Convert String containing several numbers into integers

Question

I realize that this question may have been asked several times in the past, but I am going to continue regardless.

I have a program that is going to get a string of numbers from keyboard input. The numbers will always be in the form "66 33 9" Essentially, every number is separated with a space, and the user input will always contain a different amount of numbers.

I'm aware that using 'sscanf' would work if the amount of numbers in every user-entered string was constant, but this is not the case for me. Also, because I'm new to C++, I'd prefer dealing with 'string' variables rather than arrays of chars.

Answer 1

//I modified this example from the link to make it... right
#include <iostream>

int main() {

    int p, q, r;

    std::cout << "Enter 3 integers separated by space: \n";
    std::cin >> p >> q >> r;
    std::cout << "Sum of the " << p << "," << q << 
                 " and " << r << " is = " << (p+q+r) << endl;

    return 0;

}

http://www.tenouk.com/Module18.html

cin>> is capable of doing formatted input.
If you have three ints, p, q, and r,
you just put those in that sequence after cin like in the example.
It'll skip the spaces for you and convert the "numbers" in the string into ints.

Answer 2

Here is how to split your string into strings along the spaces. Then you can process them one-by-one.

Answer 3

try strtoken to separate the string first, the you will deal with each string

Answer 4

I assume you want to read an entire line, and parse that as input. So, first grab the line:

std::string input;
std::getline(std::cin, input);

Now put that in a stringstream:

std::stringstream stream(input);

and parse

while(1) {
   int n;
   stream >> n;
   if(!stream)
      break;
   std::cout << "Found integer: " << n << "\n";
}

Remember to include

#include <string>
#include <sstream>

Answer 5

#include <string>
#include <vector>
#include <sstream>
#include <iostream>
using namespace std;

int ReadNumbers( const string & s, vector <int> & v ) {
    istringstream is( s );
    int n;
    while( is >> n ) {
     v.push_back( n );
    }
    return v.size();
}

int main() {
    string s;
    vector <int> v;
    getline( cin, s );
    ReadNumbers( s, v );
    for ( int i = 0; i < v.size(); i++ ) {
     cout << "number is " <<  v[i] << endl;
    }
}

Answer 6

#include <string>
#include <vector>
#include <iterator>
#include <sstream>
#include <iostream>

int main() {
   std::string input;
   while ( std::getline( std::cin, input ) )
   {
      std::vector<int> inputs;
      std::istringstream in( input );
      std::copy( std::istream_iterator<int>( in ), std::istream_iterator<int>(),
         std::back_inserter( inputs ) );

      // Log process: 
      std::cout << "Readed " << inputs.size() << " integers from string '" 
         << input << "'" << std::endl;
      std::cout << "\tvalues: ";
      std::copy( inputs.begin(), inputs.end(), 
         std::ostream_iterator<int>( std::cout, " " ) );
      std::cout << std::endl;
   }
 }

Answer 7

// get string
std::string input_str;
std::getline( std::cin, input_str );

// convert to a stream
std::stringstream in( input_str );

// convert to vector of ints
std::vector<int> ints;
copy( std::istream_iterator<int, char>(in), std::istream_iterator<int, char>(), back_inserter( ints ) );

Answer 8

Generic solution for unsigned values (dealing with prefix '-' takes an extra bool):

template<typename InIter, typename OutIter>
void ConvertNumbers(InIter begin, InIter end, OutIter out)
{
    typename OutIter::value_type accum = 0;
    for(; begin != end; ++begin)
    {
        typename InIter::value_type c = *begin;
        if (c==' ') {
            *out++ = accum; accum = 0; break;
        } else if (c>='0' && c <='9') {
            accum *= 10; accum += c-'0';
        }
    }
    *out++ = accum;
       // Dealing with the last number is slightly complicated because it
       // could be considered wrong for "1 2 " (produces 1 2 0) but that's similar
       // to "1  2" which produces 1 0 2. For either case, determine if that worries
       // you. If so: Add an extra bool for state, which is set by the first digit,
       // reset by space, and tested before doing *out++=accum.
}

Answer 9

#include <....>
int main()
{
   std::string s = "1 23 456 7890";
   std::deque<int> lst;
   strtk::parse(s," ",lst);
   std::copy(lst.begin(),lst.end(),std::ostream_iterator<int>(std::cout,"\t"));
   return 0;
}