问题
A pointer * and a reference & in Rust share the same representation (they both represent the memory address of a piece of data).
What's the practical differences when writing code though?
When porting C++ code to Rust can they be replaced safely (c++ pointer --> rust pointer, c++ reference --> rust reference ) ?
回答1:
Use references when you can, use pointers when you must. If you're not doing FFI or memory management beyond what the compiler can validate, you don't need to use pointers.
Both references and pointers exist in two variants. There are shared references & and mutable references &mut. There are const pointers *const and mut pointers *mut (which map to const and non-const pointers in C). However, the semantics of references is completely different from the semantics of pointers.
References are generic over a type and over a lifetime. Shared references are written &'a T in long form (where 'a and T are parameters). The lifetime parameter can be omitted in many situations. The lifetime parameter is used by the compiler to ensure that a reference doesn't live longer than the borrow is valid for.
Pointers have no lifetime parameter. Therefore, the compiler cannot check that a particular pointer is valid to use. That's why dereferencing a pointer is considered unsafe.
When you create a shared reference to an object, that freezes the object (i.e. the object becomes immutable while the shared reference exists), unless the object uses some form of interior mutability (e.g. using Cell, RefCell, Mutex or RwLock). However, when you have a const pointer to an object, that object may still change while the pointer is alive.
When you have a mutable reference to an object, you are guaranteed to have exclusive access to that object through this reference. Any other way to access the object is either disabled temporarily or impossible to achieve. For example:
let mut x = 0;
{
let y = &mut x;
let z = &mut x; // ERROR: x is already borrowed mutably
*y = 1; // OK
x = 2; // ERROR: x is borrowed
}
x = 3; // OK, y went out of scope
Mut pointers have no such guarantee.
A reference cannot be null (much like C++ references). A pointer can be null.
Pointers may contain any numerical value that could fit in a usize. Initializing a pointer is not unsafe; only dereferencing it is.
If you have a *const T, you can freely cast it to a *const U or to a *mut T using as. You can't do that with references. However, you can cast a reference to a pointer using as, and you can "upgrade" a pointer to a reference by dereferencing the pointer (which, again, is unsafe) and then borrowing the place using & or &mut. For example:
use std::ffi::OsStr;
use std::path::Path;
pub fn os_str_to_path(s: &OsStr) -> &Path {
unsafe { &*(s as *const OsStr as *const Path) }
}
In C++, references are "automatically dereferenced pointers". In Rust, you often still need to dereference references explicitly. The exception is when you use the . operator: if the left side is a reference, the compiler will automatically dereference it (recursively if necessary!). Pointers, however, are not automatically dereferenced. This means that if you want to dereference and access a field or a method, you need to write (*pointer).field or (*pointer).method(). There is no -> operator in Rust.
回答2:
Rust references are just a pointer, but the compiler endows them with borrowing semantics. When you take an immutable reference to an object, the compiler ensures that you can't modify that object until after the derived reference is gone.
来源：https://stackoverflow.com/questions/62232753/differences-between-a-pointer-and-a-reference-in-rust