s1_C-sizeof.md

CASS Notes --- C Programming --- sizeof(x)

Sometimes, you want to do address math for arrays in C. At some point, you'll come across the sizeof(x) builtin function, which has surprisingly strange functionality. On the face of it, it returns the amount of bytes used for the argument x. However, working with arrays and sizeof(), four observations can be made:

1. sizeof() is merely a compile-time trick.

   Inside a function, you cannot dynamically calculate sizeof() for an array that was passed as an argument, since the compiler can't know the (albeit fixed) length of the passed array beforehand.

2. sizeof() only uses the type of its argument.

   Indeed, this means that inside a function, C would default to returning the size of a passed array pointer (4 bytes if the argument is of type int[]) instead of the total amount of bytes used for the entire array (length times element size). For example:

   int getArraySize(int unknown[]) {
       return sizeof(unknown);
   }
   int known[8] = {};
   int total_bytes = getArraySize(known);  // => int total_bytes = sizeof(int*);   =>   int total_bytes = 4;

3. If the compiler is able to statically determine the array argument x of sizeof(x), it will specify its array subtype as much as possible.

   In the case of an array of size 8 being initialised in the same scope, the compiler won't just specify "int*", but actually, "int[8]" as type for sizeof() to parse. Indeed, int[1] (size 4 B), int[2] (size 8 B) ... are all array subtypes, which sizeof() uses only if such a subtype is mentioned in the scope of the sizeof() call. For example:

   int known[8] = {};
   int total_bytes = sizeof(known);  // => int total_bytes = sizeof(int[8]);   =>   int total_bytes = 32;

4. sizeof() not only doesn't use the value of its argument, but it doesn't even resolve its argument.

   This means we could, for example, call sizeof(a[0]) for an empty array a, sizeof(a[13]) for a 10-element array a etc ...