How to have a node structure that holds an array in C++?

Question

I can't find anything online about making a linked list with individual nodes that hold arrays.

struct node{
    string list[]; 
    string var;
    node* next = NULL; 
    node* previous = NULL; 

}; 

void insertion(node*& start, string name, string words[]){
  node* temp = new node;
  temp->var = name;
  temp->list = words;

  if (!start) { 
    start = temp;
    return;
  } else { 
    node* tail = start;
    while(tail->next) {
      tail=tail->next;
    }
    tail->next = temp;
    temp->previous = tail;
 }

The code above gives me the error:

web.cpp: In function ‘void insertion(variable*&, std::__cxx11::string, std::__cxx11::string*)’: web.cpp:18:16: error: incompatible types in assignment of ‘std::__cxx11::string* {aka std::__cxx11::basic_string*}’ to ‘std::__cxx11::string [0] {aka std::__cxx11::basic_string [0]}’ temp->list = words;

Use a `std::vector` instead of a raw array. Arrays need to have a definite size and cannot be assigned directly. — user0042, Nov 21 '17 at 21:26
@user0042 Raw arrays are more efficient and can be just as easy to use as vectors. — Cpp plus 1, Nov 21 '17 at 21:27
@Cppplus1 _"Raw arrays are more efficient"_ Can you elaborate please how so? At least for fixed size array a `std::array` should be used. — user0042, Nov 21 '17 at 21:28
@user0042 Whether or not Vectors are as efficient as arrays depends much on the compiler. — Cpp plus 1, Nov 21 '17 at 21:29
Maybe if you are using turbo c++. Any modern compiler the efficiency will be the same. — drescherjm, Nov 21 '17 at 21:30
@Cppplus1 _"depends much on the compiler"_ Sorry to tell you, but that's just nonsense. — user0042, Nov 21 '17 at 21:31
Would it be possible to use another linked list instead of an array? And if so how would that be possible? — Zachary Collins, Nov 21 '17 at 21:31
@user0042 They can change in size. Whenever you change its size, a new array is created, which is inefficient. It is much better to create an array with the correct size to begin with. — Cpp plus 1, Nov 21 '17 at 21:32
***Whenever you change its size, a new array is created*** That is not how vectors are implemented. — drescherjm, Nov 21 '17 at 21:32
They usually have grow algorithms. And only grow / shrink when a threshold is met. — drescherjm, Nov 21 '17 at 21:35
@drescherjm But when the threshold is met, a new set of memory needs to be allocated. — Cpp plus 1, Nov 21 '17 at 21:35
Yes however that is at least as good as you would do it yourself with new / delete. — drescherjm, Nov 21 '17 at 21:36
@drescherjm Only if you absolutely needed to change the size. Without using Vectors, you might find that you can initialize an array with a constant, large size. — Cpp plus 1, Nov 21 '17 at 21:37
@Cppplus1 Here's a [recommended read](https://stackoverflow.com/questions/46991224/are-there-any-valid-use-cases-to-use-new-and-delete-with-modern-c) if you are seriously interested to dive deeper into that topic. — user0042, Nov 21 '17 at 21:41

score 1 · Answer 1 · answered Nov 21 '17 at 21:40

First of all, string list[] is not allowed as a data member definition, since it is an incomplete type. So your code actually should give you an error already in the definition of struct node.

In addition to that issue, in C++, array types are not assignable (regardless of their size). So even if you wrote:

struct node{
    string list[4];
    string var;
    node* next = NULL;
    node* previous = NULL;

};

int main(){

    node n1;
    node n2;

    n1.list = n2.list;  // error: array types are not assignable
}

you'd get an error indicating that array types cannot be assigned in the sense of array1 = array2.

If you used a vector<string>, in contrast, you could assign as well as letting your vector dynamically grow as needed.

How to have a node structure that holds an array in C++?

1 Answers1