存储数组的地址列表，以解析原始的非终止文本

Question

我刚刚开始编程，但是我有很多关于如何使我的生活更容易解析文件时，通过制作一个程序映射数据的地址，当从一个文件读取到内存

注意:我把墙上的文字剪掉了，这里是一个简单的问题

如何解析没有空结束符的字符数组，但单词都以大写字母开头，因此可以使用大写作为分隔符?

基本上我想解析的文本文件，只是'WordWordWord'和发送每个单词到它自己的单独的字符串变量，然后能够写每个单词到一个文本文件添加换行符。

我想做一些更高级的东西，但我被要求剪掉文本墙，所以现在就可以了:)

//pointers and other values like file opening were declared
int len = (int) strlen( words2 );
cout << "nSize of Words2 is  : " << len << " bytesn";
// Loops through array if uppercase then...     
for (int i = 0; i < len; i++)
    {
        if (isupper(words2[i]))
        {
        // Output the contents of words2
    cout << "n Words2 is upper : " << words2[i] << "n";
        b1 = &words2[i];
    //output the address of b1 and the intvalue of words2[var]
    cout << "nChar address is  " << &b1 << " word address is " << (int) words2[i] << "n";
        cout << "nChar string is  " << b1 << " address +1 "<< &b1+1 <<"n  and string is " << b1+1 << "n";
        }
        cout << "nItem I is : i " << i << " and words2 is  " << words2[i] << "n";
    }

    fin.clear();
    fin.close();
    fout.close();

Answer 1

简单。使用增加。Tokenizer，使用char_separator("", "ABCDEFGHIJKLMNOPQRSTUVWXYZ")。""为丢弃的分隔符集合，A-Z为保留的分隔符集合。(如果你使用A-Z作为省略分隔符，你会得到ord ord ord，因为你省略了w)

Answer 2

既然你也

想做一些更高级的东西

我想看看Boost。从一开始就使用正则表达式。

Answer 3

vector<char *> parsedStrings;
char * words = "HelloHelloHello";
int stringStartAddress = 0;
for (int i = 0; i <= strlen(words); i++)
{
    /* Parses word if current char is uppercase or
     if it's the last char and an uppercase char was previously matched */
    if (isupper(words[i]) || ((i == strlen(words)) && (stringStartAddress != 0)))
    {
        // Current char is first uppercase char matched, so don't parse word
        if (stringStartAddress == 0)
        {
            stringStartAddress = ((int)(words + i));
            continue;
        }
        int newStringLength = ((int)(words + i)) - stringStartAddress;
        char * newString = new char[newStringLength + 1];
        // Copy each char from previous uppercase char up to current char 
        for (int j = 0; j < newStringLength; j++)
        {
            // Cast integer address of char to a char pointer and then get the char by dereferencing the pointer
            // Increment address to that of the next char
            newString[j] = *((char *)stringStartAddress++);
        }
        newString[newStringLength] = ''; // add null-terminator to string
        parsedStrings.push_back(newString);
    }
}