Why does little-endian vs. big-endian matter here in this code
Q: class TenNums { private: int *p; public: TenNums
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Q: Why does little-endian vs. big-endian matter here in this code
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A:
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#include <stdio.h>
struct Single
{
int num;
};
void printSingle(int f)
{
int binaryNum[33];
int i = 0;
while(f>0)
{
binaryNum[i] = f % 2;
f = f/2;
i++;
}
for (int j=i-1; j>= 0; j--)
{
printf("%d",binaryNum[j]);
}
}
int main()
{
struct Single single;
single.num = 33;
printf("Number: %d\n",single.num);
printSingle(single.num);
return 0;
}
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- C programming #include <stdio.h>int main() {int i, j, n ;printf("height? ") ;scanf("%2d", &n) ;for (i = 1 ; i <= n ; i++) {// printf("%d: ", i) ;for (j = 1 ; j <= i ; j++) {// Pick *one* of the following// printf("%d", j % 10) ;printf("*") ;}printf("\n") ;}return 0 ;} How can I get output like this by editing the given programming -- height? 5************************* and then how can I get the following?- height? 7 1 123 12345 1234567 123456789 12345678901 1234567890123#include <stdio.h>#include <stdlib.h> typedef struct Number_struct { int num;} Number; void Swap(Number* numPtr1, Number* numPtr2) { /* Your code goes here */} int main(void) { Number* num1 = NULL; Number* num2 = NULL; num1 = (Number*)malloc(sizeof(Number)); num2 = (Number*)malloc(sizeof(Number)); int int1; int int2; scanf("%d", &int1); scanf("%d", &int2); Thank you so much but i forgot to put this how i would fit in into this. num1->num = int1; num2->num = int2; Swap(num1, num2); printf("num1 = %d, num2 = %d\n", num1->num, num2->num); return 0;}Q1 #include <stdio.h> int arrC[10] = {0}; int bSearch(int arr[], int l, int h, int key); int *joinArray(int arrA[], int arrB[]) { int j = 0; if ((arrB[0] + arrB[4]) % 5 == 0) { arrB[0] = 0; arrB[4] = 0; } for (int i = 0; i < 5; i++) { arrC[j++] = arrA[i]; if (arrB[i] == 0 || (bSearch(arrA, 0, 5, arrB[i]) != -1)) { continue; } else arrC[j++] = arrB[i]; } for (int i = 0; i < j; i++) { int temp; for (int k = i + 1; k < j; k++) { if (arrC[i] > arrC[k]) { temp = arrC[i]; arrC[i] = arrC[k]; arrC[k] = temp; } } } for (int i = 0; i < j; i++) { printf("%d ", arrC[i]); } return arrC; } int bSearch(int arr[], int l, int h, int key) { if (h >= l) { int mid = l + (h - l) / 2; if…
- #include using namespace std; int main() (double degree [6] [2]= (30,40,10,70, 20, 30, 60, 70, 30, 10, 10, 85); int buffer=degree[0][0]; for (int i=1;ibuffer) buffer=degree[i][0]; cout<For the following code #include <stdio.h> #include <conio.h> int powerof(int base, int pow); int _tmain(int argc, _TCHAR* argv[]) { int x,y,z; line 3 calledint* p: int a[3]{1, 2, 3}; p = a; What is the value of *(p+2)?NoneQuestion 37 public static void main(String[] args) { Dog[] dogs = { new Dog(), new Dog()}; for(int i = 0; i >>"+decision()); } class Counter { private static int count; public static void inc() { count++;} public static int getCount() {return count;} } class Dog extends Counter{ public Dog(){} public void wo(){inc();} } class Cat extends Counter{ public Cat(){} public void me(){inc();} } The Correct answer: Nothing is output O 2 woofs and 5 mews O 2 woofs and 3 mews O 5 woofs and 5 mews O#include<bits/stdc++.h>#include<math.h>using namespace std; class TotalResistance{double series_res,parallel_res,sp_res;public:TotalResistance(){series_res=parallel_res=sp_res=0;}void seriesResistance(double resistance[],int n);void parallelResistance(double resistance[],int n);void spResistance(double resistance[],int n);};void TotalResistance::seriesResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];cout<<"Total Resistance in series is: "<<series_res<<endl;}void TotalResistance::parallelResistance(double resistance[],int n){double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in parallel is: "<<parallel_res<<endl;}void TotalResistance::spResistance(double resistance[],int n){for(int i=0;i<n;i++)series_res += resistance[i];double temp=0;for(int i=0;i<n;i++)temp += (1/resistance[i]);parallel_res = 1/temp;cout<<"Total Resistance in…class TenNums {private: int *p; public: TenNums() { p = new int[10]; for (int i = 0; i < 10; i++) p[i] = i; } void display() { for (int i = 0; i < 10; i++) cout << p[i] << " "; }};int main() { TenNums a; a.display(); TenNums b = a; b.display(); return 0;} Continuing from Question 4, let's say I added the following overloaded operator method to the class. Which statement will invoke this method? TenNums TenNums::operator+(const TenNums& b) { TenNums o; for (int i = 0; i < 10; i++) o.p[i] = p[i] + b.p [i]; return o;} Group of answer choices TenNums a; TenNums b = a; TenNums c = a + b; TenNums c = a.add(b);#include <stdio.h> int arrC[10] = {0}; int bSearch(int arr[], int l, int h, int key); int *joinArray(int arrA[], int arrB[]) { int j = 0; if ((arrB[0] + arrB[4]) % 5 == 0) { arrB[0] = 0; arrB[4] = 0; } for (int i = 0; i < 5; i++) { arrC[j++] = arrA[i]; if (arrB[i] == 0 || (bSearch(arrA, 0, 5, arrB[i]) != -1)) { continue; } else arrC[j++] = arrB[i]; } for (int i = 0; i < j; i++) { int temp; for (int k = i + 1; k < j; k++) { if (arrC[i] > arrC[k]) { temp = arrC[i]; arrC[i] = arrC[k]; arrC[k] = temp; } } } for (int i = 0; i < j; i++) { printf("%d ", arrC[i]); } return arrC; } int bSearch(int arr[], int l, int h, int key) { if (h >= l) { int mid = l + (h - l) / 2; if…void fun(int i) { do { if (i % 2 != 0) cout =1); cout << endl; } int main() { int i = 1; while (i <= 8) { fun(i); it; } cout <SEE MORE QUESTIONSRecommended textbooks for youComputer Networking: A Top-Down Approach (7th Edi…Computer EngineeringISBN:9780133594140Author:James Kurose, Keith RossPublisher:PEARSON
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