The program below terminate at
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- Can you answer it based on the code below: #include #define MAX_EMP 5 struct Employee { int id; int age; float salary; }; int count = 0; struct Employee emp[MAX_EMP]; void display() { printf("\n---=== EMPLOYEE DATA ===---\n"); printf("EMP ID\tEMP AGE\tEMP SALARY\n"); printf("======\t=======\t==========\n"); for(int i = 0; i < count; i++) { printf("%d\t%d\t%.2f\n", emp[i].id, emp[i].age, emp[i].salary); } } void add() { if(count == MAX_EMP) { printf("\nERROR!!! Maximum Number of Employees Reached\n"); return; } printf("\nAdding Employee ===============\n"); printf("Enter Employee ID: "); scanf("%d", &emp[count].id); printf("Enter Employee Age: "); scanf("%d", &emp[count].age); printf("Enter Employee Salary: "); scanf("%f", &emp[count].salary); count++; } void update() { int id, found = 0; float newSalary; printf("\nUpdate Employee Salary…#include<iostream>using namespace std;void main(){double pi = 0, denominator = 1;int counter = 999999;for (int x = 0; x < counter; x++){if (x % 2 != 0){pi = pi - (1 / denominator);}else{pi = pi + (1 / denominator);}denominator = denominator + 2;}pi = pi * 4;cout << " So the computed value of a PI is = " << pi << endl;cout << " ";//return 0;system("pause");} Note: This a program called ComputePI to compute the value of π Tutor just have to Modify This program to use nested-if (if ... else if ... else if ... else) instead. Explain by applying a double line commentVoid Do1 (int: &, a. int &b) { a = 5; a = a + b; b = a + 2; } Int main() { Int x = 10; Do1 (x,x); Cout << x << endl; } The output of this program is
- #include <stdio.h>void cubeByReference( int *nPtr ); // function prototypeint main( void ){ int number = 5; // initialize number printf("The original value of number is %d", number ); // pass address of number to cubeByReference cubeByReference( &number ); printf("\nThe new value of number is %d\n", number );} // end main void cubeByReference( int *nPtr ){ *nPtr = *nPtr* *nPtr* *nPtr;} passing argument by reference - We modify the code above 1- define a second argument (example "int number2 = 9") and a pointer to it 2- define a second function (addByReference) that adds number2 to number - passing both arguments by reference 3- print-out the result (that is in number) Upload the output and .c codeint func(int a, int b) { return (aIn Java:نقطة واحدة Let A = {a; b; c; d} and R= {(a; a); (b; c); (c; b); (d; d)} then R is Transitive Equivalent not transitivec++, print the outputMatch the C-function on the left to the Intel assemble function on the right. W: cmpl $4 movl %edi , %edi jmp .L4(,%rdi,8) %edi .L3: movl $17, %eax ret .15: movl $3, %eax int A ( int x , int y) { int a ; if ( x == 0 ) else i f ( x == 1 ) a = 3 ; else i f ( x == 2 ) a = 2 0 ; else i f ( x == 3 ) a = 2 ; else i f ( x == 4 ) a = 1 ; ret .L6: a = 17; movl $20, %eax ret .L7: movl $2, %eax ret else a = 0; .L8: return a ; movl $1, %eax .L2: ret . section .rodata . L4: .quad .L3 .quad .L5 .quad .L6 .quad .L7 .quad .L8 X: testl %edi, %edi je cmpl je cmpl je стр1 je cmpl .L16 $1, %edi .L17 $2, %edi .L18 $3, %edi int B (int x, int y) { int a; switch (x) { .L19 $4, %edi %al movzbl %al, %eax case 0: a = 17; break; sete break; case 1: a = 3; case 2: a = 20; break; case 3: a = 2; break; case 4: a = 1; a = 0; } return a; ret .L16: break; movl $17, %eax ret .L17: movl $3, %eax } ret .L18: movl $20, %eax ret .L19: movl ret $2, %eaxA return; else #include using namespace std; void division (int num, int denom); int main() { division (5, -1); division (10, 2); return 0; } void division (int num, int denom) {if (denom using namespace std; int main() (char List[4][10] = {"kareem", "ahmed", "ali", "mohamed"}; cout<Pass the first parameter by reference and the second parameter by value. *1#include 2 #include #include 4 #include 3 5 sem_t fork[3]; 6 7 void eat (int phil) { 8 printf("Philosopher %d is eating\n", phil); sleep (2); //time eating 9 10} 11 void philosopher (void * num) { int phil-*(int *)num; 12 13 printf("Philosopher %d wants to eat\n", phil); 14 sem_wait (&fork[phil]); 15 sem_wait (&fork [(phil+1)]); 16 eat (phil); 17 18 printf("Philosopher %d has finished eating\n",phil); sem_post(&fork [(phil+1)]); sem_post(&fork[phil]); 19 20} 21 int main(){ 22 int i, a[3]; 23 pthread_t tid[3]; for(i=0;i<3; i++) 24 25 sem_init(&fork[i],0,1); 26 for(i=0;i<3; i++){ 27 a[i]=i; 28 pthread_create(&tid[i], NULL, philosopher, (void *)&a[i]); 29 } 30 for(i=0;i<3;i++) 31 pthread_join(tid[i],NULL); 32} Fig.1 Figure 1 shows a program that tries to solve the dining philosophers' problem given that it only includes 3 philosophers who want to eat, where only 3 forks are available and each philosopher needs two forks. The outputs after running the program are given in Figure 2. In the…SEE MORE QUESTIONSRecommended textbooks for youDatabase System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. SudarshanPublisher:McGraw-Hill EducationStarting Out with Python (4th Edition)Computer ScienceISBN:9780134444321Author:Tony GaddisPublisher:PEARSONDigital Fundamentals (11th Edition)Computer ScienceISBN:9780132737968Author:Thomas L. FloydPublisher:PEARSONC How to Program (8th Edition)Computer ScienceISBN:9780133976892Author:Paul J. Deitel, Harvey DeitelPublisher:PEARSONDatabase Systems: Design, Implementation, & Manag…Computer ScienceISBN:9781337627900Author:Carlos Coronel, Steven MorrisPublisher:Cengage LearningProgrammable Logic ControllersComputer ScienceISBN:9780073373843Author:Frank D. PetruzellaPublisher:McGraw-Hill EducationDatabase System ConceptsComputer ScienceISBN:9780078022159Author:Abraham Silberschatz Professor, Henry F. Korth, S. SudarshanPublisher:McGraw-Hill EducationStarting Out with Python (4th Edition)Computer ScienceISBN:9780134444321Author:Tony GaddisPublisher:PEARSONDigital Fundamentals (11th Edition)Computer ScienceISBN:9780132737968Author:Thomas L. FloydPublisher:PEARSONC How to Program (8th Edition)Computer ScienceISBN:9780133976892Author:Paul J. Deitel, Harvey DeitelPublisher:PEARSONDatabase Systems: Design, Implementation, & Manag…Computer ScienceISBN:9781337627900Author:Carlos Coronel, Steven MorrisPublisher:Cengage LearningProgrammable Logic ControllersComputer ScienceISBN:9780073373843Author:Frank D. PetruzellaPublisher:McGraw-Hill Education