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C++ prime program Changing to assembly..Please help

Posted on 2004-03-24
Medium Priority
Last Modified: 2011-04-14
I am VERY new to assembly and am intrerested in trying to get this program to work in assembly. I can write it in C++ fine but I don't understand it when I try to convert it to assembly. Please HELP!!!!
I am using Irvine32 and Masm615 here is the C++ code....

#include <iostream>
using namespace std;
void main()
    int i,j;
    double prime;
    cout << "This will caluculate if a number is prime or not.\n";
      cout << "If you would like to quit enter a negative number\n";
             cout << "Please enter a number:";
             cin >> i;

      for (j=2 ; j<i ; j++)
              if (prime==0)
                 cout << i << " Is not a prime number\n";
              if ((i-j)==1)
                     cout << i  << " Is a prime number!!\n";
      }while(i > -1);

Question by:ribaldry
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Expert Comment

ID: 10675439
If you'd supply your version of the solutions, we'd gladly comment it or help to solve any specific errors.
However, Experts doesn't solve school assignments, neither will do the programming instead of you.

Just as a first hint, I'd suggest to take the assembly output from your C++ compiler and try to understand that one.
And a basic problem: after you check for 2, there is no need to check for even numbers. This small idea will speed up your code by a factor of 2.

Author Comment

ID: 10678554
This will caluculate if a number is prime or not.
If you would like to quit enter a negative number
Please enter a number:4
4 Is not a prime number
Please enter a number:5
5 Is a prime number!!
Please enter a number:-1
Press any key to continue

That was the solution....not a homework asssigment though. I understand if you can't do the work so let me try to understand what I want better. I think I could do the program but there are a few lines in the code that I am not sure how to make an assembly equivelant. here are the lines.

  for (j=2 ; j<i ; j++)   ;how would I state this and have it cover what it does with the c++ program?

  prime = i%j              ;how do I do a mod in assembly

one more thing.

How to I get and view  the assembly output with vs .Net?  Is that option /Fa?



Expert Comment

ID: 10678706
Notice the difference between the /FA and /Fa switches. /FAs puts the source code in the assembly file so you can see what instructions are generated from a given C line.
However, I don't see why don't you read/learn the VERY basics of assembly programming before doing anything. I mean the Intel Software Developer Manual, for example. You can download it from http://developer.intel.com/

There isn't modulus in x86 assembly so you have to use the following formula:
A mod B = A - B * INT(A / B)
so checking whether it's divisible by B
if (A == B * INT(A / B)) ...

You can divide using the DIV instruction.
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Accepted Solution

millsoft earned 1500 total points
ID: 10683700
BTW, just from a mathematical standpoint, the line:

for (j=2 ; j<i ; j++)  

only needs to go as far as the square root of i.

It may not seem like a big deal, but it's a large efficiency gain if i is big, and/or you want to check a lot of values.

Of course, performing SQRT in assembly will be fun.  I don't know if there is a FPU instruction for it or not. :)



Expert Comment

ID: 10685240
Of course there is FPU instruction for square root, it's called FSQRT. It's quite easy. However, if someone has problems even with the very basics of integer operations in assembly, I wouldn't lead him/her into dealing with floating point operations. Especially that FPU instructions can generate a lot of exceptions that would be hard for him/her to handle at the moment.

Expert Comment

ID: 10705349
The mod testing in assembler would be the "div" instruction.
If a number is relativly prime to another number then the remainder after a divide is non
zero. With the "div" instuction it takes eax/edx for a 64 bit number,ax/dx for a 32 bit number,
and al/ah for a 16 bit number. The remainder comes back in edx/dx/ah. If these are zero
then the number is relativly prime to the divisor. If you check all odd numbers(ie. even numbers are divisible by 2) less than the square root of the number. You have shown the number to be prime. Yes there is a square root instruction on all pentium processors ,486DX-,and some 386 machines with a coprocessor. It is FSQRT, then just change it to an integer.

Expert Comment

ID: 10731065
Dear ribaldry, please tell me what did help you in the accepted answer regarding your original question.

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