Merging two adjacent string slices [duplicate] - string

Question

This question already has answers here:
Rust: Is there an opposite for split_at_mut (i.e. join_mut)?
(2 answers)
Closed 8 months ago.
When writing a parser I ran into the problem that there are two string slices that come from the same origin string and are next to each other in memory. Of course it would be possible to simply copy the strings and merge them back into one, but that would require unnecessary computational resources. Is there a clean way to solve this in rust without unsafe code? For better illustration, here is an example of how I would like to solve it:
fn main() {
//This is the owned string.
//(Of course, this is also just a slice of a static string, but that makes no difference here).
let origin: &str = "Hello World";
//Substrings which borrow data from the original and should be adjacent in memory
let a: &str = &origin[0..5];
let b: &str = &origin[5..11];
//If the representation of a and b on the stack is:
// a: { ptr: PointerA, len: LenA }
// b: { ptr: PointerB, len: LenB }
//Then PointerA + LenA should be PointerB
//From this I conclude that there must be a way to combine these strings into c,
//which in turn would have this representation on the stack:
// c: { ptr: PointerA, len: LenA + LenB }
//The merge method doesn't actually exist, it's just a example of how I would imagen the api to look like.
let c = a.merge(b).unwrap();
assert!(c == origin)
}
Contrast this with the more inefficient current solution:
fn main() {
let origin: &str = "Hello World";
let a: &str = &origin[0..5];
let b: &str = &origin[5..11];
//Here both strings are simply copied to another location in the heap
//and need unnecessarily more memory, because the stored data exactly matches the data in origin
let c = a.to_owned() + b;
assert!(c == origin)
}
EDIT: This is a example of how i would implemented this with unsafe code, but i really don't know if it is actually safe
fn main() {
let origin: &str = "Hello World";
let a: &str = &origin[0..5];
let b: &str = &origin[5..11];
let c = merge(a, b).unwrap();
assert!(c == origin)
}
fn merge<'a>(one: &'a str, two: &'a str) -> Option<&'a str> {
unsafe {
let one: [usize; 2] = std::intrinsics::transmute(one);
let two: [usize; 2] = std::intrinsics::transmute(two);
if let Some(len) = one[1].checked_add(two[1]) {
if one[0] + one[1] == two[0] {
Some(std::intrinsics::transmute([one[0], len]))
} else {
None
}
} else {
None
}
}
}

Answer 1

You can chain the character within the splitted string:
fn main() {
//This is the owned string.
//(Of course, this is also just a slice of a static string, but that makes no difference here).
let origin: &str = "Hello World";
//Substrings which borrow data from the original and should be adjacent in memory
let a: &str = &origin[0..5];
let b: &str = &origin[5..11];
let c = a.chars().chain(b.chars());
assert_eq!(c.collect::<String>().as_str(), origin);
}
Playground
But notice that for most operation requiring &str, you would have to create a new string anyway.
So it bring us to unsafe life, and as pointed by #chayimfriedman it is UB:
use std::{slice, str};
fn main() {
//This is the owned string.
//(Of course, this is also just a slice of a static string, but that makes no difference here).
let origin: &str = "Hello World";
//Substrings which borrow data from the original and should be adjacent in memory
let a: &str = &origin[0..5];
let b: &str = &origin[5..11];
let c: &str = merge(a, b).unwrap();
assert_eq!(c, origin);
}
fn merge<'a>(a: &'a str, b: &'a str) -> Result<&'a str, String> {
let a_len = a.len();
let a_ptr = a.as_ptr();
let b_ptr = b.as_ptr();
let b_len = b.len();
if a_ptr as usize + a_len != b_ptr as usize {
return Err("Strings are not alighned".to_string());
}
Ok(unsafe {
str::from_utf8_unchecked(slice::from_raw_parts(a_ptr as *const u8, a_len + b_len))
})
}
Playground
Note that eventually you could avoid the pointers casting and instead use ptr::addr, which as for rust 1.62 it is still on nightly.