I am wondering why list[0]
and number_list[0]
have the same address.
To my understanding, we passed in a reference to number_list
in for_question4
, so list supposed to be a reference of number_list
. When we called list[0]
in for_question4
, it is supposed to be the same as &number_list[0]
.
But when I print the address, list[0]
and number_list[0]
are on the same address.
&list[0]
and &number_list[0]
are on the same address.
Does function auto-dereference the vector passed in to it? Is auto-dereference a thing in Rust? If so, in what condition would it do that?
fn for_question4(list: &[&str]) {
println!("call index without reference {:p}", list[0]); // 0x103d23d5d
println!("call index with reference {:p}", &list[0]); // 0x7fea9c405de0
}
fn main() {
let number_list = vec!["1", "2", "3"];
let result = for_question4(&number_list);
println!("call index without reference {:p}", number_list[0]); // 0x103d23d5d
println!("call index with reference {:p}", &number_list[0]); // 0x7fea9c405de0
println!("call index with two reference {:p}", &&number_list[0]); // 0x7ffeebf46f80
}
CodePudding user response:
The []
operator is evaluated before the &
operator. The []
operator auto-dereferences.
Here are some examples:
fn main() {
let n = vec!["1", "2", "3"];
println!(" n[0] {:p}", n[0]);
println!(" (&n)[0] {:p}", (&n)[0]);
println!(" (&&n)[0] {:p}", (&&n)[0]);
println!(" &n[0] {:p}", &n[0]);
println!(" &(n[0]) {:p}", &(n[0]));
println!(" &((&n)[0]) {:p}", &((&n)[0]));
println!("&((&&n)[0]) {:p}", &((&&n)[0]));
}
n[0] 0x5597b8ccf002
(&n)[0] 0x5597b8ccf002
(&&n)[0] 0x5597b8ccf002
&n[0] 0x5597b9f3fad0
&(n[0]) 0x5597b9f3fad0
&((&n)[0]) 0x5597b9f3fad0
&((&&n)[0]) 0x5597b9f3fad0
CodePudding user response:
Just figured it out thanks to @Finomnis
It was index operator ([]) that did the auto-dereference. According to reference:
For other types an index expression a[b] is equivalent to *std::ops::Index::index(&a, b), or *std::ops::IndexMut::index_mut(&mut a, b) in a mutable place expression context. Just as with methods, Rust will also insert dereference operations on a repeatedly to find an implementation.