Over the last few weeks, I’ve been devoting my free time to fleshing out the theory behind non-lexical lifetimes (NLL). I think I’ve arrived at a pretty good point and I plan to write various posts talking about it. Before getting into the details, though, I wanted to start out with a post that lays out roughly how today’s lexical lifetimes work and gives several examples of problem cases that we would like to solve.

The basic idea of the borrow checker is that values may not be mutated or moved while they are borrowed. But how do we know whether a value is borrowed? The idea is quite simple: whenever you create a borrow, the compiler assigns the resulting reference a lifetime. This lifetime corresponds to the span of the code where the reference may be used. The compiler will infer this lifetime to be the smallest lifetime that it can that still encompasses all the uses of the reference.

Note that Rust uses the term lifetime in a very particular way. In everyday speech, the word lifetime can be used in two distinct – but similar – ways:

The lifetime of a reference, corresponding to the span of time in which that reference is used. The lifetime of a value, corresponding to the span of time before that value gets freed (or, put another way, before the destructor for the value runs).

This second span of time, which describes how long a value is valid, is of course very important. We refer to that span of time as the value’s scope. Naturally, lifetimes and scopes are linked to one another. Specifically, if you make a reference to a value, the lifetime of that reference cannot outlive the scope of that value, Otherwise your reference would be pointing into free memory.

To better see the distinction between lifetime and scope, let’s consider a simple example. In this example, the vector data is borrowed (mutably) and the resulting reference is passed to a function capitalize . Since capitalize does not return the reference back, the lifetime of this borrow will be confined to just that call. The scope of data, in contrast, is much larger, and corresponds to a suffix of the fn body, stretching from the let until the end of the enclosing scope.

fn foo () { let mut data = vec! [ 'a' , 'b' , 'c' ]; // --+ 'scope capitalize ( & mut data [ .. ]); // | // ^~~~~~~~~~~~~~~~~~~~~~~~~ 'lifetime // | data .push ( 'd' ); // | data .push ( 'e' ); // | data .push ( 'f' ); // | } // <---------------------------------------+ fn capitalize ( data : & mut [ char ]) { // do something }

This example also demonstrates something else. Lifetimes in Rust today are quite a bit more flexible than scopes (if not as flexible as we might like, hence this RFC):

A scope generally corresponds to some block (or, more specifically, a suffix of a block that stretches from the let until the end of the enclosing block) [1].

until the end of the enclosing block) [1]. A lifetime, in contrast, can also span an individual expression, as this example demonstrates. The lifetime of the borrow in the example is confined to just the call to capitalize , and doesn’t extend into the rest of the block. This is why the calls to data.push that come below are legal.

So long as a reference is only used within one statement, today’s lifetimes are typically adequate. Problems arise however when you have a reference that spans multiple statements. In that case, the compiler requires the lifetime to be the innermost expression (which is often a block) that encloses both statements, and that is typically much bigger than is really necessary or desired. Let’s look at some example problem cases. Later on, we’ll see how non-lexical lifetimes fixes these cases.

Problem case #1: references assigned into a variable

One common problem case is when a reference is assigned into a variable. Consider this trivial variation of the previous example, where the &mut data[..] slice is not passed directly to capitalize , but is instead stored into a local variable:

fn bar () { let mut data = vec! [ 'a' , 'b' , 'c' ]; let slice = & mut data [ .. ]; // <-+ 'lifetime capitalize ( slice ); // | data .push ( 'd' ); // ERROR! // | data .push ( 'e' ); // ERROR! // | data .push ( 'f' ); // ERROR! // | } // <------------------------------+

The way that the compiler currently works, assigning a reference into a variable means that its lifetime must be as large as the entire scope of that variable. In this case, that means the lifetime is now extended all the way until the end of the block. This in turn means that the calls to data.push are now in error, because they occur during the lifetime of slice . It’s logical, but it’s annoying.

In this particular case, you could resolve the problem by putting slice into its own block:

fn bar () { let mut data = vec! [ 'a' , 'b' , 'c' ]; { let slice = & mut data [ .. ]; // <-+ 'lifetime capitalize ( slice ); // | } // <------------------------------+ data .push ( 'd' ); // OK data .push ( 'e' ); // OK data .push ( 'f' ); // OK }

Since we introduced a new block, the scope of slice is now smaller, and hence the resulting lifetime is smaller. Of course, introducing a block like this is kind of artificial and also not an entirely obvious solution.

Problem case #2: conditional control flow

Another common problem case is when references are used in only match arm. This most commonly arises around maps. Consider this function, which, given some key , processes the value found in map[key] if it exists, or else inserts a default value:

fn process_or_default < K , V : Default > ( map : & mut HashMap < K , V > , key : K ) { match map .get_mut ( & key ) { // -------------+ 'lifetime Some ( value ) => process ( value ), // | None => { // | map .insert ( key , V :: default ()); // | // ^~~~~~ ERROR. // | } // | } // <------------------------------------+ }

This code will not compile today. The reason is that the map is borrowed as part of the call to get_mut , and that borrow must encompass not only the call to get_mut , but also the Some branch of the match. The innermost expression that encloses both of these expressions is the match itself (as depicted above), and hence the borrow is considered to extend until the end of the match. Unfortunately, the match encloses not only the Some branch, but also the None branch, and hence when we go to insert into the map in the None branch, we get an error that the map is still borrowed.

