Split string by a character?

Question

How can I split a string such as "102:330:3133:76531:451:000:12:44412 by the ":" character, and put all of the numbers into an int array (number sequence will always be 8 elements long)? Preferably without using an external library such as boost.

Also, I'm wondering how I can remove unneeded characters from the string before it's processed such as "$" and "#"?

Answer 1

stringstream can do all these.

1. Split a string and store into int array:

   string str = "102:330:3133:76531:451:000:12:44412";
   std::replace(str.begin(), str.end(), ':', ' ');  // replace ':' by ' '
   
   vector<int> array;
   stringstream ss(str);
   int temp;
   while (ss >> temp)
       array.push_back(temp); // done! now array={102,330,3133,76531,451,000,12,44412}
   

2. Remove unneeded characters from the string before it's processed such as $ and #: just as the way handling : in the above.

PS: The above solution works only for strings that don't contain spaces. To handle strings with spaces, please refer to here based on std::string::find() and std::string::substr().

Answer 2

The standard way in C is to use strtok like others have answered. However strtok is not C++-like and also unsafe. The standard way in C++ is to use std::istringstream

std::istringstream iss(str);
char c; // dummy character for the colon
int a[8];
iss >> a[0];
for (int i = 1; i < 8; i++)
    iss >> c >> a[i];

In case the input always has a fixed number of tokens like that, sscanf may be another simple solution

std::sscanf(str, "%d:%d:%d:%d:%d:%d:%d:%d", &a1, &a2, &a3, &a4, &a5, &a6, &a7, &a8);

Answer 3

I had to write some code like this before and found a question on Stack Overflow for splitting a string by delimiter. Here's the original question: link.

You could use this with std::stoi for building the vector.

std::vector<int> split(const std::string &s, char delim) {
    std::vector<int> elems;
    std::stringstream ss(s);
    std::string number;
    while(std::getline(ss, number, delim)) {
        elems.push_back(std::stoi(number));
    }
    return elems;
}

// use with:
const std::string numbers("102:330:3133:76531:451:000:12:44412");
std::vector<int> numbers = split(numbers, ':');

Here is a working ideone sample.

Answer 4

True ! there's no elven magic

Its kinda answered here too

#include <cstring>
#include <iostream>
#include<cstdlib>
#include<vector>

int main() 
{
    char input[100] = "102:330:3133:76531:451:000:12:44412";
    char *token = std::strtok(input, ":");
    std::vector<int> v;

    while (token != NULL) {
        v.push_back( std::strtol( token, NULL, 10 ));
        token = std::strtok(NULL, ":");
    }

    for(std::size_t i =0 ; i < v.size() ; ++i)
        std::cout << v[i] <<std::endl;
}

Demo Here

Answer 5

To remove characters '#' and '$' you can use standard algorithm std::remove_if. However take into account that if there is for example the following string "12#34" then after removing '#' you will ge "1234". If you need that the resulted string will look as "12 34" or "12:34" then instead of std::remove_if it is better to use std::replace_if.

Below there is a sample code that performs the task. You need to include headers

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <algorithm>

int main()
{
    char s[] = "102:$$330:#3133:76531:451:000:$12:44412";

    std::cout << s << std::endl;

    char *p = std::remove_if( s, s + std::strlen( s ), 
        []( char c ) { return ( c == '$' || c == '#' ); } );
    *p = '\0';

    std::cout << s << std::endl;

    const size_t N = 8;
    int a[N];

    p = s;
    for ( size_t i = 0; i < N; i++ )
    {
        a[i] = strtol( p, &p, 10 );
        if ( *p == ':' ) ++p;
    }

    for ( int x : a ) std::cout << x << ' ';
    std::cout << std::endl;
}

The output is

102:$$330:#3133:76531:451:000:$12:44412
102:330:3133:76531:451:000:12:44412
102 330 3133 76531 451 0 12 44412

Answer 6

Here's one way... not the cleverest but fast to write (8 repetition's verging on warranting a loop though). This approach to parsing is quite widely useful so good to learn. !(iss >> c) ensures there's no trailing non-whitespace characters in the string.

std::istringstream iss(the_string);
char c;
int n[8];
if (iss >> n[0] >> c && c == ':' &&
    iss >> n[1] >> c && c == ':' &&
    iss >> n[2] >> c && c == ':' &&
    iss >> n[3] >> c && c == ':' &&
    iss >> n[4] >> c && c == ':' &&
    iss >> n[5] >> c && c == ':' &&
    iss >> n[6] >> c && c == ':' &&
    iss >> n[7] && !(iss >> c))
    ...

Answer 7

You can use strtok() for split your string, perhaps in while loop.

When you get individual string then can use atoi(xxx) for conversion in ints.

Answer 8

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

int main ()
{
  char str[] ="102:330:3133:76531:451:000:12:44412";
  char * pch;
  printf ("Splitting string \"%s\" into tokens:\n",str);
  pch = strtok (str,":");
  while (pch != NULL)
  {
    printf ("%s\n",pch);
    pch = strtok (NULL, ":");
  }
  return 0;
}

Answer 9

Another solution using the regular expression features in C++11.

#include <algorithm>
#include <iostream>
#include <iterator>
#include <ostream>
#include <regex>
#include <sstream>
#include <string>
#include <vector>

int main()
{
    const std::string s = "102:330:3133:76531:451:000:12:44412";

    // Replace each colon with a single space
    const std::regex pattern(":");
    const std::string r = std::regex_replace(s, pattern, " ");

    std::istringstream ist(r);

    std::vector<int> numbers;
    std::copy(std::istream_iterator<int>(ist), std::istream_iterator<int>(),
        std::back_inserter(numbers));

    // We now have a vector of numbers

    // Print it out
    for (auto n : numbers)
    {
        std::cout << n << " ";
    }
    std::cout << std::endl;

    return 0;
}