Is it possible to make an array of struct node type?

Question

I was just wondering. Say you will need to store these values:

Hours Minutes

inside an array, is the following implementation logically possible?

struct node
{
int hour;
int minutes;
};


int main()
{
int numOfLanding, minGap, hour, minutes;
cin>>numOfLanding;
cin>>minGap;
cout<<endl;

struct node *arr[numOfLanding];

for (int i=0; i<numOfLanding; i++)
{
    cin>>hour;
    cin>>minutes;
    arr[i]->hour=hour;
    arr[i]->minutes=minutes;
}

I am still trying very hard to understand struct node logic. Any help would be much appreciated!

It looks like you may be following a C guide or tutorial. The C++ way would be to use std::vector<node> arr;. — François Andrieux
– François Andrieux, Commented Apr 30, 2018 at 17:14
You are declaring an array of pointers, not an array of nodes, which is one problem. Another is that arrays are expected to have a constant size, which doesn't work when it is received from user input. Skip a few chapters ahead in your C++ tutorial to get to the container classes, they can be used to create variable-sized "arrays". — Ulrich Eckhardt
– Ulrich Eckhardt, Commented Apr 30, 2018 at 17:14

Fabio says Reinstate Monica · Accepted Answer · 2018-04-30 17:31:41Z

3

What you are asking is: is it possible to make an array of a size that is not known at compile time, but only at runtime? Yes, you can, but the way you have chosen is not standard C++: declaring an array like

struct node *arr[numOfLanding];

means using a "variable-length array", which is not (and has never been) part of the C++ standard. It was part of C, however, in C99, but then the committee decided to make it optional in C11. It is anyway possible to find a C++ compiler that supports this feature as an extension: gcc, for example. But if you use it, keep in mind that your code is not portable.

The standard way of doing it in C++ is to use new[]:

node* arr = new node[numOfLanding];

(note that using the keyword struct every time is what you would do in C; in C++ it is not required)

At this point, you access each element using the ., not the ->:

arr[i].hour=hour;
arr[i].minutes=minutes;

After you are finished using the array you have to delete it, by using:

delete[] arr;

Anyway, this style is old, and nowadays considered bad. The preferred approach is to use a container that automatically deals with the size for you, and that will manage memory so that you don't need to worry about new[] and delete[]. The best container for this is the std::vector. To use it, first you have to #include <vector>, and then you can use it like this:

std::vector<node> arr(numOfLanding);

for (auto& curr_node : arr) {
    cin>>hour;
    cin>>minutes;
    curr_node.hour=hour;
    curr_node.minutes=minutes;
}

answered Apr 30, 2018 at 17:31

Fabio says Reinstate Monica

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3 Comments

Nicholas Chen Over a year ago

Wow I didn't expect such an elaborate answer! Thank you for enlightening me with current practices :) And the vector implementation, just blew my mind. What does the declaration vector<node> mean? I've never seen a for loop being used in that way, would you mind explaining what it means?

R Sahu Over a year ago

The for loop used is called range-for loop. It is supported in C++11 and higher. Take a look at en.cppreference.com/w/cpp/language/range-for.

Fabio says Reinstate Monica Over a year ago

@NicholasChen I suppose you haven't seen templates yet. Well, in short, you can have a vector of int (and then you write vector<int>), of double (vector<double>), and so on. For node, you use vector<node>. You have to specify the type of data stored in the vector, and you do that by writing it between angle brackets. The for loop, as R Sahu told you, is called range-for and it should be easier to manage than the classic for loop, since you don't have to worry about the index and counting the elements.

Thomas Matthews · Accepted Answer · 2018-04-30 17:18:08Z

1

Yes, you can have an array of nodes.

However, if you insist on an array, and you don't know the capacity at compile time, you'll have to allocate it at run-time:

struct node
{
  int hour;
  int minutes;
};


int main()
{
  int numOfLanding, minGap, hour, minutes;
  cin>>numOfLanding;
  cin>>minGap;
  cout<<endl;

  node *arr = new node[numOfLanding];

  for (int i=0; i<numOfLanding; i++)
  {
      cin>>hour;
      cin>>minutes;
      arr[i].hour=hour;
      arr[i].minutes=minutes;
  }

  // Remember to delete the array.
  delete[] arr;
  return EXIT_SUCCESS;
}

A safer alternative is to use std::vector<node>.

Note: Since it is an array, use the '.' for access, not ->.

answered Apr 30, 2018 at 17:18

Thomas Matthews

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4 Comments

Vamsidhar Reddy Gaddam Over a year ago

This is a better answer.

Nicholas Chen Over a year ago

Thank you for your answer! So to conclude, for array of nodes, to access it, we have to use '.' instead of '->' for it to work? Why is using vectors the safer alternative as well? I also noticed you put an EXIT_SUCCESS at the end, does it have the same value as return 0?

Thomas Matthews Over a year ago

A std::vector will dynamically grow as you append items. An array is fixed in size.

Thomas Matthews Over a year ago

The EXIT_SUCCESS is one of two standard values returned by main. The other is EXIT_FAILURE. You can look them up to see their values.

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