Correct indentation on documentation comment.

This commit adjusts the indentation of code within a documentation
comment so that it is correctly highlighted as code by rustdoc.

diff --git a/src/librustc/mir/mod.rs b/src/librustc/mir/mod.rs
index c662ed6a6bf06f23553c4686250fe7ded28c3cc7..36bc2edcf584eb6293b686f6c3a7ab4b5b7720dc 100644 (file)
--- a/src/librustc/mir/mod.rs
+++ b/src/librustc/mir/mod.rs
@@ -506,25 +506,25 @@ pub enum BorrowKind {
     /// implicit closure bindings. It is needed when the closure is
     /// borrowing or mutating a mutable referent, e.g.:
     ///
-    ///    let x: &mut isize = ...;
-    ///    let y = || *x += 5;
+    ///     let x: &mut isize = ...;
+    ///     let y = || *x += 5;
     ///
     /// If we were to try to translate this closure into a more explicit
     /// form, we'd encounter an error with the code as written:
     ///
-    ///    struct Env { x: & &mut isize }
-    ///    let x: &mut isize = ...;
-    ///    let y = (&mut Env { &x }, fn_ptr);  // Closure is pair of env and fn
-    ///    fn fn_ptr(env: &mut Env) { **env.x += 5; }
+    ///     struct Env { x: & &mut isize }
+    ///     let x: &mut isize = ...;
+    ///     let y = (&mut Env { &x }, fn_ptr);  // Closure is pair of env and fn
+    ///     fn fn_ptr(env: &mut Env) { **env.x += 5; }
     ///
     /// This is then illegal because you cannot mutate an `&mut` found
     /// in an aliasable location. To solve, you'd have to translate with
     /// an `&mut` borrow:
     ///
-    ///    struct Env { x: & &mut isize }
-    ///    let x: &mut isize = ...;
-    ///    let y = (&mut Env { &mut x }, fn_ptr); // changed from &x to &mut x
-    ///    fn fn_ptr(env: &mut Env) { **env.x += 5; }
+    ///     struct Env { x: & &mut isize }
+    ///     let x: &mut isize = ...;
+    ///     let y = (&mut Env { &mut x }, fn_ptr); // changed from &x to &mut x
+    ///     fn fn_ptr(env: &mut Env) { **env.x += 5; }
     ///
     /// Now the assignment to `**env.x` is legal, but creating a
     /// mutable pointer to `x` is not because `x` is not mutable. We