1_ What is the output?
private void btnMethods_Click(object sender, EventArgs e)
{
int arg1 = 2;
double val = ValReturnMethod(arg1, 2.00);
MessageBox.Show(val.ToString());
}
private void ValReturnMethod(int val, double val2)
{
return val1 * val2 *25.50;
}
_____________________________
2_ What is the output?
private void btnMethods_Click(object sender, EventArgs e)
{
int arg1 = 4;
double val = ValReturnMethod(ref arg1);
MessageBox.Show(arg1.ToString());
}
private double ValReturnMethod(ref int val1)
{
return val1 *25.50;
}
______________________________
3_ What is the output?
private void btnMethods_Click(object sender, EventArgs e)
{
int arg1 = 4;
double arg2 = 10.50;
MessageBox.Show(ValReturnMethod(arg1,arg2).ToString());
}
private double ValReturnMthod (int val1,double val2)
{
return val1 * 25.50;
}
In: Computer Science
What is the signed decimal equivalent of the following signed-magnitude binary value?
11101010.1001
What is the binary equivalent (in two's complement binary representation) of the following signed decimal value? Represent the integer part of the binary value in 8 bits.
-58.1875
In: Computer Science
For each of the following situations provide the type of variable(s) that you would use to model it and give a kind of probability distribution that would be appropriate for modeling that variable.
In: Computer Science
C++
Text message decoder
Use getline() to get a line of user input into a string:
Enter text: IDK if I'll go. It's my BFF's birthday.
Search the string using find() for common abbreviations and replace() them with their long form. In addition, use a for loop to iterate through each character of the string and replace any occurences of '.' with '!':
Enter text: IDK if I'll go. It's my BFF's birthday. I don't know if I'll go! It's my best friend forever's birthday!
Use loops for each abbreviation to replace multiple occurrences:
Enter text: Hi BFF. IDK where I'm going but it's not going to be with you! JK you know you're my BFF. IDK why I just said that. Anyway, TMI. Why do I always give people TMI? IDK. Alright TTYL bye BFF! Hi best friend forever! I don't know where I'm going but it's not going to be with you! just kidding you know you're my best friend forever! I don't know why I just said that! Anyway, too much information! Why do I always give people too much information! I don't know! Alright talk to you later bye best friend forever!
Support these abbreviations:
Use the predefined constants for each abbreviation BFF, IDK, etc. and each long form BFF_LONG, IDK_LONG, etc. rather than hardcoding strings inside the loop.
In: Computer Science
What is the unsigned hexadecimal equivalent of the following unsigned octal value?
Do NOT include in your answer ANY insignificant zeros.
107655.643
What is the binary equivalent (in signed-magnitude binary representation) of the following signed decimal value? Represent the integer part of the binary value in 8 bits.
-116.6875
In: Computer Science
JAVA Palindrome Detector
A palindrome is any word, phrase, or sentence that reads the same forward or backward. Here are some well-known palindromes:
Able was I, ere I saw Elba
A man, a plan, a canal, Panama
Desserts, I stressed
Kayak
Write a boolean method that users recursion to determine where a String argument is a palindrome. The method should return true if the argument reads the same forward and backward. Demonstrate the method in a program.
Include the following modifications:
In: Computer Science
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
String::String(const char *cs) : cs(new char[strlen(cs)+1) { // the +1 is for the null terminator
#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;
};
#endifmystring_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;
};
#endifIn: Computer Science
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?
In: Computer Science
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;
}
In: Computer Science
Please type, I will rate you well. Thank you.
In Linux:
In: Computer Science
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.
In: Computer Science
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
In: Computer Science
In: Computer Science
In this project you will create a basic console based calculator program.
The calculator can operate in two modes: Standard and Scientific modes.
The Standard mode will allow the user to perform the following operations:
(+, -, *, /) add, subtract, multiply, and divide
The Scientific mode will allow the user to perform the same functionality as the Standard
add, subtract, multiply, and divide (+, -, *, / ) plus the following:
sin x, cos x, tan x. (sin x, cos x, tan x)
The calculator should be able to perform addition, subtraction, multiplication, and division of two or more numbers but perform sin x, cos x, and tan x of one number only (the number is in radians).
1. The calculator program will first ask the user for the mode to operate in (Standard or Scientific)
Sample Output:
Enter the calculator mode: Standard/Scientific?
Standard
2. The program should then ask the user for the operation to execute (+, -, *, /, sin x, cos x, tan x)
Sample Output for Scientific mode:
Enter '+' for addition, '-' for subtractions, '*' for multiplication, '/' for division, 'sin' for sin x, 'cos' for cos x, 'tan' for tan x:
sin
Sample Output for Scientific mode:
Enter '+' for addition, '-' for subtractions, '*' for multiplication, '/' for division, 'sin' for sin x, 'cos' for cos x, 'tan' for tan x:
sin
2a. If the user enters an invalid operation, output a message telling the user the input is invalid and re-prompt the user for the operation again.
Enter '+' for addition, '-' for subtractions, '*' for multiplication, '/' for division, 'sin' for sin x, 'cos' for cos x, 'tan' for tan x:
division
Invalid operation entered
Enter '+' for addition, '-' for subtractions, '*' for multiplication, '/' for division, 'sin' for sin x, 'cos' for cos x, 'tan' for tan x:
/
3. In order to know how many times the user will need to perform the operation, prompt the user for the number of double values they want to enter (All numbers in this program are double), then ask the user to enter the numbers.
Sample Output:
How many numbers do you want to subtract:
3
Enter 3 numbers:
4
35
9
In this example the calculator will calculate 4 + 35 + 9. The result will be 48.
The calculator should be able to perform the following operations:
2+3 = 5
3+6+1+1+1 = 12
1-2-90 = -91
10*2*3 = 60
10/2/5 = 1
cos(0) = 1
sin(0) = 0
tan(0) = 0
Note: This calculator does NOT support multiple operations in the expressions like: 10+2-8
Note: Multiple numbers are only acceptable for the operations of (+, -, *, / ) not for (sin,cos,tan)
4. Display the result
Sample Output:
Result: 50.0
5. Finally, output the result to the user and ask the user if he/she want to start over.
Sample Output:
Start over? Y/N
Y
Full Sample Output:
Enter the calculator mode: Standard/Scientific?
Standard
Enter '+' for addition, '-' for subtractions, '*' for multiplication, '/' for division, 'sin' for sin x, 'cos' for cos x, 'tan' for tan x:
addition
Invalid operation entered
Enter '+' for addition, '-' for subtractions, '*' for multiplication, '/' for division, 'sin' for sin x, 'cos' for cos x, 'tan' for tan x:
+
How many numbers do you want to subtract:
3
Enter 3 numbers:
4
35
9
Result: 48.0
Start over? Y/N
Y
Enter the calculator mode: Standard/Scientific?
Scientific
Enter '+' for addition, '-' for subtractions, '*' for multiplication, '/' for division, 'sin' for sin x, 'cos' for cos x, 'tan' for tan x:
sin
Enter number in radians to find sin:
1.5708
Result: 1
Start over? Y/N
NIn: Computer Science
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.
In: Computer Science