This particular example is relatively easy to workaround. One can (frequently) move the code for None out from the match like so:

fn process_or_default1 < K , V : Default > ( map : & mut HashMap < K , V > , key : K ) { match map .get_mut ( & key ) { // -------------+ 'lifetime Some ( value ) => { // | process ( value ); // | return ; // | } // | None => { // | } // | } // <------------------------------------+ map .insert ( key , V :: default ()); }

When the code is adjusted this way, the call to map.insert is not part of the match, and hence it is not part of the borrow. While this works, it is of course unfortunate to require these sorts of manipulations, just as it was when we introduced an artificial block in the previous example.

Problem case #3: conditional control flow across functions

While we were able to work around problem case #2 in a relatively simple, if irritating, fashion. there are other variations of conditional control flow that cannot be so easily resolved. This is particularly true when you are returning a reference out of a function. Consider the following function, which returns the value for a key if it exists, and inserts a new value otherwise (for the purposes of this section, assume that the entry API for maps does not exist):

fn get_default < 'm , K , V : Default > ( map : & 'm mut HashMap < K , V > , key : K ) -> & 'm mut V { match map .get_mut ( & key ) { // -------------+ 'm Some ( value ) => value , // | None => { // | map .insert ( key , V :: default ()); // | // ^~~~~~ ERROR // | map .get_mut ( & key ) .unwrap () // | } // | } // | } // v

At first glance, this code appears quite similar the code we saw before. And indeed, just as before, it will not compile. But in fact the lifetimes at play are quite different. The reason is that, in the Some branch, the value is being returned out to the caller. Since value is a reference into the map, this implies that the map will remain borrowed until some point in the caller (the point 'm , to be exact). To get a better intuition for what this lifetime parameter 'm represents, consider some hypothetical caller of get_default : the lifetime 'm then represents the span of code in which that caller will use the resulting reference:

fn caller () { let mut map = HashMap :: new (); ... { let v = get_default ( & mut map , key ); // -+ 'm // +-- get_default() -----------+ // | // | match map.get_mut(&key) { | // | // | Some(value) => value, | // | // | None => { | // | // | .. | // | // | } | // | // +----------------------------+ // | process ( v ); // | } // <--------------------------------------+ ... }

If we attempt the same workaround for this case that we tried in the previous example, we will find that it does not work:

fn get_default1 < 'm , K , V : Default > ( map : & 'm mut HashMap < K , V > , key : K ) -> & 'm mut V { match map .get_mut ( & key ) { // -------------+ 'm Some ( value ) => return value , // | None => { } // | } // | map .insert ( key , V :: default ()); // | // ^~~~~~ ERROR (still) | map .get_mut ( & key ) .unwrap () // | } // v

Whereas before the lifetime of value was confined to the match, this new lifetime extends out into the caller, and therefore the borrow does not end just because we exited the match. Hence it is still in scope when we attempt to call insert after the match.

The workaround for this problem is a bit more involved. It relies on the fact that the borrow checker uses the precise control-flow of the function to determine what borrows are in scope.

fn get_default2 < 'm , K , V : Default > ( map : & 'm mut HashMap < K , V > , key : K ) -> & 'm mut V { if map .contains ( & key ) { // ^~~~~~~~~~~~~~~~~~ 'n return match map .get_mut ( & key ) { // + 'm Some ( value ) => value , // | None => unreachable! () // | }; // v } // At this point, `map.get_mut` was never // called! (As opposed to having been called, // but its result no longer being in use.) map .insert ( key , V :: default ()); // OK now. map .get_mut ( & key ) .unwrap () }

What has changed here is that we moved the call to map.get_mut inside of an if , and we have set things up so that the if body unconditionally returns. What this means is that a borrow begins at the point of get_mut , and that borrow lasts until the point 'm in the caller, but the borrow checker can see that this borrow will not have even started outside of the if . So it does not consider the borrow in scope at the point where we call map.insert .

This workaround is more troublesome than the others, because the resulting code is actually less efficient at runtime, since it must do multiple lookups.

It’s worth noting that Rust’s hashmaps include an entry API that one could use to implement this function today. The resulting code is both nicer to read and more efficient even than the original version, since it avoids extra lookups on the “not present” path as well:

fn get_default3 < 'm , K , V : Default > ( map : & 'm mut HashMap < K , V > , key : K ) -> & 'm mut V { map .entry ( key ) .or_insert_with (|| V :: default ()) }

Regardless, the problem exists for other data structures besides HashMap , so it would be nice if the original code passed the borrow checker, even if in practice using the entry API would be preferable. (Interestingly, the limitation of the borrow checker here was one of the motivations for developing the entry API in the first place!)

Conclusion

This post looked at various examples of Rust code that do not compile today, and showed how they can be fixed using today’s system. While it’s good that workarounds exist, it’d be better if the code just compiled as is. In an upcoming post, I will outline my plan for how to modify the compiler to achieve just that.

Endnotes

1. Scopes always correspond to blocks with one exception: the scope of a temporary value is sometimes the enclosing statement.