Answered You can hire a professional tutor to get the answer.

QUESTION

QUESTION 1 Convert the following C program into an x86 assembly language program.Make sure to use your "Chapter 8" understanding of advanced functions, parameter passing, and local variables.Post ON

QUESTION 1

  1. Convert the following C program into an x86 assembly language program.Make sure to use your "Chapter 8" understanding of advanced functions, parameter passing, and local variables.Post ONLY your ASM file here to Blackboard when complete.#include using namespace std;int IsSemifauxtrifactored(int value){ // Return 1 if a number's factors/divisors from (value - 1) to 1 sum up to half the number value // Return 0 otherwise // A number is called "semifauxtrifactored" if its summed factors/divisors equal half the number itself // Integer division is used, so remainders on the halving can be lost // That's why... // 9 is a semifauxtrifactored number // 9 cut in half with integer division is (9 / 2) = 4 // 9 % 8 -> 1 // 9 % 7 -> 2 // 9 % 6 -> 3 // 9 % 5 -> 4 // 9 % 4 -> 1 // 9 % 3 -> 0 FACTOR! // 9 % 2 -> 1 // 9 % 1 -> 0 FACTOR! // 9 is a semifauxtrifactored number since its factors (1 3) equal half the number (4) // 6 is a normal number // 6 cut in half with integer division is (6 / 2) = 3 // 6 % 5 -> 1 // 6 % 4 -> 2 // 6 % 3 -> 0 FACTOR! // 6 % 2 -> 0 FACTOR! // 6 % 1 -> 0 FACTOR! // 6 is a normal number since its factors (1 2 3) do not equal half the number (3)}int main(){ cout int value; cin >> value; value = IsSemifauxtrifactored(value); if (value == 1) {cout } else {cout } cout system("PAUSE");}
Show more
LEARN MORE EFFECTIVELY AND GET BETTER GRADES!
Ask a Question