Question: Is there any particular reason that if the complier knew an int goes into an address, then it cant remember that an int is there when reading?
As I understand, line 0 tells the compiler to allocate 16 bits for whatever value will be placed there upon assignment in line 3.
And I understand that one of the reasons for declaring the pointer type at line 1 is to tell the compiler how many bytes to read when dereferencing the pointer p with *p on line 6.
0: int i;
1: int *p;
2:
3: i = 5;
4: p = &i;
5:
6: printf("%d",*p);
For simplicity, using bit-width of 8 in mem table.
Say i is written at 0x04 and p is written at 0x02.
addr | storedVal
-----------------
0x01 |
0x02 | 00000100
0x03 |
0x04 | 00000000
0x05 | 00000101
It could be if the language allowed it; in C++, you could delay the declaration of p until the point of assignment and just do:
auto p = &i;
and it would infer that p is an int*. C doesn't do that though. It has no syntactic facility for deriving the type of an object, so you need to declare the type explicitly.
If you're talking about printf, printf is a varargs function; it doesn't know the types of the variables passed to it (some compilers special case it for checks at compile time, but that's an implementation detail, not a language guarantee; at runtime it doesn't know). It only knows to look for an int because the %d placeholder told it to check for that (and no pointers are involved inside printf; it was passed the raw int read from the pointer, not a pointer to the int).