Question

Using c:

  char ptr[n];
  free(ptr);

In my opinion: when “char ptr[n];” is used, the memory is allocated, and ptr is pointed to it, free(ptr) should work.
And the program failed, why?(n == 5 e.g.)
Any deep analysis?

Answer 1

Because you called free on a variable not allocated with malloc.
This causes Undefined Behavior. Luckily for you it crashes and you can detect it, else it can crash at most awkward times.

You call free for deallocating memory of heap allocated variables, What you have is an array on local storage(assuming it to be in a function) and it automatically deallocates when the scope({,}) in which it was created ends.