How can I assign an int array to linkedlist struct? - c++

Question

Hi I'm a beginner at c++ and I'm having trouble solving this issue. I have a linkedlist of nodes and the node contains data of int array and a pointer pointing to the next node.
struct node {
unsigned int numbers[6];
node* next;
};
I also have a class:
private:
ticket* ticketListHead;
ticket* ticketListTail;
and in a public method:
public:
void newNode() {
int arr[6];
for(int i = 0; i < 6; ++i) {
arr[i] = ( std::rand() % 49 ) + 1;
}
node *temp = new node;
temp->numbers=arr;
temp->next=NULL;
}
The problem I believe is with the temp->numbers=arr line as I believe arrays cannot be assigned like that in C++. I'm not sure how to solve the problem in this case and I've tried looking online. Some help would be appreciated!

Answer 1

You're correct about your suspicion that you cannot assign arrays. You could instead use a wrapper type that allows assignment and define
using array = std::array<unsigned,6>; // convenient short hand
struct node {
array numbers;
node* next = nullptr; // ensure next defaults to null
};
when your newNode() method may look like
node* newNode() {
array arr;
for(auto&x: arr)
x = ( std::rand() % 49 ) + 1; // std::rand() cannot be recommended
auto temp = new node;
temp->numbers = arr;
return temp;
}
but you can avoid the temporary object arr altogether by directly writing into the new node's data:
node* newNode() {
auto temp = new node;
for(auto&x: temp->numbers)
x = ( std::rand() % 49 ) + 1; // std::rand() cannot be recommended
return temp;
}
Btw, you shouldn't use std::rand() (see this post and this presentation for reasons why). Instead, use the methods provided by <random>, when your code becomes (see also this answer)
template<typename Generator>
node* newNode(Generator&rng) {
std::uniform_int_distribution<unsigned> uni(1,49);
auto temp = new node;
for(auto&x: temp->numbers)
x = uni(rng); // guaranteed unbiased in [1,49] inclusive
return temp;
}

Answer 2

Welcome to C++, it's awesome that you're taking your first steps into a brighter world! What you'll want to look into are C++ container classes, they'll get you out of the business of managing memory on your own. The closest to your code is std::list and std::vector. I'll neglect random number seeding, that's a complex topic better discussed elsewhere.
#include <vector>
#include <list>
#include <iostream>
#include <random>
#include <limits>
using namespace std;
struct node {
node() : num(6) {
numbers.reserve(num);
for (int i = 0; i < num; ++i)
numbers.push_back(random() % numeric_limits<int>::max());
}
vector<int> numbers;
const int num;
};
int main(int argc, char** argv) {
list<node> nodes;
nodes.push_back(node()); // this replaces your newNode function
nodes.push_back(node()); // another call to better illustrate
for (auto mynode : nodes) {
for (auto mynum : mynode.numbers)
cout << mynum << " ";
cout << endl;
}
}

Answer 3

// Change node struct to this
struct node{
unsigned int *numbers;
node* next;
};
/*
Remember that normal arrays are
implemented through pointers. So you
can use a pointer named numbers in
your node struct and it would behave
as an array.
*/
// Change newNode() to this
void newNode() {
int * arr = new int[6]; //Dynamic memory allocation
for(int i = 0; i < 6; ++i)
{
arr[i] = ( std::rand() % 49 ) + 1;
}
node *temp = new node;
temp->numbers = arr;
temp->next = NULL;
}
U got it ?