Variable size of an array

[GeorgeGVD]

My first thought was

int size = 0;
 cout << "Enter size of the array" << endl;
 cin >> size;
 int a[size];

Now I know that an array has to have a constant value, but I need that the user enters an integer and use that integer as the size of the array. Any idea how?

[gerard4143]

You could create a dynamic array with the operator new.

int * a = new int[size];

[Narue]

Ideally you would use the std::vector class instead of trying to manage dynamic memory and pointers.

[sergent]

G++ and all C compilers support it. But it is not C++ standard so use what gerard4143 said.

int size;
cin >> size;
int *array = new int[size];

You can now treat it almost like a real array with indexing.

[gerard4143]

Ideally you would use the std::vector class instead of trying to manage dynamic memory and pointers.

Yeah, the standards committee spent all that time creating the standard temple libraries maybe we should consider the vector.