if array size is taken as user input, then when is memory allocated for it?

Question

if we take size of array as user input as it is now allowed in new versions of C++ and C, then when will the memory allocated of this array compile time OR run time

#include <stdio.h>
int main(void)
{
  int size, i;
  printf("Enter the size of the arrays:\n");
  scanf("%d", &size);
  int arr1[size];
}

*the memory allocated of this array compile time* - how do you imagine it is possible? — 273K, Jun 04 '22 at 18:42
"As it is now allowed in new versions of C++" This is [not true](https://stackoverflow.com/questions/1887097/why-arent-variable-length-arrays-part-of-the-c-standard). It is allowed as a GCC _extension_ to C++ that is enabled by default, it is not allowed by C++ per se. — Nathan Pierson, Jun 04 '22 at 18:44
In contrast to ISO C, ISO C++ does not support [variable-length arrays](https://en.wikipedia.org/wiki/Variable-length_array). Even in ISO C, they are only an optional feature since C11. A compliant compiler is not required to support them. — Andreas Wenzel, Jun 04 '22 at 18:44
How can the memory be allocated at compile time if the length is user-input? — Roi, Jun 04 '22 at 19:17

score 1 · Answer 1 · answered Jun 04 '22 at 18:55

When the size is unknown at compile time, it is impossible to allocate memory for the unknown size at compile time.

If you allocate memory for a size determined during program execution, whether that is by some method specified by the language standard or by some other method, the memory is allocated during program execution.

if array size is taken as user input, then when is memory allocated for it?

1 Answers1