CODEWITHSUNDEEP
stringrustconcatenationmoveownership
cr7pt0pl4gu3
cr7pt0pl4gu3

Reputation: 33

How to concatenate two strings into the third string without copying?

I'm new to Rust and I'm currently reading The Rust Programming Language book.

I'm curious about this example:

fn main() {
    let s1 = String::from("Hello, ");
    let s2 = String::from("world!");
    let s3 = s1 + &s2; // note s1 has been moved here and can no longer be used
}

Is it possible to take ownership not only of s1, but also of s2, so s2 gets invalidated as well as s1, making s3 the only possible variable that remains usable?

Upvotes: 3

Views: 4053

Answers (1)

pretzelhammer
pretzelhammer

Reputation: 15155

Updated Answer

This is not possible. A String has to be a single contiguous allocation in memory. If you want to roll your own simple solution you can probably define a type like this:

struct MyString {
    parts: Vec<String>,
}

impl MyString {
    fn concat(&mut self, other: String) {
        self.parts.push(other);
    }
}

However re-implementing every useful String method for this custom type is going to be tedious and error-prone. You can find a Rust crate which implements a Rope data structure, and an-rope seems to be such a crate, but only a tiny fraction of String methods are supported.

I gave this answer when I interpreted OP's question as one about invalidating and moving variables:

Original Answer

You can invalidate any non-Copy variable by moving it somewhere else, like by passing it to the drop function:

fn main() {
    let s1 = String::from("Hello, ");
    let s2 = String::from("world!");
    let s3 = s1 + &s2; // s1 invalidated
    drop(s2); // s2 invalidated
    // now only s3 is usable
}

If this is a common pattern in your application you can just write a function which takes ownership of 2 Strings and returns the concatenated result:

fn concat(s1: String, s2: String) -> String {
    s1 + &s2
}

fn main() {
    let s1 = String::from("Hello, ");
    let s2 = String::from("world!");
    let s3 = concat(s1, s2); // s1 & s2 invalidated
    // now only s3 is usable
}

Upvotes: 5

Related Questions