CODEWITHSUNDEEP
c++arraysstring
Piotr Wasilewicz
Piotr Wasilewicz

Reputation: 1821

How to convert string into array of integers with determined width

I need to convert string (for instance "1234567890") to array of integers but width of element is determined by user. So:

if user passed 1 it will be: 1 2 3 4 5 6 7 8 9 0

for 2 I will have an array: 12 34 56 78 90

3: 1 234 567 890

4: 12 3456 7890

etc.

What I tried:

#include <iostream>

using namespace std;

int main()
{
    string textNumber;
    int size;
    cin >> textNumber;
    cin >> size;

    int length = textNumber.length();
    int lenghtOfArray = length / size + (length % size ? 1 : 0);
    int myArray[lenghtOfArray] = {0};
    int move = lenghtOfArray- (size * lenghtOfArray - length);
    int copySize = size;
    int k = 0;

    for(int i=0; i < length;i++) {
        if(--copySize && !move){
            myArray[k] += (int)textNumber[i]-48;
        } else {
            myArray[k] += (int)textNumber[i]-48;
            ++k;
            copySize = size;
            if(move) --move;
            continue;
        }
        myArray[k] *= 10;
    }

    myArray[k] += (int)(textNumber[0]-48);

    for(int i=0; i < lenghtOfArray; i++) {
        cout << myArray[i] << " ";
    }
}

but It doesn't work for all cases ( I̶t̶ ̶w̶o̶r̶k̶s̶ ̶o̶n̶l̶y̶ ̶f̶o̶r̶ ̶s̶i̶z̶e̶=̶2̶).

Upvotes: 0

Views: 68

Answers (1)

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726479

One approach is to use std::stoi along with substr member function of std::string to break string into pieces and parse the results:

cin >> s;
cin >> size;
int len = s.length();
int count = (len+size-1) / size;
vector<int> res(count);
int pos = count-1;
while (s.length() > size) {
    res[pos--] = stoi(s.substr(s.length()-size));
    s = s.substr(0, s.length()-size);
}
if (s.length()) {
    res[0] = stoi(s);
}

Note that you need to use std::vector<int> instead of an array, because C++ standard does not allow variable-length arrays (g++ offers it as a popular extension).

Demo.

Upvotes: 1

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