declaring an dynamic size array

Question

If I want to declare a dynamic size array in the main function, I can do:-

 int m;
 cin>>m;
 int *arr= new int[m];

The following cannot be done as while compiling the compiler has to know the size of the every symbol except if it is an external symbol:-

 int m;
 cin>>m;
 int arr[m];

My questions are:

1. Why does the compiler have to know the size of arr in the above code? It is a local symbol which is not defined in the symbol table. At runtime, the stack takes care of it(same way as m). Is it because the compiler has to ascertain the size of main() (a global symbol) which is equal to the size of all objects defined in it?

2. If I have a function:

   int func(int m)
   

   Could I define int arr[m] inside the function or still I would have to do

   int *a= new int[m]
   

Answer 1

For instance :

int MyArray[5]; // correct

or

const int ARRAY_SIZE = 6;
int MyArray[ARRAY_SIZE]; // correct

but

int ArraySize = 5;
int MyArray[ArraySize]; // incorrect

Here is also what is explained in The C++ Programming Language, by Bjarne Stroustrup :

The number of elements of the array, the array bound, must be a constant expression (§C.5). If you need variable bounds, use a vector (§3.7.1, §16.3). For example:

Answer 2

To answer your questions:

1) Q: Why does the compiler have to know the size of arr in the above code?

A: If you generate assembly output, you'll notice a "subtract" of some fixed value to allocate your array on the stack

2) Q: Could I define int arr[m] i ... inside the function?

A: Sure you could. And it will become invalid the moment you exit the function ;)

Basically, you don't want an "array". A C++ "vector" would be a good alternative:

std::vector<A> v(5, A(2));

Here are a couple of links you might enjoy:

• http://www.parashift.com/c++-faq/arrays-are-evil.html

• http://blogs.msdn.com/b/ericlippert/archive/2008/09/22/arrays-considered-somewhat-harmful.aspx