According to "The Rust Programming Language" ch05-03
Here’s how it works: When you call a method with object.something() , Rust automatically adds in & , &mut , or * so that object matches the signature of the method. In other words, the following are the same:

My question is about this code

let value = Rc::new(RefCell::new(5));
*value.borrow_mut() += 10;

I can understand why value.borrow_mut() will automatic compile to (*value).borrow_mut()
But why I should add * before value.borrow_mut() for the next add operation

This is the docs about += operator

fn add_assign(&mut self, rhs: Rhs);

In my understanding *value.borrow_mut() += 10; equals to

*value.borrow_mut().add_assign(10);

It still call a method,so why the compiler doesn't automatic add * before value.borrow_mut()?

It just like this example

let mut a = 3;
let mut b = &mut a;
*b += 1;
// Why B should add * before

If I am wrong, please point it out.
Thanks for your answer

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