Thread: Elementary question - shifting memory adresses

When you declare an array, you are basically allocating memory automatically, which is saved on the internal memory. If the array is declared in global code or with the static storage class modifier, then the array will have static storage, on the stack. Every data type is stored in memory so respectively they have a memory address. There is one exception for data, declared on the register, but this occurs in rare cases if you want to perform hardcore hardware operations. The offset address, for little-endian system is the name identifier.

Code:

char arr [50];

arr will be the offset address. arr+1 will be the next address. You are basically "dereferencing" that address, using the index operator [ ] and the index itself.
arr[5] is the value located on arr + 5; You can dereference it just like a regular pointer. *(arr + 5); At the end, array address decays to a pointer after all.

Usually you can only declare array with a fixed length (number of items). Which means you can only pass a constant as an index (in the example above, 50), when declaring.
But in C99 you can also use variable-length index.

No, you can not mess with the addresses of the array. One array can not point to any other address of memory.
You can copy values, but not redirect or reposition (unless it isn't an array of pointers). The array is very similar to any other regular data type.
char arr[4]; can be considered as a 4 byte integer for example. You could treat like so, using type-casting.
int arr[1]; is almost same as int arr. The difference is in the usage.