Scrabble point counter c++

Question

#include <iostream>
#include <string>

using namespace std;
int score(string s);
char scrabbleLetters[] = {'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};
int scrabblePoints[] = {1, 3, 3, 2, 1, 4, 2, 4, 1, 8, 5, 1, 3, 1, 1, 3, 10, 1, 1, 1, 1, 4, 4, 8, 4, 10};

int main()
{
    string sWord;
    cout << "Enter the scrabble word you'd like to score.";
    cin >> sWord;
    cout << "You scored " << score(sWord)<< " points for that word!";

}

int score(string s)
{   int points = 0;
    for (int i = 0; i < s.length(); i++)
    {
        for (int j = 0; j < scrabbleLetters.length(); j++)
        {
            if (s[i] == scrabbleLetters[j])
                points += scrabblePoints[j];
        }
    }
    return points;
}

I can't seem to figure out why my code isn't compiling. The program is supposed to ask a user for a word and then score the word based on points per letter.

The current error I'm receiving is: "error: request for member 'length' in 'scrabbleLetters', which is of non-class type 'char [26]'|"

This use case really begs for a `std::map` rather than keeping two parallel arrays. — Cory Kramer, Nov 05 '14 at 20:51
You appear to be confusing Java with C++. Go re-read the chapter of your textbook that introduces arrays. It should include code that demonstrates how to use them in loops. — Rob Kennedy, Nov 05 '14 at 20:52
Simple arrays don't have .length members. Maybe try `std::array` instead? — Jonathan Potter, Nov 05 '14 at 20:52

score 2 · Answer 1 · answered Nov 05 '14 at 20:54

2

C++ built-in arrays don't have a length() member function. One way to find the size is to use

for (int i = 0; i < std::distance(std::begin(s), std::end(s)); ++i) {
    ...
}

Given that the above approach is a bit ugly, it is possible to package it up a function, e.g.:

template <typename T, std::size_t Size>
constexpr std::size_t length(T const (&)[Size]) {
    return Size;
}
...
for (std::size_t i(0); i != length(s); ++i) {
    ...
}

Specifically for an array of char (or, in general for any type T with sizeof(T) == 1) is to use sizeof(s). Note, however, that this does not work for types where sizeof(T) != 1. You may be better off not to use a built-in array but rather use a std::vector<char>:

std::vector<int> s{'a', 'b' /*...*/ };
for (std::size_t i(0); i != s.size(); ++i) {
    ...
}

answered Nov 05 '14 at 20:54

Dietmar Kühl

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Your answer is much better than what I was going to put lol. +1 for you when my upvote recharges. – John Odom Nov 05 '14 at 20:55
This answer will suffer if the array is decayed, by, i.e., passing it to a function as a parameter – LoPiTaL Nov 05 '14 at 21:15
@LoPiTaL: you mean, the first two code samples will - correctly - not compile in that case, alerting the user to the problem? ... motivating further to change to use `std::vector`? – Dietmar Kühl Nov 05 '14 at 21:19
Yes, sorry. I have not commented here the compile-time error (see my comment in Sahu answer), but it is true. It will fail to compile, which, at least, tells you that something happened. `std::vector` would be a really better solution than the others. – LoPiTaL Nov 05 '14 at 21:23

score 0 · Answer 2 · answered Nov 05 '14 at 21:08

You could eliminate the search and use direct access.

Convert the string to all lowercase
Subtract 'a' from the letter to get a relative offset.
Use the relative offset as an index into the points array

Here's some code snippet examples:

const unsigned int scrabblePoints[] =
{1, 3, 3,  2, 1, 4, 2, 4, 1, 8, 5, 1,  3,
 1, 1, 3, 10, 1, 1, 1, 1, 4, 4, 8, 4, 10};

int main()
{
    string sWord;
    cout << "Enter the scrabble word you'd like to score.";
    cin >> sWord;

    // Transform the word into all lowercase.
    std::transform(sWord.begin(), sWord.end(), sWord.begin, std::tolower);

    unsigned int points = 0;
    for (unsigned int i = 0; i < sWord.length(); ++i)
    {
       const char c = sWord[i];

       // Check if the character is a letter,
       //    it could be something like '?'.
       if (isalpha(c))
       {
         // Since the point's array starts with the letter 'a',
         // the index can be calculated by subtracting 'a' from
         // the character.
         unsigned int index = c - 'a';
         points += scrabblePoints[index];
       }
    }
    cout << "You scored "
         << points
         << " points for that word!"
         << "\n";
  return 0; // Since main() returns a value.    
}

R Sahu · Answer 3 · 2014-11-05T21:31:01.300

-1

Couple of more ways to solve the problem (in addition to the answer by Dietmar Kühl)

Compute the length of the array holding the scrabble letters before looping starts.
```
int score(string s)
{
   int points = 0;
   int len = sizeof(scrabbleLetters);
   for (int i = 0; i < s.length(); i++)
   {
      for (int j = 0; j < len; j++)
      {
         if (s[i] == scrabbleLetters[j])
            points += scrabblePoints[j];
      }
   }
   return points;
}
```
Word of caution: This approach is brittle. The definition of scrabbleLetters must be visible to the function for this to work. Otherwise, sizeof(scrabbleLetters) will end up being sizeof(char*), which will not work.

A much better approach - avoid the inner loop altogether.

int score(string s)
{
   int points = 0;
   for (int i = 0; i < s.length(); i++)
   {
      char ch = s[i];
      points += scrabblePoints[ch-'a'];
   }
   return points;
}

edited Nov 05 '14 at 21:31

answered Nov 05 '14 at 21:02

R Sahu

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Your first solution will suffer if the array is decayed, by, i.e., passing it to a function as a parameter. That solution is even worse than Dietman Kuhl's one, since his would throw a compile-time error, while yours will silently compile sizeof(scrabbleLetters) as sizeof(char *) – LoPiTaL Nov 05 '14 at 21:18
@LoPiTaL, I realize that. The place where it is being used, the full declaration of the variable is visible. Hence `sizeof` will evaluate to the number of elements in the array since its elements are of type `char`. – R Sahu Nov 05 '14 at 21:22

Scrabble point counter c++

3 Answers3