拼字游戏点计数器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; 
} 

我似乎无法弄清楚为什么我的代码没有编译。该程序应该向用户询问一个单词，然后根据每个字母的点数对单词进行评分。

我收到的当前错误是：“错误：'scrabbleLetters'中的成员'length'的请求，它是非类类型'char [26]'|”

Answer 1

C++内置阵列没有length()成员函数。找到尺寸的一种方法是使用

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

鉴于上述方法是有点难看，有可能将它打包了一个函数，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) { 
    ... 
} 

具体地为阵列char（或者通常对于任何类型的T与sizeof(T) == 1）是使用sizeof(s)。然而，请注意，这不是而是适用于类型为sizeof(T) != 1。您可能会关闭不更好地使用内置阵列，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

的更多的方式来解决这个问题（除了迪特马尔库尔答案）

夫妇

1. 在循环开始之前计算包含拼字游戏字母数组的长度。

   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; 
   } 
   

   字谨慎：这种方法是易碎的。该函数的定义scrabbleLetters必须对此功能可见。否则，sizeof(scrabbleLetters)将最终为sizeof(char*)，这将不起作用。

2. 一个更好的方法 - 完全避免内部循环。

   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; 
   } 
   

Answer 3

您可以消除搜索和使用直接访问。

1. 将字符串到所有小写
2. 减“一”从信中来获得相对偏移。
3. 使用相对作为索引偏移量点阵列

这里的一些代码段的例子：

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.  
}