C ++中的字符串标记化包括定界符字符 [英] string tokenization in C++ including delimiter characters

问题描述

我有以下形式的字符串
a = x + y 或 abc = xyz + 5 或 6 + 5 或 f（p）

I have strings of following form a = x + y or abc = xyz + 5 or 6 + 5 or f(p)

我需要的是标记字符串，以便我读每个运算符和操作数
所以对于 a = x + y 令牌返回应为 a，=，x，+，y c> abc = xyz + 5 它应该返回 abc，=，xyz，+，5 。请注意运算符和操作数之间可能有或可能没有空格

What i need is to tokenize the string such that I read each operator and operand so for a = x + y tokens returns should be a,=,x,+,y and in case of abc=xyz+5 it should return abc,=,xyz,+,5. please note that there may or may not be spaces between operator and operands

这是我尝试过的

void tokenize(std::vector<std::string>& tokens, const char* input, const char* delimiters) {
    const char* s = input;
    const char* e = s;
    while (*e != 0) {
        e = s;
        while (*e != 0 && strchr(delimiters, *e) == 0) {
            ++e;
        }
        if ( *e != ' ' && strchr(delimiters, *e) != 0 ){
            std::string op = "";
            op += *e;
            tokens.push_back(op);
        }
        if (e - s > 0) {
            tokens.push_back(std::string(s,e - s));
        }
        s = e + 1;
    }
}

推荐答案

可以使用这个实现。
第一个参数是要标记的std :: string，第二个参数是要使用的分隔符。它返回一个字符串向量的符号化。非常简单而高效。

You can use this implementation. First argument is the std::string you want to tokenize, second argument is the delimiter you want to use. It returns a vector of strings tokenized. Very simple yet efficient.

vector<string> tokenizeString(const string& str, const string& delimiters)
{  
   vector<string> tokens;
   // Skip delimiters at beginning.
   string::size_type lastPos = str.find_first_not_of(delimiters, 0);
   // Find first "non-delimiter".
   string::size_type pos = str.find_first_of(delimiters, lastPos);

   while (string::npos != pos || string::npos != lastPos)
    {  // Found a token, add it to the vector.
      tokens.push_back(str.substr(lastPos, pos - lastPos));
      // Skip delimiters.  Note the "not_of"
      lastPos = str.find_first_not_of(delimiters, pos);
      // Find next "non-delimiter"
      pos = str.find_first_of(delimiters, lastPos);
   }
    return tokens;
}

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