pushing strings into multidimensional vector c++ - c++

Question

I am currently trying to separate a string by the number 2 and pushing the sub characters that i get into a 2d vector, the problem is that every time I try, I get a segmentation fault when I try to push the second row of characters in the vector. After trying a few things, I think the problem lies in the vector "some" after I clear the content of the vector. It seems to me that after the clearing, I am no longer able to push values into the vector. I hope that somebody has any suggestions because I am stuck.
std::string str = "11121112111";
std::vector<int> some;
std::vector<std::vector<int> > somemore;
for (unsigned int i = 0; i < str.length(); i++)
{
if (str[i] == '2')
{
somemore.push_back(some);
some.clear();
}
else
{
some.push_back(1);
}
}
for (unsigned int i = 0; i <= 3; i++)
{
for (unsigned int j = 0; j <= 3; j++)
{
std::cout << somemore[i][j] << std::endl;
}
}

Answer 1

With c++11 and Boost you can make a much more elegant solution, without the need for loops with an incrementing index.
#include <vector>
#include <string>
#include <iostream>
#include <boost/algorithm/string/split.hpp>
#include <boost/algorithm/string/classification.hpp>
int main()
{
std::string str = "11121112111";
std::vector<std::string> string_vector;
boost::split(string_vector, str, boost::is_any_of("2"));
std::vector<std::vector<int>> int_vector;
for (auto& s : string_vector)
int_vector.push_back(std::vector<int>(s.size(), 1));
for (auto& v : int_vector)
for (auto& i : v)
std::cout << i << std::endl;
return 0;
}

Answer 2

I would change the last part:
for(unsigned int i = 0; i <=3; i++)
{
for(unsigned int j = 0; j <=3; j++)
{
std::cout << somemore[i][j] << std::endl;
}
}
Into this:
for(unsigned int i = 0; i < somemore.size(); i++)
{
for(unsigned int j = 0; j < some.size(); j++)
{
std::cout << somemore[i][j] << std::endl;
}
}
It is much safer.

Answer 3

As I already mentioned in comments, you have two problems in your code:
You are not pushing the last some into somemore because there is no 2 at the end of str.
Your last loops are too large - You have a 3x3 matrix but you expect a 4x4 since you go from 0 to 3.
By the way, since you are only counting ones, you don't need some:
std::string str = "11121112111";
std::vector<std::vector<int>> somemore;
size_t count = 0;
for (size_t = 0; i < str.length(); i++) {
if (str[i] == '2') {
somemore.push_back(std::vector<int>(count, 1));
count = 0;
}
else {
++count;
}
}
for (size_t i = 0; i < somemore.size(); ++i) {
for (size_t j = 0; j < somemore[i].size(); ++j) {
std::cout << somemore[i][j] << std::endl;
}
}
You could also replace the last two loops with iterators, or if you have c++11 available:
for (const auto &s: somemore) {
for (const auto &v: s) {
std::cout << v << std::endl;
}
}

Answer 4

In this loop
for(unsigned int i = 0; i < str.length(); i++)
{
if(str[i] == '2')
{
sommore.push_back(som);
som.clear();
}
else
{
som.push_back(1);
}
}
where it is not clear whether som is the vector declared like
std::vector<int> some;
^^^^^
the last part of the string
std::string str = "11121112111";
^^^
is ignored by vector sommore. So the vector contains only two elements that corresponds to two 2(s) in the string.
As result these loops
for(unsigned int i = 0; i <=3; i++)
{
for(unsigned int j = 0; j <=3; j++)
{
std::cout << sommore[i][j] << std::endl;
}
}
that use the magic number 3 have undefined behaviour.
Even if the vector and its sub-vectors contain 3 elements even in this case conditions i <=3 and j <=3 are wrong.
You can adopt the following approach shown in this demonstratrive program
#include <iostream>
#include <string>
#include <vector>
int main()
{
std::vector<std::vector<int>> v;
std::string str = "11121112111";
for ( std::string::size_type pos = 0; ( pos = str.find_first_not_of( "2", pos ) ) != std::string::npos; )
{
auto n = str.find( "2", pos );
if ( n == std::string::npos ) n = str.size();
v.push_back( std::vector<int>( n - pos, 1 ) );
pos = n;
}
for ( const auto &row : v )
{
for ( int x : row ) std::cout << x;
std::cout << std::endl;
}
}
The program output is
111
111
111