light)

- the size of each message = 1500 octets.

- the speed of the link between host and router = 1Mb/s

- the delay of treatment of a message in the router = 10ms

1) What is is the total delay to send a message from A to B ?

2) If we use the Stop and Wait protocol to send a message from A to B and the message size ACK = 64 octets, what is the utilization rate ?

3) If we use the Go-back-N protocol, with a time frame of size N, what is the value of N that maximize the flow control?

Language:C++

Program:Visual basic

I have errors on line 102 "expression must have a constant value

line 107,108,123,124,127,128: array type int[n] not assignable

#include<iostream>
#include <fstream>
using namespace std;
#define size 10000

// Displays the current Inventory Data
void Display(int box_nums[],int nboxes[],double ppBox[],int n) // Prints Box number , number of boxes and price per box.
{
cout<<"Box number Number of boxes in stock Price per box"<<"\n";
cout<<"-------------------------------------------------------------\n";
for(int i=0; i<n; i++)
{
cout<<box_nums[i]<<" "<<nboxes[i]<<" "<<ppBox[i]<<"\n";
}
}

// sort's the inventory data according to the price per box from low to high
void sortByPrice(int box_nums[],int nboxes[],double priceBox[],int n)
{
int i, j, min_idx , temp2;
double temp;
for (i = 0; i < n-1; i++)
{
min_idx = i; // min_idx is used to store data in ascending order
for (j = i+1; j < n; j++) // used selection sort to sort the data based on price per box
{
if (priceBox[j] < priceBox[min_idx])
min_idx = j;
}
temp = priceBox[min_idx];
priceBox[min_idx] = priceBox[i]; // temp is a variable to swap data
priceBox[i] = temp;

temp2 = nboxes[min_idx];
nboxes[min_idx] = nboxes[i]; // temp2 is a variable to swap data
nboxes[i] = temp2;

temp2 = box_nums[min_idx];
box_nums[min_idx] = box_nums[i];
box_nums[i] = temp2;
  
}   
}

// sort's the inventory data according to Box number from low to high
void sortByBoxNumber(int box_nums[],int nboxes[],double priceBox[],int n)
{
int i, j, min_idx , temp2;
double temp;
for (i = 0; i < n-1; i++)
{
min_idx = i; // min_idx is used to store data in ascending order
for (j = i+1; j < n; j++)
{
if (box_nums[j] < box_nums[min_idx]) // used selection sort to sort the data based on price per box
min_idx = j;
}

temp2 = box_nums[min_idx];
box_nums[min_idx] = box_nums[i];
box_nums[i] = temp2;

temp = priceBox[min_idx];
priceBox[min_idx] = priceBox[i];
priceBox[i] = temp;

temp2 = nboxes[min_idx];
nboxes[min_idx] = nboxes[i];
nboxes[i] = temp2;
}   
}

// Searches for the price per box of the corresponding box number entered by user.
double lookUpByBoxNumber(int boxNumber,int box_nums[],double priceBox[],int n)
{
int low =0, high = n-1;
int mid;
while(low <= high) // used binary search to search for the corresponding box number and its price-per-box
{
mid = low + (high-low)/2;

if(box_nums[mid] > boxNumber)
{
high = mid - 1;
}
else if(box_nums[mid] == boxNumber)
{
return priceBox[mid];
}
else
{
low = mid + 1;
}
  
}
return -1;
}

//Reordered the data whose number of boxes are less than 100
void reorderReport(int box_nums[],int nboxes[],int n)
{
int reorderBoxNums[n],reorderBoxes[n],k=0;
for(int i=0; i<n; i++)
{
if(nboxes[i] < 100)
{
reorderBoxNums[k] = box_nums[i];
reorderBoxes[k] = nboxes[i];
k++;
}
}
int i, j, max_idx , temp2;
for (i = 0; i < k-1; i++) // sorts the data in reordered data according to number of boxes in inventory from low to high
{
max_idx = i;
for (j = i+1; j < k; j++)
{
if (reorderBoxes[j] > reorderBoxes[max_idx])
max_idx = j;
}

temp2 = reorderBoxes[max_idx];
reorderBoxes[max_idx] = reorderBoxes[i];
reorderBoxes[i] = temp2;

temp2 = reorderBoxNums[max_idx];
reorderBoxNums[max_idx] = reorderBoxNums[i];
reorderBoxNums[i] = temp2;
}
cout<<"REORDERED REPORT IS: \n";
cout<<"The boxes in invetory whose stock are less than 100 are: \n";
cout<<"Box number Number of boxes in stock"<<"\n";
cout<<"------------------------------------------"<<"\n";
for(int i=0 ; i<k; i++)
{
cout<<reorderBoxNums[i]<<" "<<reorderBoxes[i]<<"\n";
}
}
int main()
{

std::fstream myfile("inventory.txt");
int box_number[size] , numberOfBoxes[size] ;
double pricePerBox[size],sp;
int bn ,nb,i = 0;
while(myfile >> bn >> nb >> sp) // fetch data from file inventory.txt
{
box_number[i] = bn;
numberOfBoxes[i] = nb;
pricePerBox[i] = sp;
i++;
}

int n = i, bnumber ; // n stores number of records in file , bnumber is the box number which is to be searched for price-per-box by user
double val; // val is variable used for value stored from lookup by box-number
char option;
bool exit = true; // exit variable to exit the while loop

// Menu for the user
cout<<"\nChoose a option in the Menu a/b/c/d/e/f :"<<"\n";
cout<<"a. Display the data"<<"\n";
cout<<"b. Sort data by price, low to high"<<"\n";
cout<<"c. Sort data by box number, low to high"<<"\n";
cout<<"d. Look up the Price of the box given the box number"<<"\n";
cout<<"e. Generate Reorder Report"<<"\n";
cout<<"f. Exit"<<"\n";

while(exit)
{
cout<<"Enter your choice a/b/c/d/e/f : ";
cin>>option;

switch(option)
{
case 'a':
Display(box_number,numberOfBoxes,pricePerBox,n);
break;
case 'b':
sortByPrice(box_number,numberOfBoxes,pricePerBox,n);
cout<<"Data has been Successfully sorted by price"<<"\n";
cout<<"Please, choose option 'a' to display sorted data"<<"\n";
break;
case 'c':
sortByBoxNumber(box_number,numberOfBoxes,pricePerBox,n);
cout<<"Data has been Successfully sorted by Box Number"<<"\n";
cout<<"Please, choose option 'a' to display sorted data"<<"\n";
break;
case 'd':
sortByBoxNumber(box_number,numberOfBoxes,pricePerBox,n);
cout<<"Enter the box number for which you want to search the price : ";
cin>>bnumber;
val = lookUpByBoxNumber(bnumber,box_number,pricePerBox,n);
if(val < 0)
{
cout<<"There is no price of the box for the box number you are searching for\n";
}
else
{
cout<<"The price-per-box of the Box-Number you searched is "<<val<<"\n";
}
break;
case 'e':
reorderReport(box_number,numberOfBoxes,n);
break;
case 'f':
exit = false;
break;
default :   
cout<<"Invalid options , enter a valid option"<<"\n";
break;

}

}
return 0;
}

Please type, I will rate you well. Thank you.

In Linux:

• Identify three commands that use SUID mode. Discuss why this is so?
• Identify the commands that are generally causing the highest load on your system and discuss your perspectives on why they cause such a load. Discuss your perspectives on the considerations system administration must give while using these commands.

Exercise 1. Rectangle, Circle and Square Write three Python classes named Rectangle constructed by a length and width, a Circle constructed by a radius and a Square constructed by a side length. Both classes should have the methods that compute: - The area - The diagonal - The perimeter Use as much abstraction as you can. At the end of the file, use those classes to calculate the perimeter of a circle with radius the half of the diagonal of a rectangle with length 20 and width 10.

For the following problem, draw a flowchart, write a function and then execute the function with your choice of input parameter values:

For given two input parameters: If parameter 1 is smaller than parameter 2 return the remainder of parameter 1 by parameter 2 Otherwise, return the multiplication of two input parameters.

Python programming language

1. Identify three applications of information systems at the college or the university that you are attending. Write three applications, and provide an example of the type of decisions that are being improved by each application.

