why can't I store values in my 2D vector by push back?

Question

I got stuck in many problems where I was trying to store values in 2D vectors. So I have written this simple code.

I am just storing and printing my values :

int main()
{
    vector<vector<int>> vec;
    vector<int> row{1,3,5,7,9,12,34,56};
    int i,n,m,rs,vs;
    rs=row.size();
    cout<<"rs = "<<rs<<endl;
    for(i=0;i<(rs/2);i++)
    {
        vec[i].push_back(row.at(i));
        vec[i].push_back(row.at(i+4));
    }
    vs=vec.size();
    cout<<vs<<endl;
    for(n=0;n<vs;n++)
    {
        for(m=0;m<2;m++)
        {
            cout<<vec[n][m]<<" ";

        }
        cout<<endl;
    }
    return 0;

}

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Thomas Sablik · Answer 1 · 2019-09-08T12:10:44.050

First you should read Why is “using namespace std;” considered bad practice?.

Declare variables when you use them and not at the beginning of your program.

The vector vec is empty at the beginning. In the loop

for(i=0;i<(rs/2);i++)
{
    vec[i].push_back(row.at(i));
    vec[i].push_back(row.at(i+4));
}

you are taking a reference to the i-th element in vec with

vec[i]

but this element does not exist. This is undefined behavior and can result in a segmentation fault. You can fix it by constructing the vector with the needed elements

#include <iostream>
#include <vector>

int main()
{
    std::vector<int> row{1,3,5,7,9,12,34,56};
    int rs = row.size();
    std::vector<std::vector<int>> vec(rs / 2);
    std::cout << "rs = " << rs << '\n';
    for(int i = 0; i < rs / 2; ++i)
    {
        vec[i].push_back(row.at(i));
        vec[i].push_back(row.at(i + 4));
    }
    int vs = vec.size();
    std::cout << vs << '\n';
    for(int n = 0; n < vs; ++n)
    {
        for(int m = 0; m < 2; ++m)
        {
            std::cout << vec[n][m] << " ";
        }
        std::cout << '\n';
    }
    return 0;
}

In this example the line

std::vector<std::vector<int>> vec(rs / 2);

constructs a vector containing rs / 2 default constructed elements. Alternatively you can start with an empty vector and push back elements in the loop

#include <iostream>
#include <vector>

int main()
{
    std::vector<int> row{1,3,5,7,9,12,34,56};
    int rs=row.size();
    std::vector<std::vector<int>> vec;
    std::cout << "rs = " << rs << '\n';
    for(int i = 0; i < rs / 2; ++i)
    {
        vec.push_back({row.at(i), row.at(i+4)});
        //
        // is similar to:
        // vec.push_back({});
        // vec.back().push_back(row.at(i));
        // vec.back().push_back(row.at(i+4));
        //
        // is similar to:
        // vec.push_back({});
        // vec[i].push_back(row.at(i));
        // vec[i].push_back(row.at(i+4));
    }
    int vs = vec.size();
    std::cout << vs << '\n';
    for(int n = 0; n < vs; ++n)
    {
        for(int m = 0; m < 2; ++m)
        {
            std::cout << vec[n][m] << " ";
        }
        std::cout << '\n';
    }
    return 0;
}

I recommend the first solution. It's better to allocate memory for all elements and work with it instead of allocate memory in each loop iteration.

why can't I store values in my 2D vector by push back?

1 Answers1