Why does printing a pointer print the same thing as printing the dereferenced pointer?

Question

From the Rust guide:

To dereference (get the value being referred to rather than the reference itself) y, we use the asterisk (*)

So I did it:

fn main() {     let x = 1;     let ptr_y = &x;     println!("x: {}, ptr_y: {}", x, *ptr_y); } 

This gives me the same results (x=1; y=1) even without an explicit dereference:

fn main() {     let x = 1;     let ptr_y = &x;     println!("x: {}, ptr_y: {}", x, ptr_y); } 

Why? Shouldn't ptr_y print the memory address and *ptr_y print 1? Is there some kind of auto-dereference or did I miss something?

Answer 1

Rust usually focuses on object value (i.e. the interesting part of the contents) rather than object identity (memory addresses). The implementation of Display for &T where T implements Display defers directly to the contents. Expanding that macro manually for the String implementation of Display:

impl<'a> Display for &'a String {     fn fmt(&self, f: &mut Formatter) -> Result {         Display::fmt(&**self, f)     } } 

That is, it is just printing its contents directly.

If you care about object identity/the memory address, you can use the Pointer formatter, {:p}:

fn main() {     let x = 1;     let ptr_y = &x;     println!("x: {}, ptr_y: {}, address: {:p}", x, ptr_y, ptr_y); } 

Output:

x: 1, ptr_y: 1, address: 0x7fff4eda6a24 

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