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

Because generically speaking they are not the same. The first one defines one element array of SomeType and the other defines pointer to SomeType.

The first allocates memory for that one element, the other does not.

In general: sizeof(SOmeType[1]) != sizeof(SomeType*).