How to read in user entered comma separated integers?
c++, cin
Solution
All of the existing answers are excellent, but all are specific to your particular task. Ergo, I wrote a general touch of code that allows input of comma separated values in a standard way:
template<class T, char sep=','>
struct comma_sep { //type used for temporary input
T t; //where data is temporarily read to
operator const T&() const {return t;} //acts like an int in most cases
};
template<class T, char sep>
std::istream& operator>>(std::istream& in, comma_sep<T,sep>& t)
{
if (!(in >> t.t)) //if we failed to read the int
return in; //return failure state
if (in.peek()==sep) //if next character is a comma
in.ignore(); //extract it from the stream and we're done
else //if the next character is anything else
in.clear(); //clear the EOF state, read was successful
return in; //return
}
Sample usage http://coliru.stacked-crooked.com/a/a345232cd5381bd2:
typedef std::istream_iterator<comma_sep<int>> istrit; //iterators from the stream
std::vector<int> vec{istrit(in), istrit()}; //construct the vector from two iterators
Since you're a beginner, this code might be too much for you now, but I figured I'd post this for completeness.
Problem
I'm writing a program that prompts the user for: - Size of array - Values to be put into the array First part is fine, I create a dynamically allocated array (required) and make it the size the user wants. I'm stuck on the next part. The user is expected to enter in a series of ints separated by commas such as: 1,2,3,4,5 How do I take in those ints and put them into my dynamically allocated array? I read that by default cin takes in integers separated by whitespace, can I change this to commas? Please explain in the simplest manner possible, I am a beginner to programming (sorry!) EDIT: TY so much for all the answers. Problem is we haven't covered vectors...is there a method only using the dynamically allocated array I have? so far my function looks like this. I made a default array in main. I plan to pass it to this function, make the new array, fill it, and update the pointer to point to the new array. ``` int *fill (int *&array, int *limit) { cout << "What is the desired array size?: "; while ( !(cin >> *limit) || *limit < 0 ) { cout << " Invalid entry. Please enter a positive integer: "; cin.clear(); cin.ignore (1000, 10); } int *newarr; newarr = new int[*limit] //I'm stuck here } ```