Converting a pointer to different type in C

Question

int i = 42;
int *p1 = &i;
int long *p2 = (long*)p1;

Is this undefined behavior? In C++, I think it is implementation defined behavior for some reason.

I looked in C Standard:

C99 6.3.2.3/7 A pointer to an object or incomplete type may be converted to a pointer to a different object or incomplete type. If the resulting pointer is not correctly aligned 57) for the pointed-to type, the behavior is undefined. Otherwise, when converted back again, the result shall compare equal to the original pointer.

57) In general, the concept "correctly aligned" is transitive: if a pointer to type A is correctly aligned for a pointer to type B, which in turn is correctly aligned for a pointer to type C, then a pointer to type A is correctly aligned for a pointer to type C.

What does the term correctly aligned mean here in practice? How do you know if you're doing it correctly without stepping into Undefined behavior?

It's Undefined Behavior if you dereference p2. Reason: You violate strict-aliasing. — Mysticial
On 32bit arch, sizeof(int) == sizeof(long), so there is no difference. On 64 - you'll have some garbage on other half of long, or could even leave allocated memory regions, so again - there is no point to do that. Other than that, it's correct. — keltar
@Krishnabhadra: int long *p2 is a legitimate but slightly unorthodox way of writing signed long int *p2, long *p2, long int *p2 or long int signed *p2 (these are all equivalent to each other). — Jonathan Leffler
In a 64-bit environment on RISC machines, there's a decent chance that i will be aligned on an odd 32-bit address boundary but long must be aligned on a 64-bit address boundary (or even 32-bit address boundary). Therefore, on such systems, you would get a SIGBUS if you tried to read or modify memory via p2. — Jonathan Leffler

Luchian Grigore Luchian Grigore · Accepted Answer · 2012-12-04T04:47:55

It basically means that if, say, int is aligned to 4 bytes, and int long is aligned to 8 bytes, the behavior is undefined. Say you have something like:

 0x04     0x08    0x0C    0x10
+------+-------+-------+-------+
|      |       |   i   |       |
+------+-------+-------+-------+

In this case, &i == 0x0C (which is valid because int is aligned to 4 bytes). When you cast to int long*, the pointer is converted to an aligned one: p2 == 0x08, because our theoretical system aligns int long to 8 bytes, so you'd be basically reading off an address you don't own if you dereference p1, ergo the undefined behavior.

Converting a pointer to different type in C

1 Answers