{"id":1004,"date":"2016-07-21T00:38:04","date_gmt":"2016-07-20T22:38:04","guid":{"rendered":"https:\/\/blog.mi.hdm-stuttgart.de\/?p=1004"},"modified":"2023-06-07T14:59:37","modified_gmt":"2023-06-07T12:59:37","slug":"rust-fast-and-secure","status":"publish","type":"post","link":"https:\/\/blog.mi.hdm-stuttgart.de\/index.php\/2016\/07\/21\/rust-fast-and-secure\/","title":{"rendered":"Rust – fast and secure"},"content":{"rendered":"

Rust, a fairly new programming language promises to be fast and secure. The following blog entry discusses how Rust tries to achieve these two goals.<\/p>\n

The key concept is that every resource always belongs to exactly one variable. More precisely one lifetime, which is normally automatically created on variable creation. The concept of lifetimes would go into too much detail, so in this blog entry we assume that lifetimes and variables are the same. If you want to learn more about lifetimes or any other topic of Rust, you can read \u201cthe book\u201d \u2013 the main resource when learning Rust: https:\/\/doc.rust-lang.org\/book\/<\/a><\/p>\n

<\/p>\n

If the programmer creates a new resource, for example a vector (which is a dynamic growable array) this vector always belongs to one variable. The following line creates a new vector in Rust:<\/p>\n

let v = vec![1, 2, 3, 4];<\/pre>\n
If we would assign this vector to another variable then the vector is \u201cmoved\u201d, which means that the original variable isn\u2019t usable anymore. The following lines would compile:<\/p>\n
let v = vec![1, 2, 3, 4];\n\nlet v2 = v;<\/pre>\n
However, the next lines would not:<\/p>\n
let v = vec![1, 2, 3, 4];\n\nlet v2 = v;\n\nv.push(5);<\/pre>\n
Because the vector was moved from v to v2, the variable v is not usable anymore. There are two different behaviors in Rust when the programmer assigns a variable to another:<\/p>\n
    \n
  1. moving\n