Access vector of vector pointers

Question

I wanted to create a matrix with vectors. In the below code, I created a vector with each entry containing a pointer to another vector(myvector) that acts as columns. I push random values to the myvector (i.e. columns). But when I try to access the values of arrays, it pops an compile error saying "error: no match for 'operator*' (operand type is 'std::vector<int>') at the cout statement. I wonder how do I access the values. I'm pretty sure this is a naive question.

#include <iostream>
#include <vector>
using namespace std;
int main ()
{
  std::vector<vector<int>*> main;
  
  for(int j=0; j<3; j++){
        vector<int> *myvector = new vector<int>;
        main.push_back(myvector);
  }
  
  main[0]->push_back(1);
  main[0]->push_back(4);
  main[1]->push_back(6);
  main[1]->push_back(7);
  main[1]->push_back(8);
  main[2]->push_back(3);
   
  for(int j=0; j<3; j++){
    for(uint32_t i=0; i<main[j]->size(); i++)   
        std::cout<<main[j][i]<<" ";
    cout<<"\n";
  }
  
  return 0;
}

`main[j]` is a pointer, so the `[i]` in `main[j][i]` iterates the pointer, not the inner vector. I think `(*main[j])[i]` is what you are after, but I already horribly misread the question once. — user4581301, Oct 24 '21 at 04:14
Side question, Why a pointer to a `vector`? This is usually a bad idea. — user4581301, Oct 24 '21 at 04:15
You could use `std::cout << main[j]->at(i)` but before you do that ask yourself why you're using pointers. There's no good reason for it in this code. — Retired Ninja, Oct 24 '21 at 04:16
don't new your vector's its totally not necessary. Vectors manage their own memory dynamically already. The only thing you're doing is just adding a level of indirection and making things harder for yourself. Also have a look here : http://isocpp.github.io/CppCoreGuidelines/CppCoreGuidelines#Rr-newdelete — Pepijn Kramer, Oct 24 '21 at 05:17

score 3 · Answer 1 · answered Oct 24 '21 at 04:22

3

You could just have done

vector<vector<int>> main;

each index of vector main represents another vector

so to push a number in the first column you can do

main[0].push_back = (_number_)

To access a number in first row in first column we can do main[0][0]

answered Oct 24 '21 at 04:22

Akki

96
1
7

While I agree with this answer in the information-vacuum, the asker might have one of those incredibly rare, legit cases for a pointer to a `vector` that their MRE doesn't express. – user4581301 Oct 24 '21 at 04:27
This is a fine suggestion but you neglected to answer the original question – Patrick Parker Oct 24 '21 at 05:12

Pepijn Kramer · Accepted Answer · 2021-10-24T05:45:29.753

This example shows both the syntax you where looking for, and also an example of how you should use std::vector without new/delete.

#include <iostream>
#include <vector>
#include <memory>

// using namespace std; <== teach yourself NOT to do this.
// https://stackoverflow.com/questions/1452721/why-is-using-namespace-std-considered-bad-practice

void access_pointers_in_2d_vector()
{
    std::vector<std::vector<int>*> values; // don't call your variables main!

    for (int j = 0; j < 3; j++)
    {
        std::vector<int>* myvector = new std::vector<int>;
        values.push_back(myvector);
    }

    values[0]->push_back(1);
    values[0]->push_back(4);
    values[1]->push_back(6);
    values[1]->push_back(7);
    values[1]->push_back(8);
    values[2]->push_back(3);

    for (int j = 0; j < 3; j++)
    {
        for (uint32_t i = 0; i < values[j]->size(); i++)
        {
            //==================================================================
            // this is the syntax you're looking for
            // first dereference the pointer then use operator[] 
            std::cout << (*values[j])[i] << " ";
            //==================================================================
        }

        std::cout << "\n";
    }

    // don't forget to cleanup your memory!
    // if you typed new somewhere then there should
    // ALWAYS be a matching delete in your code too!
    for (int j = 0; j < 3; j++)
    {
        delete values[j]; // <<==== !!!!!!!
    }
}

// for dynamic memory managment new/delete aren't recommended anymore.
// use std::unique_pointer (or if your design really requires it std::shared_ptr)
void using_unique_pointer()
{
    // If you really need pointers, use std::unique_ptr
    // it will prevent you from introducing memory leaks
    const std::uint32_t size = 3ul;
    std::vector<std::unique_ptr<std::vector<int>>> values(size);
    for (auto& p : values)
    {
        p = std::make_unique<std::vector<int>>();
    }

    values[0]->push_back(1);
    values[0]->push_back(4);
    values[1]->push_back(6);
    values[1]->push_back(7);
    values[1]->push_back(8);
    values[2]->push_back(3);

  // output loop is same as for normal pointers.
   // no need to call delete, std::unique_ptr will do that for you
}

void without_pointers()
{
    // However your whole code in idiomatic c++ should look like this.
    // https://en.cppreference.com/w/cpp/container/vector/vector constructor (10)
    // https://en.cppreference.com/w/cpp/language/range-for these loops avoid bugs related to 
    // letting indices go out of bounds.

    std::cout << "\nusing (nested) initializer list and range based for loops : \n";
    std::vector<std::vector<int>> rows{ {1,4}, {6,7,8}, {3} };
    for (const auto& row : rows)
    {
        for (const auto& value : row)
        {
            std::cout << value << " ";
        }
        std::cout << "\n";
    }
}


int main()
{
    access_pointers_in_2d_vector();
    using_unique_pointer();
    without_pointers();
    return 0;
}

Access vector of vector pointers

2 Answers2