问题：

Question

基于这个很快关闭的问题：
Trying to create a program to read a users input then break the array into seperate words are my pointers all valid?

而不是关闭我认为一些额外的工作可能会帮助OP澄清问题。

问题：

我想标记用户输入并将标记存储到一个单词数组中我想使用标点符号（。， - ）作为分隔符，从而将其从令牌流中删除。

在C中，我会使用strtok()将数组分解为标记，然后手动构建数组像这样：

主要功能：

char **findwords(char *str);

int main()
{
    int     test;
    char    words[100]; //an array of chars to hold the string given by the user
    char    **word;  //pointer to a list of words
    int     index = 0; //index of the current word we are printing
    char    c;

    cout << "die monster !";
    //a loop to place the charecters that the user put in into the array  

    do
    {
        c = getchar();
        words[index] = c;
    }
    while (words[index] != '\n');

    word = findwords(words);

    while (word[index] != 0) //loop through the list of words until the end of the list
    {
        printf("%s\n", word[index]); // while the words are going through the list print them out
        index ++; //move on to the next word
    }

    //free it from the list since it was dynamically allocated
    free(word);
    cin >> test;

    return 0;
}

行标记器：

char **findwords(char *str)
{
    int     size = 20; //original size of the list 
    char    *newword; //pointer to the new word from strok
    int     index = 0; //our current location in words
    char    **words = (char **)malloc(sizeof(char *) * (size +1)); //this is the actual list of words

    /* Get the initial word, and pass in the original string we want strtok() *
     *   to work on. Here, we are seperating words based on spaces, commas,   *
     *   periods, and dashes. IE, if they are found, a new word is created.   */

    newword = strtok(str, " ,.-");

    while (newword != 0) //create a loop that goes through the string until it gets to the end
    {
        if (index == size)
        {
            //if the string is larger than the array increase the maximum size of the array
            size += 10;
            //resize the array
            char **words = (char **)malloc(sizeof(char *) * (size +1));
        }
        //asign words to its proper value
        words[index] = newword;
        //get the next word in the string
        newword = strtok(0, " ,.-");
        //increment the index to get to the next word
        ++index;
    }
    words[index] = 0;

    return words;
}

对上述代码的任何评论都将不胜感激但是，另外，在C ++中实现这一目标的最佳技术是什么？

Answer 1

看一下boost tokenizer在C ++上下文中比strtok()更好的东西。

Answer 2

许多问题已经涵盖了如何在C ++中对流进行标记化示例：How to read a file and get words in C++

但更难找到的是如何获得与strtok（）相同的功能：

基本上，strtok（）允许您将字符串拆分为一大堆用户定义的字符，而C ++流只允许您使用white space作为分隔符。幸运的是，white space的定义是由语言环境定义的，因此我们可以修改语言环境以将其他字符视为空格，这样我们就可以以更自然的方式对流进行标记化。

#include <locale>
#include <string>
#include <sstream>
#include <iostream>

// This is my facet that will treat the ,.- as space characters and thus ignore them.
class WordSplitterFacet: public std::ctype<char>
{
    public:
        typedef std::ctype<char>    base;
        typedef base::char_type     char_type;

        WordSplitterFacet(std::locale const& l)
            : base(table)
        {
            std::ctype<char> const&  defaultCType  = std::use_facet<std::ctype<char> >(l);

            // Copy the default value from the provided locale
            static  char data[256];
            for(int loop = 0;loop < 256;++loop) { data[loop] = loop;}
            defaultCType.is(data, data+256, table);

            // Modifications to default to include extra space types.
            table[',']  |= base::space;
            table['.']  |= base::space;
            table['-']  |= base::space;
        }
    private:
        base::mask  table[256];
};

然后我们可以在这样的地方使用这个方面：

    std::ctype<char>*   wordSplitter(new WordSplitterFacet(std::locale()));

    <stream>.imbue(std::locale(std::locale(), wordSplitter));

问题的下一部分是如何将这些单词存储在数组中。好吧，在C ++中你不会。您可以将此功能委托给std :: vector / std :: string。通过阅读代码，您将看到代码在代码的同一部分中执行两项主要操作。

管理记忆。
标记数据。

基本原则Separation of Concerns，您的代码应该只尝试做两件事之一。它应该进行资源管理（在这种情况下是内存管理），还是应该进行业务逻辑（数据的标记化）。通过将这些代码分成代码的不同部分，您可以更方便地使用代码并更容易编写代码。幸运的是，在这个例子中，所有资源管理都已经由std :: vector / std :: string完成，因此我们可以专注于业务逻辑。

正如已经多次展示的那样，标记化流的简单方法是使用运算符＆gt;＆gt;和一个字符串。这会将流分解为单词。然后，您可以使用迭代器自动循环流标记流。

std::vector<std::string>  data;
for(std::istream_iterator<std::string> loop(<stream>); loop != std::istream_iterator<std::string>(); ++loop)
{
    // In here loop is an iterator that has tokenized the stream using the
    // operator >> (which for std::string reads one space separated word.

    data.push_back(*loop);
}

如果我们将其与一些标准算法结合起来以简化代码。

std::copy(std::istream_iterator<std::string>(<stream>), std::istream_iterator<std::string>(), std::back_inserter(data));

现在将上述所有内容合并到一个应用程序中

int main()
{
    // Create the facet.
    std::ctype<char>*   wordSplitter(new WordSplitterFacet(std::locale()));

    // Here I am using a string stream.
    // But any stream can be used. Note you must imbue a stream before it is used.
    // Otherwise the imbue() will silently fail.
    std::stringstream   teststr;
    teststr.imbue(std::locale(std::locale(), wordSplitter));

    // Now that it is imbued we can use it.
    // If this was a file stream then you could open it here.
    teststr << "This, stri,plop";

    cout << "die monster !";
    std::vector<std::string>    data;
    std::copy(std::istream_iterator<std::string>(teststr), std::istream_iterator<std::string>(), std::back_inserter(data));

    // Copy the array to cout one word per line
    std::copy(data.begin(), data.end(), std::ostream_iterator<std::string>(std::cout, "\n"));
}

如何标记（单词）将标点符号分类为空格

问题：

2 个答案: