Initialise the array length via a function

Question

Dear forum,
II need a number of integers as an array, but I don't know how many in advance. An I need this array only in the scope, then do waya with it. So I used
#include <iostream>
using namespace::std;
const int f ();
int main () {
int array[f ()];
return 0;
const int f () {
return 42;
. It worked fine under gcc, but CC compains with
Error: An integer constant expression is required within th\
e array subscript operator.
Is there a switch which makes CC swallow this program?
Cheers, Peter.

Answer 1

You are posting this question in the C forum, but
your question is about the C++ compiler.Hello Clamage,
this I realised 1e-6 s after clicking submit. It shall not happen again. Anyway, thanks for the quick response!
>
In standard C++, the dimension of an array allocated
statically or in a block must be a constant integer
expression. Reference: C++ Standard, section 8.3.4.
If g++ allows a non-constant value, that would be
non-standard extension.... which are getting at you after a while...
>
You can get the effect by allocating the array on the
heap:
int* array = new int[f()];Indeed. But what made me make the array static is that with new[], you also have to delete[] (which one always forgets).
>
But in C++ you have alternatives to C-style arrays.
You can use the standard vector class template. It
manages memory for you, allows a vector to be
expanded, optionally checks subscripts for being in
range. When you create a vector, you can specify the
starting capacity when you know the sizeI should try this.
Once again, thanks for your patience!
Cheers, Peter.