Why use SomeType[1] instead of SomeType* as last member in structs

Question

I saw in the code next statement:

SomeType someVar[1];

Later someVar is used as a pointer to SomeType. Why would one

Answer 1

1. SomeType someVar[1]; allocates memory on the stack and it gives you a block/function scoped variable. So it will be automatically destroyed when it is out of the block/function.

2. SomeType* someVar; is a pointer (to nothing meaningful yet), so it doesn't allocate any for SomeType. However if you have something like this:

SomeType* someVar = malloc(sizeof(SomeType));

of equivalent:

SomeType* someVar = new SomeType(...);

Then that is memory allocation on the heap. So it is not destroyed when out of scope and it needs to be destroyed manually by free or delete.