Prime numbers program

Question

I\'m currently trying out some questions just to practice my programming skills. ( Not taking it in school or anything yet, self taught ) I came across this problem which re

Answer 1

There are a two approaches to testing for primality you might want to consider:

1. The problem domain is small enough that just looping over the numbers until you find the Nth prime would probably be an acceptable solution and take less than a few milliseconds to complete. There are a number of simple optimizations you can make to this approach for example you only need to test to see if it's divisible by 2 once and then you only have to check against the odd numbers and you only have to check numbers less than or equal to the aquare root of the number being tested.
2. The Sieve of Eratosthenes is very effective and easy to implement and incredibly light on the math end of things.

As for why you code is crashing I suspect changing the line that reads

for( int d=0; d<=b; d++) 

to

for( int d=0; d<b; d++) 

will fix the problem because you are trying to read from a potentially uninitialized element of the array which probably contains garbage.

Answer 2

I'm trying to improve my functional programming at the moment so I just coded up the sieve quickly. I figure I'll post it here. If you're still learning, you might find it interesting, too.

#include <iostream>
#include <list>
#include <math.h>
#include <functional>
#include <algorithm>

using namespace std;

class is_multiple : public binary_function<int, int, bool>
{
    public:
        bool operator()(int value, int test) const
        {
            if(value == test) // do not remove the first value
                return false;
            else
                return (value % test) == 0;
        }
};

int main() 
{
    list<int> numbersToTest;
    int input = 500;

    // add all numbers to list
    for(int x = 1; x < input; x++)
        numbersToTest.push_back(x);

    // starting at 2 go through the list and remove all multiples until you reach the squareroot
    // of the last element in the list
    for(list<int>::iterator itr = ++numbersToTest.begin(); *itr < sqrt((float) input); itr++)
    {
        int tmp = *itr;
        numbersToTest.remove_if(bind2nd(is_multiple(), *itr));  
        itr = find(numbersToTest.begin(), numbersToTest.end(), tmp); //remove_if invalidates iterator 
                                                                 // so find it again. kind of ugly
    }

    // output primes
    for(list<int>::iterator itr = numbersToTest.begin(); itr != --numbersToTest.end(); itr++)
        cout << *itr << "\t";

    system("PAUSE");

    return 0;
}

Any advice on how to improve this would be welcome by the way.

Answer 3

I haven't looked at your code, but your array must be large enough to contain all the values you will store in it. 100 certainly isn't going to be enough for most input for this problem.

E.g. this code..

int someArray[100];
someArray[150] = 10;

Writes to a location large than the array (150 > 100). This is known as a memory overwrite. Depending on what happened to be at that memory location your program may crash immediately, later, or never at all.

A good practice when using arrays is to assert in someway that the element you are writing to is within the bounds of the array. Or use an array-type class that performs this checking.

For your problem the easiest approach would be to use the STL vector class. While you must add elements (vector::push_back()) you can later access elements using the array operator []. Vector will also give you the best iterative performance.

Here's some sample code of adding the numbers 0-100 to a vector and then printing them. Note in the second loop we use the count of items stored in the vector.

#include <vector> // std::vector

...

const int MAX_ITEMS = 100;

std::vector<int> intVector;

intVector.reserve(MAX_ITEMS); // allocates all memory up-front

// add items
for (int i = 0; i < MAX_ITEMS; i++)
{
  intVector.push_back(i);  // this is how you add a value to a vector;
}

// print them
for (int i = 0; i < intVector.size(); i++)
{
  int elem = intVector[i]; // this access the item at index 'i'
  printf("element %d is %d\n", i, elem);
}

Answer 4

Since this is for pedagogical purposes, I would suggest implementing the Sieve of Eratosthenes.

Answer 5

Since your question is about programming rather than math, I will try to keep my answer that way too.

The first glance of your code makes me wonder what on earth you are doing here... If you read the answers, you will realize that some of them didn't bother to understand your code, and some just dump your code to a debugger and see what's going on. Is it that we are that impatient? Or is it simply that your code is too difficult to understand for a relatively easy problem?

To improve your code, try ask yourself some questions:

1. What are a, b, c, etc? Wouldn't it better to give more meaningful names?
2. What exactly is your algorithm? Can you write down a clearly written paragraph in English about what you are doing (in an exact way)? Can you modify the paragraph into a series of steps that you can mentally carry out on any input and can be sure that it is correct?
3. Are all steps necessary? Can we combine or even eliminate some of them?
4. What are the steps that are easy to express in English but require, say, more than 10 lines in C/C++?
5. Does your list of steps have any structures? Loops? Big (probably repeated) chunks that can be put as a single step with sub-steps?

After you have going through the questions, you will probably have a clearly laid out pseudo-code that solves the problem, which is easy to explain and understand. After that you can implement your pseudo-code in C/C++, or, in fact, any general purpose language.

Answer 6

Running your code through a debugger, I've found that it crashes with a floating point exception at "if ((a % c) == 0)". The reason for this is that you haven't initialized anything in num, so you're doing "a % 0".