How to read in user entered comma separated integers?

Question

I\'m writing a program that prompts the user for:

1. Size of array
2. Values to be put into the array

First part is fine, I create a dynamica

Answer 1

Victor's answer works but does more than is necessary. You can just directly call ignore() on cin to skip the commas in the input stream.

What this code does is read in an integer for the size of the input array, reserve space in a vector of ints for that number of elements, then loop up to the number of elements specified alternately reading an integer from standard input and skipping separating commas (the call to cin.ignore()). Once it has read the requested number of elements, it prints them out and exits.

#include <iostream>
#include <iterator>
#include <limits>
#include <vector>

using namespace std;

int main() {
    vector<int> vals;
    int i;
    cin >> i;
    vals.reserve(i);
    for (size_t j = 0; j != vals.capacity(); ++j) {
        cin >> i;
        vals.push_back(i);
        cin.ignore(numeric_limits<streamsize>::max(), ',');
    }
    copy(begin(vals), end(vals), ostream_iterator<int>(cout, ", "));
    cout << endl;
}

Answer 2

A priori, you should want to check that the comma is there, and declare an error if it's not. For this reason, I'd handle the first number separately:

std::vector<int> dest;
int value;
std::cin >> value;
if ( std::cin ) {
    dest.push_back( value );
    char separator;
    while ( std::cin >> separator >> value && separator == ',' ) {
        dest.push_back( value );
    }
}
if ( !std::cin.eof() ) {
    std::cerr << "format error in input" << std::endl;
}

Note that you don't have to ask for the size first. The array (std::vector) will automatically extend itself as much as needed, provided the memory is available.

Finally: in a real life example, you'd probably want to read line by line, in order to output a line number in case of a format error, and to recover from such an error and continue. This is a bit more complicated, especially if you want to be able to accept the separator before or after the newline character.

Answer 3

The code can be simplified a bit with new std::stoi function in C+11. It takes care of spaces in the input when converting and throws an exception only when a particular token has started with non-numeric character. This code will thus accept input

" 12de, 32, 34 45, 45 , 23xp,"

easily but reject

" de12, 32, 34 45, 45 , 23xp,"

One problem is still there as you can see that in first case it will display " 12, 32, 34, 45, 23, " at the end where it has truncated "34 45" to 34. A special case may be added to handle this as error or ignore white space in the middle of token.

wchar_t in;
std::wstring seq;
std::vector<int> input;
std::wcout << L"Enter values : ";

while (std::wcin >> std::noskipws >> in)
{
    if (L'\n' == in || (L',' == in))
    {
        if (!seq.empty()){
            try{
                input.push_back(std::stoi(seq));
            }catch (std::exception e){ 
                std::wcout << L"Bad input" << std::endl;
            }
            seq.clear();
        }
        if (L'\n' == in) break;
        else continue;
    }
    seq.push_back(in);
}

std::wcout << L"Values entered : ";
std::copy(begin(input), end(input), std::ostream_iterator<int, wchar_t>(std::wcout, L", "));
std::cout << std::endl;

Answer 4

You can use scanf instead of cin and put comma beside data type symbol

#include<bits/stdc++.h>
using namespace std;
int main()
 {
    int a[10],sum=0;
    cout<<"enter five numbers";
    for(int i=0;i<3;i++){
    scanf("%d,",&a[i]);
    sum=sum+a[i];
    }
    cout<<sum;
}

Answer 5

First, take the input as a string, then parse the string and store it in a vector, you will get your integers.

vector<int> v;
string str;
cin >> str;
stringstream ss(str);
for(int i;ss>>i;){
    v.push_back(i);
    if(ss.peek() == ','){
        ss.ignore();
    }
}
for(auto &i:v){
    cout << i << " ";
}

Answer 6

#include <iostream>
using namespace std;

int main() {
    int x,i=0;
    char y;               //to store commas
    int arr[50];
    while(!cin.eof()){
        cin>>x>>y;
        arr[i]=x;
        i++;
    }

    for(int j=0;j<i;j++)
        cout<<arr[j];     //array contains only the integer part
    return 0;
}