Need a code written in Armv7 Assembly language
Budget: $10 – $30 USD
i just need this simple code written in assembly language ARMv7 using visUAL Arm emulator, i can provide the c++ code as well
Here is the assignment
"You will be given an integer array of positive numbers. Your program should process the array and do the following for each entry:
Determine if the number is odd or even.
Add to a counter for each type, i.e. an even counter and an odd counter.
Add the number to a sum for each type, i.e. a total of the odd numbers and a total of the even numbers.
A value of -999 indicates the end of the array. Your program should end at that point.
The input will be the following variable in your program:
L3Array DCD 22,9,333,47,72,128,111,44,-999
The actual number of array entries and their values will be different when your program is graded. The last number will still be -999.
The output of your program will be the counts and totals placed in the following variables:
L3OddCt DCD 0 ; count of odd numbers
L3OddTot DCD 0 ; total of odd numbers
L3EvCt DCD 0 ; count of even numbers
L3EvTot DCD 0 ; total of even numbers
Your program must use the variable names L3Array, L3OddCt, L3OddTot, L3EvCt, and L3EvTot.
The expected results for the array contents shown above are:
L3OddCt = 4
L3OddTot = 500
L3EvCt = 4
L3EvTot = 266
Your program will be tested with a different set of values in L3Array. The expected output will vary based on that input. Do not assume that the array will contain eight values."
and here is the C++ code that does this assignment
"
#include <iostream>
using namespace std;
int main()
{
int L3Array[256];
int i = 0, number = 0, L3OddCt = 0, L3OddTot = 0, L3EvCt = 0, L3EvTot = 0;
while (number != -999)
{
cin >> number;
L3Array[i++] = number;
}
i = 0;
while (L3Array[i] != -999)
{
if (L3Array[i] % 2 == 0)
{
L3EvCt++;
L3EvTot += L3Array[i];
}
else
{
L3OddCt++;
L3OddTot += L3Array[i];
}
i++;
}
cout << endl;
cout << "L3OddCt = " << L3OddCt << endl;
cout << "L3OddTot = " << L3OddTot << endl;
cout << "L3EvCt = " << L3EvCt << endl;
cout << "L3EvTot = " << L3EvTot << endl;
return 0;
}
"
Here is the assignment
"You will be given an integer array of positive numbers. Your program should process the array and do the following for each entry:
Determine if the number is odd or even.
Add to a counter for each type, i.e. an even counter and an odd counter.
Add the number to a sum for each type, i.e. a total of the odd numbers and a total of the even numbers.
A value of -999 indicates the end of the array. Your program should end at that point.
The input will be the following variable in your program:
L3Array DCD 22,9,333,47,72,128,111,44,-999
The actual number of array entries and their values will be different when your program is graded. The last number will still be -999.
The output of your program will be the counts and totals placed in the following variables:
L3OddCt DCD 0 ; count of odd numbers
L3OddTot DCD 0 ; total of odd numbers
L3EvCt DCD 0 ; count of even numbers
L3EvTot DCD 0 ; total of even numbers
Your program must use the variable names L3Array, L3OddCt, L3OddTot, L3EvCt, and L3EvTot.
The expected results for the array contents shown above are:
L3OddCt = 4
L3OddTot = 500
L3EvCt = 4
L3EvTot = 266
Your program will be tested with a different set of values in L3Array. The expected output will vary based on that input. Do not assume that the array will contain eight values."
and here is the C++ code that does this assignment
"
#include <iostream>
using namespace std;
int main()
{
int L3Array[256];
int i = 0, number = 0, L3OddCt = 0, L3OddTot = 0, L3EvCt = 0, L3EvTot = 0;
while (number != -999)
{
cin >> number;
L3Array[i++] = number;
}
i = 0;
while (L3Array[i] != -999)
{
if (L3Array[i] % 2 == 0)
{
L3EvCt++;
L3EvTot += L3Array[i];
}
else
{
L3OddCt++;
L3OddTot += L3Array[i];
}
i++;
}
cout << endl;
cout << "L3OddCt = " << L3OddCt << endl;
cout << "L3OddTot = " << L3OddTot << endl;
cout << "L3EvCt = " << L3EvCt << endl;
cout << "L3EvTot = " << L3EvTot << endl;
return 0;
}
"