1. Type compatibility and pointer operation
Kei Hasegawa

2. Type compatibility
Refer to ISO IEC C98/C99 for restrict definition. Roughly say,
Type T1 is compatible with type T2
<->
extern T1 x;
extern T2 x;
is not error.

3. Array compatibility typedef int T1[]; typedef int T2[10]; extern T1 x;
is not error. So,
Type int [] is compatible with type int [10]

4. Function compatibility
typedef void T1(int (*)(int), double (*)[]);
typedef void T2(int (*)(), double (*)[2]);
typedef void T3();
extern T1 x;
extern T2 x;
extern T3 x;
is not error. So,
Type void (int (*)(int), double (*)[]), type void (int (*)(), double (*)[2]) and type void () are compatible with for each other.

5. Tagged type compatibility
struct S;
extern struct S x; // incomplete type
struct S { int a; };
extern struct S x; // complete type
is not error. So,
incomplete struct S is compatible with complete struct S.

6. Pointer compatibility
Type T1 is compatible with type T2
<->
extern T1 *x;
extern T2 *x;
is not error.

7. Canonical rule of pointer operation
Pointers to qualified or unqualified versions of compatible types shall have the same representation and alignment requirements.
For example, int* p and const int* q shall have the same representation and alignment requirements.
So that’s why comparison int* p and const int* q or assignment like q = p are valid. But p = q is error in sense of discarding qualifier.
For int** p2 and const int** q2, any operation is not defined. i.e. it’s error.
Type of p2 is pointer to int* and type of q2 is pointer to const int*, but int* is not compatible with const int*.

8. Pointer and pointer operation – applied canonical rule
p – q
p < q, p > q, p <= q, p >= q
p == q, p != q
possible to compare void* and 0
expr ? p : q
can be void* or 0
p = q