对于rust中的元素，可以是Copy类型和非Copy类型，从而导致在元素访问时存在不同的底层逻辑。
首先，对于vec[idx]，返回的是引用&T
对于Copy类型：
let mut ve=vec![1,2,3];let it=vec[1];//copyprintln!("{}", it32);
it的类型是int32而非&int32,因为：
vec[1]是a place expression，
let it=vec[1]; is evaluated in a value expression context
If the type of that value implements Copy, then the value will be copied.
所以i32会被copy给it
https://doc.rust-lang.org/reference/expressions.html
let mut ve=vec![1,2,3];println!("{:#p}",vei32.as_ptr());let it32r = &vei32[0];//it32r 值为直接对vec第二个元素的借用println!("{:p}", it32r);//vei32.as_ptr()与it32r地址相同
it32r 值为直接对vec第二个元素的借用，原因：
The & (shared borrow) and &mut (mutable borrow) operators are unary prefix operators.
When applied to a place expression, this expressions produces a reference (pointer) to the location that the value refers to.
所以vei32[0]返回借用，&vei32[0]直接获取元素地址仍是借用。
https://doc.rust-lang.org/reference/expressions/operator-expr.html#borrow-operators
let mut ve=vec![1,2,3];let it32r = &vei32[0];vei32.push(4);//报错：//cannot borrow `vei32` as mutable because it is also borrowed as immutable //mutable borrow occurs hereprintln!("{:p}", it32r);
原因分析：对于vec,vec[idx]相当于vec的buffer的切片，所以&vei32[0]虽然是对第一个元素的借用，但也是对整个buffer的借用，所以会导致vei32.push(4);报错。
对于非Copy类型：
let mut vec=vec![String::from("a"),String::from("b"),String::from("c")];let it = vec[1];//cannot move out of index of `Vec<String>`//move occurs because value has type `String`, which does not implement the `Copy` trait
所以对于非Copy类型元素，对索引需要借用：
let mut vec=vec![String::from("a"),String::from("b"),String::from("c")];let it = &vec[1];vec.push(String::from("value"));//cannot borrow `vec` as mutable because it is also borrowed as immutable//mutable borrow occurs here//同上