How to find the power set of a given set without using left shift bit?

Question

I'm trying to figure out how to implement an algorithm to find a power set given a set, but I'm having some trouble. The sets are actually vectors so for example I am given Set<char> set1{ 'a','b','c' }; I would do PowerSet(set1); and I would get all the sets but if I do Set<char> set2{ 'a','b','c', 'd' }; I would do PowerSet(set2) and I would miss a few of those sets.

Set<Set<char>> PowerSet(const Set<char>& set1)
{
    Set<Set<char>> result;
    Set<char> temp;
    result.insertElement({});
    int card = set1.cardinality();
    int powSize = pow(2, card);
    
    for (int i = 0; i < powSize; ++i)
    {
        for (int j = 0; j < card; ++j)
        {
            if (i % static_cast<int> ((pow(2, j)) + 1))
            {
                temp.insertElement(set1[j]);
                result.insertElement(temp);
            }
        }
        temp.clear();
    }
    return result;
}

For reference:

• cardinality() is a function in my .h where it returns the size of the set.
• insertElement() inserts element into the set while duplicates are ignored.
• Also the reason why I did temp.insertElement(s[j]) then result.insertElement(temp) is because result is a set of a set and so I needed to create a temporary set to insert the elements into then insert it into result.
• clear() is a function that empties the set.
• I also have removeElem() which removes that element specified if it exists, otherwise it'll ignore it.

Answer 1

Your if test is nonsense -- it should be something like

if ((i / static_cast<int>(pow(2,j))) % 2)

you also need to move the insertion of temp into result after the inner loop (just before the temp.clear()).

With those changes, this should work as long as pow(2, card) does not overflow an int -- that is up to about card == 30 on most machines.