How do I extract all the numbers in a string where the numbers are interspersed with letters?

How do I loop through a string consisting of numbers and letters and add only numbers to the vector?

For example if the input is:

> e385p336J434Y26C2Z6X5Z2

I want to get a vector of int like this:

> number = {385, 336, 434, 26, 2, 6, 5, 2}

The best I got was to iterate over the line and add all the digits like that:

#include <bits/stdc++.h>


int main(){
    string f = "e385p336J434Y26C2Z6X5Z2";
    vector<int> f_numb; 
    string sum; 
    for (int i = 0; i < f.size(); ++i){
         if (('0' <= f[i]) && (f[i] <= '9')){
            sum += (f[i]);
            
             
         }
    }
    //std::cout << sum << std::endl; 
    vector<int> m_numb; 
    for (int i = 0; i < sum.size(); ++i){
        m_numb.push_back(sum[i] - '0'); 
    }
    int sm; 
    for (int i = 0; i < m_numb.size(); ++i){
        sm += m_numb[i]; 
        std::cout << m_numb[i] << " "; 
    }
    std::cout << std::endl; 
    
}

Foregoing the unstated reason you're not using std::isdigit, you can/should simply build each number as you process its digits. Note that the code below does NOT check, nor care, about unsigned overflow, and makes no attempt at processing negative numbers.

#include <iostream>
#include <string>
#include <vector>

int main()
{
    std::string f = "e385p336J434Y26C2Z6X5Z2";
    std::vector<unsigned int> f_numb;

    for (auto it = f.begin(); it != f.end();)
    {
        if ('0' <= *it && *it <= '9')
        {
            unsigned int sm = 0;
            for (;it != f.end() && '0' <= *it && *it <= '9'; ++it)
                sm = (sm * 10) + (*it - '0');

            f_numb.emplace_back(sm);
        }
        else
        {
            ++it;
        }
    }

    for (auto x : f_numb)
        std::cout << x << ' ';
    std::cout << '\n';
}

Output

385 336 434 26 2 6 5 2 

That, assuming I understand your question vs. your code, which differ highly in their apparent goals.