Question

我尝试读取一个字典文件，其中每行包含由空格分隔的字ID，字和频率。问题是用于存储单词的地图具有相同的值。如果你能帮助我，我非常感激。

typedef struct{
    int id;
    int count;
    char* word;
} WORD;

//read file
std::map<int, WORD*> readWordMap(char* file_name)
{
    std::ifstream infile(file_name, std::ifstream::in);
    std::cout<<"word map read file:"<<file_name<<std::endl;
    if (! infile) {
        std::cerr<<"oops! unable to open file "<<file_name<<std::endl;
        exit(-1);
     }
     std::map<int, WORD*> map;
     std::vector<std::string> tokens;
     std::string line;
     char word[100];
     int size;
     while (std::getline(infile, line)) {
         size =  (int)split(line, tokens, ' ');
         WORD* entry = (WORD*) malloc(sizeof(WORD*));
         entry->id = atoi(tokens[0].c_str());
         entry->count = atoi(tokens[2].c_str());
         strcpy(word, tokens[1].c_str());
         entry->word = word;

         map[entry->id] = entry;
         std::cout<< entry->id<<" "<<entry->word<<" "<<entry->count<<std::endl;

      }
      infile.close();
      std::cout<<map.size()<<std::endl;
      std::map<int, WORD*>::const_iterator it;
      for (it = map.begin(); it != map.end(); it++) {
           std::cout<<(it->first)<<" "<<(it->second->word)<<std::endl;

      }

      return map;
}

//split string by a delimiter
size_t split(const std::string &txt, std::vector<std::string> &strs, char ch)
{
    size_t pos = txt.find( ch );
    size_t initialPos = 0;
    strs.clear();

    while( pos != std::string::npos ) {
        strs.push_back( txt.substr( initialPos, pos - initialPos + 1 ) );
        initialPos = pos + 1;

        pos = txt.find( ch, initialPos );
    } 

   strs.push_back( txt.substr( initialPos, std::min( pos, txt.size() ) - initialPos + 1      ) );

   return strs.size();
}

数据文件：

2 I  1
3 gave  1
4 him  1
5 the  3
6 book  3
7 .  3
8 He  2
9 read  1
10 loved  1

结果：

2 I  1
3 gave  1
4 him  1
5 the  3
6 book  3
7 .  3
8 He  2
9 read  1
10 loved  1
map size:9
2 loved 
3 loved 
4 loved 
5 loved 
6 loved 
7 loved 
8 loved 
9 loved 
10 loved

Answer 1

您忘记在WORD::word之前为strcpy分配内存。并且您将char word[100]的地址分配给地图的所有项目，这些项目对所有项目都是相同的。

最好使用std::string而不是C风格的字符串。此外，您可以使用std::stoi将字符串转换为整数。试试这个：

struct WORD{
    int id;
    int count;
    std::string word;
};

std::map<int, WORD> readWordMap(const std::string &file_name)
{
     ...
     std::map<int, WORD> map;
     ...

     while (std::getline(infile, line)) {
         ...

         WORD entry;
         entry.id = std::stoi(tokens[0]);
         entry.count = std::stoi(tokens[2]);
         entry.word = tokens[1];

         map[entry.id] = entry;

         ...
      }
      infile.close();
      ...
}

Answer 2

WORD* entry = (WORD*) malloc(sizeof(WORD*));

分配WORD pointer而不是整个WORD结构。

编译器一直在分配条目，因为没有被任何事情搞砸（它们都指向一些随机地址，甚至可能不属于你的程序。）而你重复地将该指针添加到地图。因此，地图的所有第一个都指向同一位置（巧合）。它应该是

WORD* entry = new WORD;

这是一种更清洁的方式

struct WORD{
    int id;
    int count;
    std::string word;
};

while (std::getline(infile, line)) {
     WORD* entry = new WORD;
     std::istringstream iss(line);

     iss >> entry->id >> entry->word >> entry->count;
     map[entry->id] = entry;
     std::cout<< entry->id<<" "<<entry->word<<" "<<entry->count<<std::endl;
  }

带指针值的C ++映射

2 个答案: