27. Write a Pep/9 assembly language program that corresponds to the following C program: int width; int length; int perim; int main () ( scanf ("&d %d", &width, &length); perim printf("width printf("length printf("perim return 0: (width + length) * 2; = d\n", width); 8d\n\n", length); d\n", perim) ; %3D

27. Write a Pep/9 assembly language program that corresponds to the following C program: int width; int length; int perim; int main () ( scanf ("&d %d", &width, &length); perim printf("width printf("length printf("perim return 0: (width + length) * 2; = d\n", width); 8d\n\n", length); d\n", perim) ; %3D

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

See similar textbooks

Similar questions

Complete the following C++ programs into Pep/9 assembly language: 1) int main(){int cop; int driver; cop = 0; driver = 40; while (cop <= driver) { cop += 25; driver += 20; } cout << cop; return 0;}
Example: The Problem Input File Using C programming language write a program that simulates a variant of the Tiny Machine Architecture. In this implementation memory (RAM) is split into Instruction Memory (IM) and Data Memory (DM). Your code must implement the basic instruction set architecture (ISA) of the Tiny Machine Architecture: //IN 5 //OUT 7 //STORE O //IN 5 //OUT 7 //STORE 1 //LOAD O //SUB 1 55 67 30 55 67 1 LOAD 2- ADD 3> STORE 4> SUB 5> IN 6> OUT 7> END 8> JMP 9> SKIPZ 31 10 41 30 //STORE O 67 //OUT 7 11 /LOAD 1 //OUT 7 //END 67 70 Output Specifications Each piece of the architecture must be accurately represented in your code (Instruction Register, Program Counter, Memory Address Registers, Instruction Memory, Data Memory, Memory Data Registers, and Accumulator). Data Memory will be represented by an integer array. Your Program Counter will begin pointing to the first instruction of the program. Your simulator should provide output according to the input file. Along with…
Translate the following C++ program to Pep/9 assembly language. Code should also be commented. #include <iostream>using namespace std;int Lesser(int x, int y) { int least; if (x < y) least = x; else least = y; return least;}int main(){ int num1; int num2; cout << "Enter two integers: "; cin >> num1 >> num2 cout << "Lesser: " << Lesser(num1, num2); << endl; return 0;}
translate a c++ program into pep/9 assembly language #include <iostream> using namespace std; int myAge;void ShowVal(int age){ cout << "Age:" << age << endl;} int main() {cout << "Enter age: ";cin >> myAge;ShowVal(myAge);return 0;}
Given the following code, show step by step and explain how variables are created and eliminated in RAM memory. void p(int& a, float b) { a=1; b=4.0f; } int main(void) { float a=3; int c=4; p(c,a); }
Convert the following C++ programs into Pep/9 assembly #include <iostream> using namespace std; void times(int& prod, int mpr, int mcand) { prod = 0; while (mpr != 0) { if (mpr % 2 == 1) prod = prod + mcand; mpr /= 2; mcand *= 2; } } int main(){ int product, n, m; cout << "Enter two numbers: "; cin >> n >> m; times(product, n, m); cout << "Product: " << product << endl; return 0; } Submit: Pep/9 source code along with screen capture showing it running in the Pep simulator
15. Translate the following C program to Pep/9 assembly language. #include int main() { int numitms, j, data, sum; scanf("%d", &numItms); sum = 0; for (j = 1; j <= numitms; j++) { scanf("%d", &data); sum += data; } printf ("Sum: %d\n", sum); return 0; } Sample Input 48-376 Sample Output Sum: 18
please modify the c language program into a pep 9 assembly language #include <stdio.h> int num, answer; int divide ( int numer, int denom ) { int quotient, remain; remain = numer; quotient = 0; while ( remain >= denom ) { remain -= denom; quotient++; } // end for return quotient; } // end of divide( ) int modulus ( int n, int d ) { int quot, rem; rem = n; quot = 0; while ( rem >= d ) { rem -= d; quot++; } // end for return rem; } // end of modulus( ) int main() { printf("? "); scanf("%d", &num); answer = divide(num, 20); printf("%d / 20 = %d\n", num, answer); answer = modulus(num, 20); printf("%d mod 20 = %d\n", num, answer); answer = divide(num, 35); printf("%d / 35 = %d\n", num, answer); answer = modulus(num, 35); printf("%d mod 35 = %d\n", num, answer); return 0; } // end of main