How can I declare an array inside a function according to the size of a parameter?

Question

Here is what I have tried:

int fun1(vector s)
{ 
    const int n = s.size();
    int arr[n]; //<----want to declare an array of length s.size()         


        

Answer 1

When you declare an array like this

int arr[n];

the compiler will allocate memory on the stack for it. In this case the C++ standard requires that n is known at compile time, i.e. it must be const.

The answer to your question is to get the memory from the heap at run time like this:

int* arr = new int[n];

In this case the memory is allocated at run time and so the value of n doesn't need to be known until run time. If you use this approach don't forget to free up the memory when you're done with:

delete [] arr;

However, as the comments suggest, using a std::vector<int> would almost certainly be a better approach. Unless you've got a good reason not to, I'd go with that.

Answer 2

For this, C++ has std::vector<int>(n), which retains most of the semantics of a traditional C array but also adds a lot of benefits (dynamic allocation being one, resizing is another, algorithm support is yet another one). Even when your underlying code requires a C array, you can still use the vector and pass an address of the first element down (they are guaranteed to be contiguous).

Typically, std::vector uses heap for underlying storage, so on one hand you are better protected from stack overflows (pun intended), on the other hand, your code now uses dynamic allocation.

Answer 3

Declaring array by int arr[N]; the size N must be determined in compile-time (except for some compiler extensions which allow you to define them in run-time too). By the way, you can make it:

std::unique_ptr<int[]> arr (new int [n]);

// ... or ...

std::vector<int> arr(n);