How do I create a pointer to an array class member?

Question

I am trying to return the address of an array @element from my Hash Table using the find function. However, I am getting the compiler error:

QuadraticProbing.cpp:134:59: error: invalid conversion from ‘const char*’ to ‘char*’ [-fpermissive]
             return isActive( currentPos ) ? wordElement : ITEM_NOT_FOUND;

Basically, I just want to return a pointer to @element. So I tried creating a pointer wordElement to @element and tried returning wordElement. But that didn't work. Here is a snippet of my code, I can't figure out how to get a pointer to @element in HashEntry.

//Main
int main()
{
    QuadraticHashTable<char*> table(100);
    table.insert("HELLO WORLD");
    if (table.find(document[i]) == NULL))
        cout << "OH NO!";
}

//Class that has element that I want to return in find.
template <class HashedObj>
class QuadraticHashTable
{
  public:
    QuadraticHashTable()

    const HashedObj & find( const HashedObj & x ) const;

    enum EntryType { ACTIVE, EMPTY, DELETED };
  private:
    struct HashEntry
    {
        char element[20];
        EntryType info;


        HashEntry( const HashedObj & e = HashedObj( ), EntryType i = EMPTY )
          : info( i ) 
          {
            if (e != NULL)
                strcpy(element, e);
          }
    };
        vector<HashEntry> array;

//Find Function
    template <class HashedObj>
    const HashedObj & QuadraticHashTable<HashedObj>::find( const HashedObj & x ) const
    {
        int currentPos = findPos( x );
        const char * wordElement = array[currentPos].element;
        return isActive( currentPos ) ? wordElement : ITEM_NOT_FOUND;
    }

We have no idea what ITEM_NOT_FOUND is, but the design is not viable regardless. array[currentPos].element is a char array. The function tries to return a HashedObj &. This cannot work. And why are you designing in a special "not-a-value" value? Do you know what the billion dollar mistake is? — n. m. could be an AI
– n. m. could be an AI, Commented Feb 26, 2017 at 8:20

Yakk - Adam Nevraumont · Accepted Answer · 2017-02-26 15:44:47Z

2

QuadraticHashTable<char*> table(100);
table.insert("HELLO WORLD");

A HashedObject is a char*

You pass "HELLO WORLD" to insert, which expects a const HashedObject&.

The const there applies at the top level, so it is a char* const& not a const char*& (which would be a different error).

Given that your entries are basically char[20] and how you wrote the entry, this code only works if HashedObjects are raw C strings. The template parameter is pointless as written. So there is that.

But char const* as the template parameter is the other way to make your code compile. But really, a template that works with exactly one type is pretty pointless.

edited Feb 26, 2017 at 15:44

answered Feb 26, 2017 at 8:16

Yakk - Adam Nevraumont

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