How to put digits of an integer in a vector in C++

Question

If a user enters an integer like 4210 for example, how can I put each digit of that integer in a vector in C++?

Solution 1

It can be done like:

std::vector<int> numbers;
int x;
std::cin >> x;
while(x>0)
{
   numbers.push_back(x%10);
   x/=10;
}

std::reverse(numbers.begin(), numbers.end());

Solution 2

The Easiest way I found is this :

std::vector<int> res;

int c;
std::cin >> c;

while(c>0)

    {
    res.insert(res.begin(),c%10);
    c/=10;
    }

Solution 3

I don't understand why people advise such round about solutions as converting back and forth to int when all you want is digit by digit... for a number expressed in decimal by the user.

To transform "4321" into std::vector<int>{4, 3, 2, 1} the easiest way would be:

std::string input;
std::cin >> input;

std::vector<int> vec;

for (char const c: input) {
    assert(c >= '0' and c <= '9' and "Non-digit character!");
    vec.push_back(c - '0');
}

Comments

[berkus]

Simplest would be to input it as a string and then just iterate that string, pushing every element into a vector. After you have each digit separately it's trivial to lexically cast it to integer for vector<int>.

[Mohamed Ahmed]

Well, I will try it. It doesn't matter - for my application - if the addition was in reverse order.

[Red Alert]

change it to a stack, or treat the vector like a stack..."reverse order" is relative to how you read it later.

[Floris Velleman]

@MohamedAhmed g++ -std=c++11 -Wall -o, should work fine.

[Mohamed Ahmed]

Those numbers represent the powers in an equation of the taps in a linear feedback shift register, I will use the numbers as addresses for bits to be xored in a bitset. So, I don't think that conversion into a string is correct if I want to do that, I just want them integers as they are. Thank you.

[dixit_chandra]

use 'numbers.emplace(numbers.begin(), x%10);' instead of 'push_back()' to avoid overhead of reversing the vector.

[Florian Winter]

Using emplace(begin()) does not improve performance, but instead makes performance worse, because each emplace at the beginning requires all previously added elements to be moved, giving the entire loop complexity O(n^2), whereas std::reverse has complexity O(1).