C++ static array like dynamic array

Question

It seems to be remembered that static array in C++ can initialize only from const expression, but if you write:

#include <iostream>

int main() {
  int n;
  std::cin >> n;
  int a[n];
  std::cout << sizeof(a);

  return 0;
}

this program successfully compiles (gcc C++17) and print n * sizeof(int). But why is this so?

Gcc (unfortunately) allows VLAs by default as an extension. Compile with `-std=c++17` and/or `-Wvla -Werror` and it won't. — Jesper Juhl, Dec 25 '17 at 12:38
But how is VLAs work? Is there any reservation of the stack? — Mano, Dec 25 '17 at 12:40
Compiler can do the trick. For example, it would allocate memory space in heap, and delete it at every exit point (e.g. all return statements in the function). — Stephen, Dec 25 '17 at 12:53

Ron · Accepted Answer · 2017-12-28T22:40:34.387

Variable-length arrays are not part of the standard. They can be present as compiler extensions which is the case with GCC. When compiling you are likely to receive a warning of:

warning: ISO C++ forbids variable length array 'a' [-Wvla]

When applied to arrays, the sizeof operator returns the size of the entire array which is the size of the underlying type times the number of elements. The reference states, emphasis mine:

The size of each VLA instance does not change during its lifetime, but on another pass over the same code, it may be allocated with a different size.

The official GCC documentation titled 6.19 Arrays of Variable Length states:

These arrays are declared like any other automatic arrays, but with a length that is not a constant expression. The storage is allocated at the point of declaration and deallocated when the block scope containing the declaration exits.

That being said prefer std::vector or std::array to raw (C style) arrays.

C++ static array like dynamic array

1 Answers1