Rust lifetime for database connection bundle - rust

Question

I'm trying to design a struct to carry around a Postgres connection, transaction, and a bunch of prepared statements, and then execute the prepared statements repeatedly. But I'm running into lifetime problems. Here is what I've got:
extern crate postgres;
use postgres::{Connection, TlsMode};
use postgres::transaction::Transaction;
use postgres::stmt::Statement;
pub struct Db<'a> {
conn: Connection,
tx: Transaction<'a>,
insert_user: Statement<'a>,
}
fn make_db(url: &str) -> Db {
let conn = Connection::connect(url, TlsMode::None).unwrap();
let tx = conn.transaction().unwrap();
let insert_user = tx.prepare("INSERT INTO users VALUES ($1)").unwrap();
Db {
conn: conn,
tx: tx,
insert_user: insert_user,
}
}
pub fn main() {
let db = make_db("postgres://paul#localhost/t");
for u in &["foo", "bar"] {
db.insert_user.execute(&[&u]);
}
db.tx.commit().unwrap();
}
Here is the error I'm getting (on Rust 1.15.0 stable):
error: `conn` does not live long enough
--> src/main.rs:15:14
|
15 | let tx = conn.transaction().unwrap();
| ^^^^ does not live long enough
...
22 | }
| - borrowed value only lives until here
|
note: borrowed value must be valid for the anonymous lifetime #1 defined on the body at 13:28...
--> src/main.rs:13:29
|
13 | fn make_db(url: &str) -> Db {
| ^
I've read the Rust book (I've lost count how many times), but I'm not sure how to make progress here. Any suggestions?
EDIT: Thinking about this some more I still don't understand why in principle I can't tell Rust, "conn lives as long as Db does". The issue is with moving conn, but what if I don't move it? I understand why in C you can't return a pointer to stack-allocated memory, e.g.:
#include <stdio.h>
int *build_array() {
int ar[] = {1,2,3};
return ar;
}
int main() {
int *ar = build_array();
printf("%d\n", ar[1]);
}
And I get how that is similar to in Rust returning a &str or returning a vec slice.
But in Rust you can do this:
#[derive(Debug)]
struct S {
ar: Vec<i32>,
}
fn build_array() -> S {
let v = vec![1, 2, 3];
S { ar: v }
}
fn main() {
let s = build_array();
println!("{:?}", s);
}
And my understanding is that Rust is smart enough so that returning S doesn't actually require a move; essentially it is going straight to the caller's stack frame.
So I don't understand why it can't also put Db (including conn) in the caller's stack frame. Then no moves would be required, and tx would never hold an invalid address. I feel like Rust should be able to figure that out. I tried adding a lifetime hint, like this:
pub struct Db<'a> {
conn: Connection<'a>,
tx: Transaction<'a>,
insert_user: Statement<'a>,
}
But that gives an "unexpected lifetime parameter" error. I can accept that Rust can't follow the logic, but I'm curious if there is a reason why in principle it couldn't.
It does seem that putting conn on the heap should solve my problems, but I can't get this to work either:
pub struct Db<'a> {
conn: Box<Connection>,
tx: Transaction<'a>,
insert_user: Statement<'a>,
}
Even with a let conn = Box::new(Connection::connect(...));, Rust still tells me "conn does not live long enough". Is there some way to make this work with Box, or is that a dead end?
EDIT 2: I tried doing this with macros also, to avoid any extra stack frames:
extern crate postgres;
use postgres::{Connection, TlsMode};
use postgres::transaction::Transaction;
use postgres::stmt::Statement;
pub struct Db<'a> {
conn: Connection,
tx: Transaction<'a>,
insert_user: Statement<'a>,
}
macro_rules! make_db {
( $x:expr ) => {
{
let conn = Connection::connect($x, TlsMode::None).unwrap();
let tx = conn.transaction().unwrap();
let insert_user = tx.prepare("INSERT INTO users VALUES ($1)").unwrap();
Db {
conn: conn,
tx: tx,
insert_user: insert_user,
}
}
}
}
pub fn main() {
let db = make_db!("postgres://paul#localhost/t");
for u in &["foo", "bar"] {
db.insert_user.execute(&[&u]);
}
db.tx.commit().unwrap();
}
But that still tells me that conn does not live long enough. It seems that moving it into the struct should really not require any real RAM changes, but Rust still won't let me do it.

Answer 1

Starting with this function:
fn make_db(url: &str) -> Db {
unimplemented!()
}
Due to lifetime elision, this is equivalent to:
fn make_db<'a>(url: &'a str) -> Db<'a> {
unimplemented!()
}
That is, the lifetimes of all the references inside the Db struct must live as long as the string slice passed in. That only makes sense if the struct is holding on to the string slice.
To "solve" that, we can try to separate the lifetimes:
fn make_db<'a, 'b>(url: &'a str) -> Db<'b> {
unimplemented!()
}
Now this makes even less sense because now we are just making up a lifetime. Where is that 'b coming from? What happens if the caller of make_db decides that the concrete lifetime for the generic lifetime parameter 'b should be 'static? This is further explained in Why can't I store a value and a reference to that value in the same struct?, search for "something is really wrong with our creation function".
We also see the part of the question with "Sometimes, I'm not even taking a reference of the value" in the other question, which says in the answer:
the Child instance contains a reference to the Parent that created it,
If we check out the definition for Connection::transaction:
fn transaction<'a>(&'a self) -> Result<Transaction<'a>>
or the definition if you don't believe the docs:
pub struct Transaction<'conn> {
conn: &'conn Connection,
depth: u32,
savepoint_name: Option<String>,
commit: Cell<bool>,
finished: bool,
}
Yup, a Transaction keeps a reference to its parent Connection. Now that we see that Transaction has a reference to Connection we can return to the other question to see how to solve the problem: split apart the structs so that the nesting mirrors the lifetimes.
This was a very long-winded way of saying: no, you cannot create a single structure that contains a database and a transaction of that database due to the implementation of the postgres crate. Presumably the crate is implemented in this fashion for maximum performance.
I don't see why [returning Db<'b>] makes less sense. Normally when a function returns a thing, the thing lives as long as it is assigned to something. Why can't -> Db work the same way?
The entire point of references is that you don't own the referred-to value. You return Db and the caller of make_db would own that, but what owns the thing that Db is referring to? Where did it come from? You cannot return a reference to something local as that would violate all of Rust's safety rules. If you want to transfer ownership, you just do that.
See also
Is there any way to return a reference to a variable created in a function?
Return local String as a slice (&str)

Answer 2

Using the other answer, I put together working code that lets me bundle up the transaction and all the prepared statements, and pass them around together:
extern crate postgres;
use postgres::{Connection, TlsMode};
use postgres::transaction::Transaction;
use postgres::stmt::Statement;
pub struct Db<'a> {
tx: Transaction<'a>,
insert_user: Statement<'a>,
}
fn make_db(conn: &Connection) -> Db {
let tx = conn.transaction().unwrap();
let insert_user = tx.prepare("INSERT INTO users VALUES ($1)").unwrap();
Db {
tx: tx,
insert_user: insert_user,
}
}
pub fn main() {
let conn = Connection::connect("postgres://paul#localhost/t", TlsMode::None).unwrap();
let db = make_db(&conn);
for u in &["foo", "bar"] {
db.insert_user.execute(&[&u]);
}
db.tx.commit().unwrap();
}
As I understand it, Rust wants to guarantee that conn lives as long as db, so by keeping conn outside of the "constructor", the lexical structure ensures that it won't get removed too early.
My struct still doesn't encapsulate conn, which seems too bad to me, but at least it lets me keep everything else together.