c program to pep 9 assembly language void findLog ( int *result, int num, int b ) { int k; num--; *result = 0; while (num > 0) { *result = *result + 1; num = divide(num, b); } // end while } // end of findLog( )

Convert the following c++ code into pep9 assembly language. #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; }

Based on the following C code: { int x, y, z, sum, total; int *pt1, *pt2, *ptc; x = 99; y = 86; pt1 = &y; pt2 = pt1; ptc = &x; z = (*(&y))/2 + (*pt2)/3 + 5; sum = (x + y + *pt1 + *pt2 + z)/3; total = sum + *ptc + *pt1; } (1) What is the value of z? (2) What is the value of sum? (3) What is the value of the total? (4) What is the address of the data 86?

Write in C Language Description Please write a program that read in anexpression and give its result. Input Input will be an expression content only digit, +, -, /, * and space. Output Please output the result of expression without consider the priority of +, -, *, /. When you do thedivision, please round down to integer. Sample Input 1 1 + 2 * 3 / 4 Sample Output 1 2

Convert the following c++ code into pep9 assembly language #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; }

in C programing

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...

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Translate the following C program to Pep/9 assembly language. #include int inpNum, sum, count, min; int main() { sum = 0; count = 0; min = 32767; printf("? "); scanf("%d", &inpNum); while (inpNum != 0 ) { sum += inpNum; count++; if (inpNum < min) { min = inpNum; } printf("? "); scanf("%d", &inpNum); } // end while printf("Sum printf("Count printf("Min } // end of main = = %d\n", sum); = %d\n", count); %d\n", min);

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;}

Convert the following C++ program to pep9/Assembly language BR Main format not machine language #include <iostream>using namespace std; int times(int mpr, int mcand){int prod = 0;while (mpr != 0){if (mpr % 2 == 1)prod = prod + mcand;mpr /= 2;mcand *= 2;} return prod;} int main(){ int n, m; cout << "Enter two numbers: ";cin >> n >> m; cout << "Product: " << times(n, m) << endl; return 0;}

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