The pointer return pattern is mostly due to C limitations. Rust is smart enough to automatically output into a pointer when a structure is big enough (this is i

The pointer return pattern is mostly due to C limitations. Rust is smart enough to automatically output into a pointer when a structure is big enough (this is invisible in source code, this should be rather optimizer's decision really whether to do it) and supports multiple return arguments by use of tuples. Also, even if somebody did insist on explicit mut-output pattern in Rust anyway, they would quickly realize that they need to initialize the value before getting a mutable reference to it - Rust NEVER allows access to uninitialized memory (other than unsafe blocks where you can trick Rust into accesing uninitialized memory and cause undefined behavior by using `std::mem::uninitialized`). The noise related to it should hint that this is not the intended way to do it. Here is a code example to clarify what I mean. fn write2(out: &mut i32) { *out = 2; } fn main() { let mut two; write2(&mut two); } And error message: error[E0381]: use of possibly uninitialized variable: `two` --> src/main.rs:7:17 | 7 | write2(&mut two); | ^^^ use of possibly uninitialized `two` Fixing that requires explicitly writing a value into `two`, for instance with `let mut two = 0`, which for complex structs... yeah, you may as well return a struct directly instead of bothering with pointer output. [std::mem::uninitialized]: https://doc.rust-lang.org/std/mem/fn.uninitialized.html https://doc.rust-lang.org/std/mem/fn.uninitialized.html