Calling array with variable size as parameter

Question

I was trying to make a program in which the user decides the dimensions of a 2-D array. I'm getting an error on the function definition while compiling. Why is this wrong and what would be the correct way to do it?

I'm using the Dev-C++ 5.7.1 compiler (if that's relevant).

#include<iostream>

using namespace std;

int R=0,C=0;                 

void func(int);

int main() {
    cin>>R>>C;
    int array[C][R];
                      // DO STUFF HERE
    func(array);
                      // DO SOME MORE STUFF
    return 0;
}

void func(int arr[][R]) {  
                     // DO STUFF HERE
}

Answer 1

ISO-C++ forbids VLAs. To dynamically allocate an array you'll either need to do some raw pointer tricks or use a vector of vectors.

vector of vectors approach:

std::cin >> R >> C;
std::vector<std::vector<int>> array(R, std::vector<int>(C));

The signature for func then becomes (const correctness may be different)

void func(const std::vector<std::vector<int>>& v);

The above is the easier, more maintainable, safer, shorter solution.

---

With pointers and pointers to pointers you can do it but it becomes more complicated, and you need to delete everything that you new

int R, C;
std::cin >> R >> C;
int **array = new int*[R]; // allocates space for R row pointers
for (int i = 0; i < R; ++i) {
    array[i] = new int[C]; // for each row, allocate C columns
}

func(R, C, array);

//then delete everything
for (int i = 0; i < R; ++i) {
    delete [] array[i]; // delete all of the ints themselves
}
delete [] array; // delete the row pointers.

with the signature for func being

void func(int r, int c, int **arr);

again, vector of vectors will be a lot easier on you.