How to convert numbers into vector of integers? - c++

Question

I want my program to ask the user to input any number and then store it into a std::vector where every single digit is allocated to a separate vector index:
input: 142
output vector: [1, 4, 2]
I tried this:
int main()
{
std::vector<int> v;
int number;
cin >> number;
for(unsigned int i = 100; i > 0; i/=10)
{
v.push_back(number/i);
number -= (number/i)*i;
}
for(size_t i = 0; i < v.size(); ++i)
{
std::cout<<v[i]<<std::endl;
}
}
It works. But what should I do when the input-length is unknown?

Answer 1

Use simply std::string and for each char(which is actually integers) of string convert to integer as follows: SEE LIVE HERE
#include <iostream>
#include <vector>
#include <string>
int main()
{
std::vector<int> v;
std::string number = "123456789987654321";
for(auto& Integer: number)
v.emplace_back(static_cast<int>(Integer - '0'));
for(const auto& it: v) std::cout << it << " ";
}
Output:
1 2 3 4 5 6 7 8 9 9 8 7 6 5 4 3 2 1
In case of entering unwanted characters/ user inputs(for example some negative numbers or even this: +-1234567), you can go for something with try-catch. try to convert the char to int, otherwise skip in the catch block as follows. SEE LIVE HERE
#include <iostream>
#include <vector>
#include <string>
#include <exception>
int main()
{
std::vector<int> v;
std::string number = "+-1234567";
for(auto& Integer: number)
{
std::string Char(1, Integer); // convert to string
try { v.emplace_back(std::stoi(Char)); }
catch(...) { continue; } // any case of exceptions
/* or using much simpler std::isdigit from <cctype>
by which conversion to std::string and try-catch can be avoided.
if(std::isdigit(Integer))
v.emplace_back(static_cast<int>(Integer - '0'));
*/
}
for(const auto& it: v) std::cout << it << " ";
}
Output:
1 2 3 4 5 6 7
Edte: As per #Aconcagua suggested, included the solution with std::isdigit

Answer 2

while(number) {
v.push_back(number%10);
number /= 10;
}
std::reverse(v.begin(), v.end());

Answer 3

Change the for initilization unsigned int i = number rather than unsigned int i = 100. The re-written for statement will be:
for(unsigned int i = number; i > 0; i/=10)

Answer 4

I you want to stay with digits and std::deque is an option you could do the following:
int main()
{
std::deque<int> v;
int number;
cin >> number;
while(number != 0)
{
v.push_front(number%10);
number = number/10;
}
for(size_t i = 0; i < v.size(); ++i)
{
std::cout<<v[i]<<std::endl;
}
return 0;
}
With the modulo operator you simply take the last digit and insert it in the front of the deque. Afterwards you "cut off" that digit with /10.
I used a deque, because you can't push front with vectors.