Deletion on a non pointer array in c++

user_18806627 0
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When I have an array like this:

int* test = new int[50];
for (int i = 0; i < 50; i++)
{
    test[i] = dist4(rng);
}

(Filled with random numbers for testing)

I can free the memory like this:

delete[] test;

But when I declare the array like this:

int test[50];
for (int i = 0; i < 50; i++)
{
    test[i] = dist4(rng);
}

I can't free the memory with delete or delete[].

What's the proper way of freeing the memory here?

The "dist4" function is just a random number generator:

random_device dev;
mt19937 rng(dev());
uniform_int_distribution<mt19937::result_type> dist4(1,4); // distribution in range [1, 4]
delete-operator arrays c++
3 Answers

What's the proper way of freeing the memory here?

No need to free memory explicitly using delete or delete[] in the latter case.

Assuming int test[50]; is declared inside a function, it has automatic storage duration and when test goes out of scope it will be automatically destroyed.

user_120025700

The memory will automatically be freed when it does out of scope like any other non-dynamically allocated variable. This is due to automatic storage duration

user_123057010

Write the code like this using a local scope:

void a_function ()
{

  // some code

  {
    int test[50];
    for (int i = 0; i < 50; i++)
    {
      test[i] = dist4(rng);
    }
  }
  // here the space for test will recovered.

  // some more code...

}
user_33701240
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