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;
}

我似乎无法弄清楚为什么我的代码没有编译。该程序应该询问用户一个单词，然后根据每个字母的点数对该单词进行评分。

我收到的当前错误是：“错误：在'scrabbleLetters'中请求成员'长度'，这是非类型'char [26]'|”

Answer 1

C ++内置数组没有length()成员函数。找到大小的一种方法是使用

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

鉴于上述方法有点难看，可以将其打包成一个函数，例如：

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) {
    ...
}

具体来说，对于char（或者通常适用于T的任何类型sizeof(T) == 1）的数组，使用sizeof(s)。但请注意，不适用于sizeof(T) != 1的类型。您可能最好不使用内置数组，而是使用std::vector<char>：

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

Answer 2

您可以取消搜索并使用直接访问权限。

将字符串转换为全部小写
从字母中减去'a'以获得相对偏移。
使用相对偏移量作为点数组的索引

以下是一些代码段示例：

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.    
}

Answer 3

解决问题的方法还有很多（除了DietmarKühl的答案）

在循环开始之前计算持有拼字游戏字母的数组的长度。

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;
}

提醒：这种做法很脆弱。 scrabbleLetters的定义必须对函数可见才能实现。否则，sizeof(scrabbleLetters)将最终成为sizeof(char*)，这将无效。

更好的方法 - 完全避免内循环。

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;
}

拼字游戏点计数器c ++

3 个答案: