CODEWITHSUNDEEP
c++arraysconstructor
makeNicePlusOne
makeNicePlusOne

Reputation: 33

Array initialization by constructor

I have:

class first{
   private:
   int *array;

   public:
   first(int x){
     array = new int[x][10];
   }

I want to call this class by:

first class1 = new first(10);

Why it doesn't work ? How to inintialize array by size from constructor ??

Upvotes: 2

Views: 675

Answers (2)

Mysticial
Mysticial

Reputation: 471569

Just this is enough:

first class1(10);

new is for when you're allocating a pointer.

first *class1 = new first(10);

Furthermore, you have an incompatibility here:

array = new int[x][10];

array is an int*, but new int[x][10] is a 2D array. I'm not sure which one you want.

For the 1D array:

int *array;
array = new int[x];

For the 2D array:

int (*array)[10];
array = new int[x][10];

That said, you might be better off using std::vector.

Side Note: Since you have memory allocation in the constructor, you should also implement a destructor, copy-constructor, and copy-assignment operator.

Upvotes: 4

Mark B
Mark B

Reputation: 96301

You've indicate that you want a one-dimensional array (int*) but attempted to allocate a two-dimensional array (new [x][10]).

I'll assume you need one dimension.

The C++ way to do this is with vector.

#include <vector>

class first{
   private:
   std::vector<int> array;

   public:
   explicit first(int x) : array(x) {
   }
};

Upvotes: 2

Related Questions