Estimate the time complexity for the following C+ program: III I/ Start //I/I/ #include const int n=10; Int Fact (Int); vold BSort(int A[n][n]); int main() vold BSort(int A[n][n]) ( Int temp; for(int j=0; } A[+1]) for(int i=0;i>A[I]U]; temp=A[i]; A[i]=A[+1]; A[i+1]=temp; BSort(A); cout<<"Factorial = "<

Estimate the time complexity for the following C+ program: III I/ Start //I/I/ #include const int n=10; Int Fact (Int); vold BSort(int A[n][n]); int main() vold BSort(int A[n][n]) ( Int temp; for(int j=0; } A[+1]) for(int i=0;i>A[I]U]; temp=A[i]; A[i]=A[+1]; A[i+1]=temp; BSort(A); cout<<"Factorial = "<

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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Convert the following C++ programs into Pep/9 assembly #include <iostream> using namespace std; int minimum (int i1, int i2){if (i1 < i2)return i1;elsereturn i2;}int main (){int n, m;cin >> n >> m;cout << "Minimum: " << minimum (n, m) << endl;return 0;} Submit: Pep/9 source code along with screen capture showing it running in the Pep simulator
Translate the following C program into NASM. #include <stdio.h> int ary[] = {12, 40, -2, 89, 35, -7, 6}; int main() { int sum = 0; int highest = 0; for (int x = 0; x < 7; x++) { if (highest < ary[x]) highest = ary[x]; sum += ary[x]; } printf("Sum is %d\n", sum); printf("Highest value is %d\n", highest); } Use indexing (the [ebx+esi] or [ebx+edi] form). You can have several “dw” values on the same line. Use the “loop” command.
Convert the following C++ programs into Pep/9 assembly #include <iostream> using namespace std; void minimum (int i1, int i2){ if (i1 < i2) cout << i1 << " is less than " << i2 << endl; else if (i1 > i2) cout << i2 << " is less than " << i1 << endl; else cout << i1 << " equals " << i2 << endl;}int main (){ int n, m; cout << "Enter two integers: "; cin >> n >> m; minimum (n, m); return 0;} Submit: Pep/9 source code along with screen capture showing it running in the Pep simulator
Convert the following C++ programs into Pep/9 assembly #include <iostrea> using namespace std; int minimum (int i1, int i2){ if (i1 < i2) cout << i1 << " is less than " << i2 << endl; else if (i2 < i1) cout << i2 << " is less than " << i1 << endl; else cout << i1 << " equals " << i2 << endl;}int main (){int n, m;cin >> n >> m;cout << "Minimum: " << minimum (n, m) << endl;return 0;} Submit: Pep/9 source code along with screen capture showing it running in the Pep simulator
Consider the following recursive function: void exercise(int x) { if (x > 0 && x < 10) { cout << x << " "; exercise(x + 1); } } What is the output of the following statements? a. exercise(0); b. exercise(5); c. exercise(10); d. exercise(-5);
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Translate the following C program into NASM. #include <stdio.h> int ary[] = {12, 40, -2, 89, 35, -7, 6}; int main() { int sum = 0; int highest = 0; for (int x = 0; x < 7; x++) { if (highest < ary[x]) highest = ary[x]; sum += ary[x]; } printf("Sum is %d\n", sum); printf("Highest value is %d\n", highest); } Use indexing (the [ebx+esi] or [ebx+edi] form). You can have several “dw” values on the same line. Use the “loop” command.
Zybooks C++ 1.7 LAB: Introduction to data structures labs Step 1: Producing correct output Three commented-out lines of code exist in main(). Uncomment the lines and click the "Run program" button. Verify that the program's output is: 2 + 2 = 4 Unknown function: PrintPlus2 Secret string: "abc" Submit your code for grading. Your submission will pass the "Compare output" test only, achieving 1 of the possible 10 points. Step 2: Inspecting the LabPrinter class Inspect the LabPrinter class implemented in the LabPrinter.h file. Access LabPrinter.h by clicking on the orange arrow next to main.cpp at the top of the coding window. Member functions Print2Plus2() and PrintSecret() print strings using std::cout. Step 3: Implementing CallFunctionNamed() Remove the three uncommented lines from main(). Then implement the CallFunctionNamed() function in main.cpp to handle three cases: If functionName is "Print2Plus2", call printer's Print2Plus2() member function. If functionName is "PrintSecret",…
Fix the following c++ code : #include<iostream>using namespace std;void Ashraf(int [][],int,int);int main(){int a[2][3];Ashraf(a,2,3);}void Ashraf (int a[ ] [], int R, int C){for(int r=0; r<R; r++){for(int c=0; c<C; c++){a[r][c]=1;cout<<a[r][c]<<"\t";}cout<<endl;}
Change the code only in python or c program needed very quickly
Convert the following C++ program into Pep/9 Assembly. #include <iostream>using namespace std;void SetEqualGreater (int &num1, int &num2){ if (num1 < num2) num1 = num2; else num2 = num1;}int main() { int x, y; // Using Local Variables cin >> x >> y; SetEqualGreater(x, y); cout << "x = y = " << x << endl; return 0;} Submit: Source file along with screen captures showing the program running in the Pep/8 simulator.
Correct this code written in c. hint: the last ending (// calling function) #include<stdio.h>#include<math.h> function(float x, float y, int n){int i;float R,R_power_n=1,realpart,impart,theta;R=sqrt((x*x)+(y*y));theta = tan(y/x); // since tan inverse theta = y/x i.e., theta = tan(y/x) for(i=1;i<=n;i++) // loop for calculation of R power n value{R_power_n = R_power_n*R;} // R power n result stored in R_power_ntheta=n*theta; //calculating n thetarealpart=R_power_n*cos(theta); // Calculating real partimpart=R_power_n*sin(theta); // Calculating imaginary partprintf("\n%.2f + j%.2f", x,y);printf("\n\nPower of %d is %.2f + j%.2f\n\n", n,realpart,impart);}int main(){int main(){int n;float x;float y; // Because of real numbers, take x and y as float data typeprintf("\nEnter real values x:");scanf("%f",&x);printf("\nEnter real values y:");scanf("%f",&y);printf("\nEnter the n Value:");scanf("%d",&n);(x,y)=fuction(x,y,n); // calling function}