Converting vector<string> to char** results in segmentation fault - c++

Question

I am a absolutely begginer in C++ but i have an assignent. I have to convert a vector to char** to use it in execvp. So i wrote this test which is building ok but when i run it i get segmentation fault instead of "First". I am bad at pointer and i would like your help. Thanks
#include <iostream>
#include <vector>
#include <string>
#include <cstring>
using namespace std;
int main()
{
vector<string>* list = new vector<string>();
list->push_back("First");
list->push_back("Second");
list->push_back("Third");
char ** arr;
arr[0] = strdup(list->at(0).c_str());
cout << arr[0] << endl;
return 0;
}
EDIT 1
Ok here's the actual problem. The Command::getCommandArray() seems to work as expected but when i call it from main and try to display the arr[0] it results in segmentation fault again. Why?
// Command.cpp
char** Command::getCommandArray()
{
int i=0;
int len = args->size() + 2;
char* arr[len];
arr[0] = strdup(this->commandName->c_str());
arr[len-1] = NULL;
for (i=0; i<this->args->size(); i++)
{
arr[i+1] = strdup(this->args->at(i).c_str());
}
cout << "in: " << arr[0] << endl; // PRINTS AS EXPECTED
return arr;
}
// main.cpp
do {
char** arr = tmp->getCommandArray();
cout << arr[0] << endl; // SEGMENTATION FAULT HERE
} while((tmp = tmp->getPipeline()) != NULL);
EDIT 2 Ok i solved it. Thank you very much!
I had to change line
char* arr[len];
to
char** arr = new char*[len];

Answer 1

I see no reason to duplicate strings here, or to manually manage memory. You can just use another vector to store the char pointers, and get a pointer to that vector's memory.
int main()
{
std::vector<std::string> list;
list.push_back("First");
list.push_back("Second");
list.push_back("Third");
std::vector<char*> ptr_list;
for (std::size_t i = 0; i != list.size(); ++i) {
ptr_list.push_back(&list.at(i)[0]);
}
char** arr = &ptr_list[0];
std::cout << arr[0] << std::endl;
}

Answer 2

It should be something like:
int main()
{
std::vector<std::string> list;
list.push_back("First");
list.push_back("Second");
list.push_back("Third");
char** arr = new char*[list.size()];
for (std::size_t i = 0; i != list.size(); ++i) {
arr[i] = strdup(list.at(i).c_str());
}
std::cout << arr[0] << std::endl;
for (std::size_t i = 0; i != list.size(); ++i) {
free(arr[i]);
}
delete[] arr;
}

Answer 3

Note that the argv and envp arrays you pass in to execve() need to be null terminated (in addition to the strings in those arrays):
char ** convert(std::vector<std::string> v)
{
char ** t = new char* [v.size() + 1];
// Need this for execve
t[v.size()] = nullptr;
for (int i = 0 ; i < v.size(); ++i)
{
t[i] = strdup(v[i].c_str());
}
return t;
}