2.  RFID tags are being increasingly used by companies such as Macy's, Walmart, and Home Depot. Identify an additional company that uses RFIDs and describe the company’s specific application of RFIDs.

Write a C++ programm which

• reads an input file named 'input.txt'
• content of input.txt are not known to you
• copies content of input file into an output file named 'output.txt'
• contents of output.txt should be exactly the same as contents of input.txt
• the user should be given a choice to select input and output file in any folder
• remember to check that input file opened successfully

Please add comments next to the code, so I can learn from the solution. Also please use beginner coding, nothing too advance.

C++ question.

I want to write all output to the file "output.txt", but it will write just first character. Can you fix it?

#include

#include

#include

#include

#include

using namespace std;

using std::cin;

using std::cout;

using std::string;

// remove dashes convert letters to upper case

string normalize(const string &isbn) {

string ch;

for (char i : isbn) {

if (i == '-') {

continue; // if "-" then skip it

}

if (isalpha(i)) {

i = toupper(i); // Check uppercase

}

ch += i;

}

return ch;

}

// return the number of digits in to the string

size_t numDigits(string &str) {

size_t numDigits = 0;

for (char ch : str) {

if (isdigit(ch)) {

++numDigits;

}

}

return numDigits;

}

enum ValidationCode {

Ok, // validation passed

NumDigits, // wrong number of digits

ExtraChars,// extra characters in isbn

Done

};

enum ValidationCode validate(string &isbn) {

int Done = 4;

string normal = normalize(isbn);

size_t count = numDigits(normal);

if (normal.size() == Done)

exit(0);

if (count != 10) {

return NumDigits;

}

if (normal.size() == 10 || normal.size() == 11 && normal[10] == 'X') {

return Ok;

}

return ExtraChars;

}

int main() {

// open a file

ofstream file("output.txt");

//The following code is taken from (https://en.cppreference.com/w/cpp/io/basic_ios/fail)

// check if the file can open

if(!file) // operator! is used here

{

std::cout << "File opening failed\n";

return EXIT_FAILURE;

} //end of borrowed code

// read a file

//The following code is referenced from (https://stackoverflow.com/questions/7868936/read-file-line-by-line-using-ifstream-in-c)

std::ifstream ifs("test_data.txt");

// check if the file can read

if (ifs.fail())

{

std::cerr << "test_data.txt could not read" << std::endl;

return -1;

} //end of referenced code

std::string str;

while (ifs >> str){

switch (validate(str)) {

case Ok:

cout << str << " is a valid isbn\n";

file << str << " is a valid isbn\n";

break;

case NumDigits:

cout << str << " doesn't have 10 digits\n";

file << str << " doesn't have 10 digits\n";

break;

case ExtraChars:

cout << str << " has extra characters\n";

file << str << " has extra characters\n";

break;

default:

cout << "ERROR: validate(" << str << ") return an unknown status\n";

file << "ERROR: validate(" << str << ") return an unknown status\n";

break;

}

}

ifs.close();

file.close();

}

test_data.txt

1-214-02031-3
0-070-21604-5
2-14-241242-4
2-120-12311-x
0-534-95207-x
2-034-00312-2
1-013-10201-2
2-142-1223
3-001-0000a-4
done

Using the sample code included at the end of the document to create the tables, Please write an anonymous PL/SQL program the following problems.

Problem 1. Print out estimated charge for rental ID 1 if the customer returns the tool in time. The charge is computed by the price in the price_tool table * number of units the customer plans to rent. E.g., if a customer rents a tool hourly for 5 hours, and the hourly rate for the tool is $6, the estimated charge should be $30. [30 points]

Problem 2. [30 points] Print out name of tools rented by Susan and their due time.

