Make a C++ program with class, object, constructor, data member with access modifier and member function overloading. FileName: Data.cpp The program must have the following data members and formula. All data members are private. Data Member Formula area3 cube (initial value is Zero) cylinder (initial value is Zero) rectangle (initial value is Zero) sphere (initial value is Zero) |(pi)r²h Lwh 4(pi)r3 3 pi (value is 3.1416) None The C++ program have the fix main() function. int main (){ Formula shape; shape.compute (20);//for cube - shape.compute (12.75, 10);//for cylinder - one fractional and one whole number shape.compute (5, 7, 15);//for rectangle - three whole number shape.compute (25.50);//for sphere - one fractional number one whole number Expected Output: The

Make a C++ program with class, object, constructor, data member with access modifier and member function overloading. FileName: Data.cpp The program must have the following data members and formula. All data members are private. Data Member Formula area3 cube (initial value is Zero) cylinder (initial value is Zero) rectangle (initial value is Zero) sphere (initial value is Zero) |(pi)r²h Lwh 4(pi)r3 3 pi (value is 3.1416) None The C++ program have the fix main() function. int main (){ Formula shape; shape.compute (20);//for cube - shape.compute (12.75, 10);//for cylinder - one fractional and one whole number shape.compute (5, 7, 15);//for rectangle - three whole number shape.compute (25.50);//for sphere - one fractional number one whole number Expected Output: The

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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Still learning to create program based off of UML diagram. Need help creating order class and a test order class to execute based on the given UML diagram in java script on eclipse. Please provide with ss of the work if possible. Attatching the UML diagram and instruction and how the display result should look in with images.
/* GCC *//* gcc -Wa,-adhln -g -masm=intel -m32 "Project 2.c" > "Project 2-g.asm" *//* gcc -Wa,-adhln -O -masm=intel -m32 "Project 2.c" > "Project 2-o.asm" */ #include <stdio.h> #define NOINLINE __attribute__ ((noinline)) static NOINLINE int function1(int x, int y){ int i;int sum;int values[10];sum = 0;for (i = 0; i < 10; i++) {values[i] = 10 * i + x * y;sum += values[i];} return (sum);} static int NOINLINE function2(int *values, int valuesLen){ int i;int sum; sum = 0;for (i = 0; i < valuesLen; i++) {sum += values[i];} return (sum);} static NOINLINE int function3(int x){ int y; y = x / 10;return (y);} static NOINLINE int function4(int a, int b, int c, int d){ int r; if (a > b)r = a;else if (a > c)r = 2 * a;else if (a > d)r = 3 * a;elser = -1;return (r);} int main(int argc, char **argv){ int i;int j;int k;int values[10]; i = 1;j = 2;k = function1(i, j);printf("function1: i = %d, j = %d, k = %d\n", i, j, k); for (i = 0; i < 10; i++) {values[i] = i;}k =…
/* GCC *//* gcc -Wa,-adhln -g -masm=intel -m32 "Project 2.c" > "Project 2-g.asm" *//* gcc -Wa,-adhln -O -masm=intel -m32 "Project 2.c" > "Project 2-o.asm" */ #include <stdio.h> #define NOINLINE __attribute__ ((noinline)) static NOINLINE int function1(int x, int y){ int i;int sum;int values[10];sum = 0;for (i = 0; i < 10; i++) {values[i] = 10 * i + x * y;sum += values[i];} return (sum);} static int NOINLINE function2(int *values, int valuesLen){ int i;int sum; sum = 0;for (i = 0; i < valuesLen; i++) {sum += values[i];} return (sum);} static NOINLINE int function3(int x){ int y; y = x / 10;return (y);} static NOINLINE int function4(int a, int b, int c, int d){ int r; if (a > b)r = a;else if (a > c)r = 2 * a;else if (a > d)r = 3 * a;elser = -1;return (r);} int main(int argc, char **argv){ int i;int j;int k;int values[10]; i = 1;j = 2;k = function1(i, j);printf("function1: i = %d, j = %d, k = %d\n", i, j, k); for (i = 0; i < 10; i++) {values[i] = i;}k =…
In the study of formal languages state-transition diagrams are often usedto visualize changes in a machine’s configuration as it acts on input. To visualize amachine’s configuration think of its parts: Commonly a finite set of states, a finiteinput alphabet Σ, Perhaps storage devices (a stack, input/output tapes). The states arepictured as named circles sometimes decorated with symbols to denote special states,e.g. start and final states. Changes in configuration are denoted by labeled edges andperhaps changes in storage.1. Describe how edges are labeled and their meaning for finite state machines.2. Describe how edges are labeled and their meaning for pushdown automata.3. Describe how edges are labeled and their meaning for Turing machines.
