使用strtok函数在C中拆分字符串

Question

I'm trying to do split some strings by {white_space} symbol. btw, there is a problem within some splits. which means, I want to split by {white_space} symbol but also quoted sub-strings.

example,

char *pch;
char str[] = "hello \"Stack Overflow\" good luck!";
pch = strtok(str," ");
while (pch != NULL)
{
    printf ("%s\n",pch);
    pch = strtok(NULL, " ");
}

This will give me

hello
"Stack
Overflow"
good
luck!

But What I want, as you know,

hello
Stack Overflow
good
luck!

Any suggestion or idea please?

Answer 1

You'll need to tokenize twice. The program flow you currently have is as follows:

1) Search for space

2) Print all characters prior to space

3) Search for next space

4) Print all characters between last space, and this one.

You'll need to start thinking in a different matter, two layers of tokenization.

1. Search for Quotation Mark
2. On odd-numbered strings, perform your original program (search for spaces)
3. On even-numbered strings, print blindly

In this case, even numbered strings are (ideally) within quotes. ab"cd"ef would result in ab being odd, cd being even... etc.

The other side, is remembering what you need to do, and what you're actually looking for (in regex) is "[a-zA-Z0-9 \\t\\n]*" or, [a-zA-Z0-9]+. That means the difference between the two options, are whether it's separated by quotes. So separate by quotes, and identify from there.

Answer 2

Try altering your strategy.

Look at non-white space things, then when you find quoted string you can put it in one string value.

So, you need a function that examines characters, between white space. When you find '"' you can change the rules and hoover everything up to a matching '"' . If this function returns a TOKEN value and a value (the string matched) then what calls it, can decide to do the correct output. Then you have written a tokeniser, and there actually exist tools to generate them called "lexers" as they are used widely, to implement programming languages/config files.

Assuming nextc reads next char from string, begun by firstc( str) :

for (firstc( str); ((c = nextc) != NULL;) {
    if (isspace(c))
        continue;
    else if (c == '"')
        return readQuote;       /* Handle Quoted string */
    else
        return readWord;        /* Terminated by space & '"' */
}
return EOS;

You'll need to define return values for EOS, QUOTE and WORD, and a way to get the text in each Quote or Word.

Answer 3

Here's the code that works... in C

The idea is that you first tokenize the quote, since that's a priority (if a string is inside the quotes than we don't tokenize it, we just print it). And for each of those tokenized strings, we tokenize within that string on the space character, but we do it for alternate strings, because alternate strings will be in and out of the quotes.

#include <stdio.h>
#include <string.h>
#include <stdbool.h>

int main() {
  char *pch1, *pch2, *save_ptr1, *save_ptr2;
  char str[] = "hello \"Stack Overflow\" good luck!";
  pch1 = strtok_r(str,"\"", &save_ptr1);
  bool in = false;
  while (pch1 != NULL) {
    if(in) {
      printf ("%s\n", pch1);
      pch1 = strtok_r(NULL, "\"", &save_ptr1);
      in = false;
      continue;
    }
    pch2 = strtok_r(pch1, " ", &save_ptr2);
    while (pch2 != NULL) {
      printf ("%s\n",pch2);
      pch2 = strtok_r(NULL, " ", &save_ptr2);
    }
    pch1 = strtok_r(NULL, "\"", &save_ptr1);
    in = true;
  }
}

References

• Tokenizing multiple strings simultaneously
• http://linux.die.net/man/3/strtok_r
• http://www.cplusplus.com/reference/cstring/strtok/

Answer 4

Here it is in C++. I am sure it can be written more elegantly, but it works and is a start:

#include <iostream>
#include <stdexcept>
#include <vector>
#include <string>

using namespace std;

using Tokens = vector<string>;


Tokens split(string const & sentence) {
  Tokens tokens;
  // indexes to split on
  string::size_type from = 0, to;

  // true if we are inside quotes: we don't split by spaces and we expect a closing quote
  // false otherwise
  bool in_quotes = false;

  while (true) {
    // compute to index
    if (!in_quotes) {
      // find next space or quote
      to = sentence.find_first_of(" \"", from);
      if (to != string::npos && sentence[to] == '\"') {
        // we found an opening quote
        in_quotes = true;
      }
    } else {
      // find next quote (ignoring spaces)
      to = sentence.find('\"', from);
      if (to == string::npos) {
        // no enclosing quote found, invalid string
        throw invalid_argument("missing enclosing quotes");
      }
      in_quotes = false;
    }
    // skip empty tokens
    if (from != to) {
      // get token
      // last token
      if (to == string::npos) {
        tokens.push_back(sentence.substr(from));
        break;
      }
      tokens.push_back(sentence.substr(from, to - from));
    }
    // move from index
    from = to + 1;
  }
  return tokens;
}

test it:

void splitAndPrint(string const & sentence) {
  Tokens tokens;
  cout << "-------------" << endl;
  cout << sentence << endl;
  try {
    tokens = split(sentence);
  } catch (exception &e) {
    cout << e.what() << endl;
    return;
  }
  for (const auto &token : tokens) {
    cout << token << endl;
  }
  cout << endl;
}

int main() {
  splitAndPrint("hello \"Stack Overflow\" good luck!");
  splitAndPrint("hello \"Stack Overflow\" good luck from \"User Name\"");
  splitAndPrint("hello and good luck!");
  splitAndPrint("hello and \" good luck!");

  return 0;
}

output:

-------------
hello "Stack Overflow" good luck!
hello
Stack Overflow
good
luck!

-------------
hello "Stack Overflow" good luck from "User Name"
hello
Stack Overflow
good
luck
from
User Name

-------------
hello and good luck!
hello
and
good
luck!

-------------
hello and " good luck!
missing enclosing quotes