--------- Sample code

drop table rental cascade constraints;

drop table tool_price cascade constraints;

drop table tool cascade constraints;

drop table category cascade constraints;

drop table cust cascade constraints;

drop table time_unit cascade constraints;

create table cust(

cid int, -- customer id

cname varchar(50), --- customer name

cphone varchar(10), --- customer phone

cemail varchar(30), --- customer email

primary key(cid)

);

insert into cust values(1,'John','1234567888','[email protected]');

insert into cust values(2,'Susan','1235555555','[email protected]');

insert into cust values(3,'David','1237777777','[email protected]');

create table category(

ctid int, --- category id

ctname varchar(30), --- category name

parent int, --- parent category id

primary key(ctid),

foreign key (parent) references category

);

insert into category values(1,'mower',null);

insert into category values(2,'electric mower',1);

insert into category values(3,'gasoline mower',1);

insert into category values(4,'carpet cleaner',null);

create table tool

(tid int, --- tool id

tname varchar(50), -- tool name

ctid int, --- category id

quantity int,

primary key (tid),

foreign key (ctid) references category

);

insert into tool values(1,'21 inch electric mower',2,2);

insert into tool values(2,'30 inch large gasoline mower',3,2);

insert into tool values(3,'small carpet cleaner',4,2);

insert into tool values(4,'large carpet cleaner',4,2);

create table time_unit

(tuid int, --- time unit id

len interval day to second, --- length of unit, can be 1 hour, 1 day, etc.

min_len int, --- minimal number of units

primary key (tuid)

);

--- hourly minimal four hours.

insert into time_unit values(1, interval '1' hour, 4);

-- hourly minimal 1 day

insert into time_unit values(2, interval '1' day, 1);

--- weekly

insert into time_unit values(3, interval '7' day, 1);

create table tool_price

(tid int, --- too id

tuid int, --- time unit id

price number,

primary key(tid,tuid),

foreign key(tid) references tool,

foreign key(tuid) references time_unit

);

--- mower, $20 per 4 hours. $30 per day

insert into tool_price values(1,1,5.00);

insert into tool_price values(1,2,30);

insert into tool_price values(1,3,120);

insert into tool_price values(2,1,7.00);

insert into tool_price values(2,2,40);

insert into tool_price values(2,3,160);

insert into tool_price values(3,1,6.00);

insert into tool_price values(3,2,32);

insert into tool_price values(3,3,125);

insert into tool_price values(4,1,7.00);

insert into tool_price values(4,2,40);

insert into tool_price values(4,3,160);

create table rental

(

rid int, --- rental id

cid int, --- customer id

tid int, --- tool id

tuid int, --- time unit id

num_unit int, --- number of units, if unit = 1 hour, num_unit = 5 means 5 hours.

start_time timestamp, -- rental start time

end_time timestamp, --- suppose rental end_time

return_time timestamp,--- actual return time

credit_card varchar(20),

total number, --- total charge

primary key (rid),

foreign key(cid) references cust,

foreign key(tid) references tool,

foreign key(tuid) references time_unit

);

-- John rented a mower for 4 hours,

insert into rental values(1,1,1,1,4,timestamp '2019-08-01 10:00:00.00',null,null,'123456789',null);

-- susan rented a small carpet cleaner for one day

insert into rental values(2,2,3,2,1,timestamp '2019-08-11 10:00:00.00',null,null,'123456789',null);

--susan also rented a small mower for 5 hours, before 8 am case

insert into rental values(3,2,1,1,5,timestamp '2019-08-12 21:00:00.00',null,null,'123456789',null);

--david also rented a small carpet cleaner for 4 hours, after 10 pm case

insert into rental values(4,3,3,1,4,timestamp '2019-08-13 19:00:00.00',null,null,'12222828828',null);

commit;

Use the median of 3 partitioning algorithm (given in the next page) to implement quick sort. This algorithm chooses the pivot element as the median of the 3 elements namely: A[p], A[r], and A[(p+r)/2].(Java language)

Quicksort(A,p,r)

1 if p

2 N = r- p +1

3 m = Median3(A, p, p + N/2 , r)

4 exchange A[m] with A[r]

5 q = Partition (A,p,r)

6 Quicksort(A,p, q-1)

7 Quicksort(A,q+1,r)

Write a function of interest to you. It should that take at least 3 inputs from different types and return at least three different values.

Call your function from the main.

Print the resultant outputs in the main (not in the function).

Construct a DFA machine to recognize the language of all binary numbers which are a multiple of 5.
L = { w belongs {0,1}* : w is a binary number and is a multiple of 5}
Example: 101 belongs to L; 1010 belongs to L; 110 doesn't belong to L.