A temp object is mut!

#1

The output is 2 , is it a breaking of the rule?

fn main () {
    struct S {
        id : u8,
    }
    impl std::ops::Drop for S {
        fn drop(&mut self) {
            println!("{}",self.id);
        }
    }
    S{id:1}.id=2;
}
#2

That makes a temporary, sets a field on it, and then that temporary gets dropped. It's the same as

let mut temp = S{id:1};
temp.id = 2;
drop(temp);

What "rule" do you think this is breaking?

#3

the object is default to immutable , temp object seems an exception ?

#4

In reality that rule is about variable names, not values. The same value can even change between mutability as it is moved between variables.

// I'm immutable
let var1 = "Hello ".to_string();
// Now I'm mutable
let mut var2 = var1;
var2.push_str("world!");
// Now I'm immutable again
let var3 = var2;
5 Likes