C has many statements for decisions and loops, while MIPS has few. Which of the following do or do not explain this imbalance? Why? 1. More decision statements make code easier to read and understand. 2. Fewer decision statements simplify the task of the underlying layer that is responsible for execution. 3. More decision statements mean fewer lines of code, which generally reduces coding time. 4. More decision statements mean fewer lines of code, which generally results in the execution of fewer operations.
"C:\PROGRA~1\JetBrains\CLion 2022.2.3\bin\mingw\bin\g++.exe" -g -std=gnu++14 -MD -MT CMakeFiles/untitled.dir/sequence.cpp.obj -MF CMakeFiles\untitled.dir\sequence.cpp.obj.d -o CMakeFiles/untitled.dir/sequence.cpp.obj -c C:/Users/r1821655/CLionProjects/untitled/sequence.cppC:/Users/r1821655/CLionProjects/untitled/sequence.cpp:117:36: error: 'template<class Item> class CS3358_FA2022_A04_sequenceOfNum::sequence' used without template arguments 117 | typename sequence<Item>::size_type sequence::size() const { return used; } | ^~~~~~~~C:/Users/r1821655/CLionProjects/untitled/sequence.cpp:117:53: error: non-member function 'typename CS3358_FA2022_A04_sequenceOfNum::sequence<Item>::size_type CS3358_FA2022_A04_sequenceOfNum::size()' cannot have cv-qualifier 117 | typename sequence<Item>::size_type sequence::size() const { return used; } |…
Create a process graph for the following program. int main () { int x = 4; if (fork() == 0) printf("printf1: x=%d\n", ++x); printf("printf2: x=%d\n", --x); exit 0; } Solution (may be hand-drawn): Create a process graph for the following program. int main () { fork(); printf("hello\n"); fork(); printf("world\n"); exit 0; } Solution (may be hand-drawn):
In an interview, following code snippet was provided: first, second, third = 'red', 'blue', 'green'first, second, third = second, third, firstcomplexa, complexb = 1+4j, 4+3jprint(first)print(third)print(second)print(complexa + complexb) Mark the correct print outputs. A. red, green, blue, Error B. green, blue, red, (5 + 7j) C. blue, red, green, (5 + 7j) D. Code throws an error.
OPERATING SYSTEMS LETTER B ONLY Given the following page reference with a 4-page frame.
Processes are represented by which DFD symbol Rectangles Directional vectors Circles Parallel lines
For this homework assignment you are not allowed to use::MOVSB, MOVSW, MOVSD, CMPSB,CMPSW,CMPD,SCASB, SCASW, SCASD,STOSB, STOSW, STOSD, LODSB, LODSW, and LODSD. It is required that for each question write only one procedure that does the requested job. Only write the assembly part and avoid using directives. DO NOT USE IRVINE'S LIBRARY.Write Assembly code (only assembly code and no procedure call) that performs decryption (recovers the original string) of an encrypted string. The encryption is performed by rotating each plaintext byte a varying number of positions in different directions. For example, in the following array that represents the encryption key, a negative value indicates a rotation to the left and a positive value indicates a rotation to the right. The integer in each position indicates the magnitude of the rotation: key BYTE -4, 4, -1, 0, 3, -5, 2, -4Encryption algorithm rotates each plaintext byte by the amount indicated by its matching key array value. Then, align…
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