[FIXED] What is the usage of the asterisk symbol in Rust?

Issue

I'm new to Rust and I don't understand the following piece of code:

let mut x = 5;
{
    let y = &mut x;
    *y += 1;
}
println!("{}", x);

Explanation from the Rust site:

You'll also notice we added an asterisk (*) in front of y, making it *y, this is because y is a &mut reference. You'll need to use astrisks [sic] to access the contents of a reference as well.

If *y is a reference, why does the following code work

fn main() {
    let mut x = 5;
    {
        let y = &mut x;
        println!("{}", y);
    }
}

I know I'm not modifying the value here, but what is the difference and why would y += 1; not work?

Solution

If *y is a reference

*y is not a reference. y is a reference; *y dereferences y, allowing you access to the referred-to value.

what is the difference [between += and println!]

println! is a macro that automatically references the arguments given to it. In addition, the Display trait (used via {} in the format string) is implemented for all references to types that themselves implement Display (impl<'a, T> Display for &'a T where T: Display + ?Sized).

Thus, println!("{}", y); is actually printing out a reference to a reference to a value. Those intermediate references are automatically dereferenced due to the implementation of Display.

+=, on the other hand, is implemented via the AddAssign trait. The standard library only implements adding an integer type to itself (impl AddAssign<i32> for i32). That means that you have to add an appropriate level of dereferencing in order to get both sides to an integer.

Answered By - Shepmaster

Answer Checked By - Mildred Charles (PHPFixing Admin)