问题
Here are two function signatures I saw in the Rust documentation:
fn modify_foo(mut foo: Box<i32>) { *foo += 1; *foo }
fn modify_foo(foo: &mut i32) { *foo += 1; *foo }
Why the different placement of mut?
It seems that the first function could also be declared as
fn modify_foo(foo: mut Box<i32>) { /* ... */ }
回答1:
mut foo: T means you have a variable called foo that is a T. You are allowed to change what the variable refers to:
let mut val1 = 2;
val1 = 3; // OK
let val2 = 2;
val2 = 3; // error: re-assignment of immutable variable
This also lets you modify fields of a struct that you own:
struct Monster { health: u8 }
let mut orc = Monster { health: 93 };
orc.health -= 54;
let goblin = Monster { health: 28 };
goblin.health += 10; // error: cannot assign to immutable field
foo: &mut T means you have a variable that refers to (&) a value and you are allowed to change (mut) the referred value (including fields, if it is a struct):
let val1 = &mut 2;
*val1 = 3; // OK
let val2 = &2;
*val2 = 3; // error: cannot assign to immutable borrowed content
Note that &mut only makes sense with a reference - foo: mut T is not valid syntax. You can also combine the two qualifiers (let mut a: &mut T), when it makes sense.
回答2:
If you're coming from C/C++, it might also be helpful to think of it basically like this:
// Rust C/C++
a: &T == const T* const a; // can't mutate either
mut a: &T == const T* a; // can't mutate what is pointed to
a: &mut T == T* const a; // can't mutate pointer
mut a: &mut T == T* a; // can mutate both
You'll notice that these are inverses of each other. C/C++ take a "blacklist" approach, where if you want something to be immutable you have to say so explicitly, while Rust takes a "whitelist" approach, where if you want something to be mutable you have to say so explicitly.
来源：https://stackoverflow.com/questions/28587698/whats-the-difference-between-placing-mut-before-a-variable-name-and-after-the