---

yaml
---
r: 151188
b: refs/heads/try2
c: b7dba33
h: refs/heads/master
v: v3

Author: adrientetar

[refs]

@@ -5,7 +5,7 @@ refs/heads/snap-stage3: 78a7676898d9f80ab540c6df5d4c9ce35bb50463
 refs/heads/try: 519addf6277dbafccbb4159db4b710c37eaa2ec5
 refs/tags/release-0.1: 1f5c5126e96c79d22cb7862f75304136e204f105
 refs/heads/ndm: f3868061cd7988080c30d6d5bf352a5a5fe2460b
-refs/heads/try2: 53711461070ec8ad0d71c4bcdb35de3120e07ec5
+refs/heads/try2: b7dba3300e48e87e321dcfcc8db09dd3e69e48ba
 refs/heads/dist-snap: ba4081a5a8573875fed17545846f6f6902c8ba8d
 refs/tags/release-0.2: c870d2dffb391e14efb05aa27898f1f6333a9596
 refs/tags/release-0.3: b5f0d0f648d9a6153664837026ba1be43d3e2503

branches/try2/src/doc/rust.md

@@ -471,7 +471,7 @@ Two examples of paths with type arguments:
 # fn f() {
 # fn id<T>(t: T) -> T { t }
 type T = HashMap<int,~str>;  // Type arguments used in a type expression
-let x = id::<int>(10);         // Type arguments used in a call expression
+let x = id::<int>(10);       // Type arguments used in a call expression
 # }
 ~~~~
 

branches/try2/src/doc/tutorial.md

@@ -982,7 +982,8 @@ The obvious approach is to define `Cons` as containing an element in the list
 along with the next `List` node. However, this will generate a compiler error.
 
 ~~~ {.ignore}
-// error: illegal recursive enum type; wrap the inner value in a box to make it representable
+// error: illegal recursive enum type; wrap the inner value in a box to make it
+// representable
 enum List {
     Cons(u32, List), // an element (`u32`) and the next node in the list
     Nil
@@ -1054,10 +1055,10 @@ immutable, the whole list is immutable. The memory allocation itself is the
 box, while the owner holds onto a pointer to it:
 
 ~~~ {.notrust}
-          List box             List box           List box            List box
-        +--------------+    +--------------+    +--------------+    +--------------+
-list -> | Cons | 1 | ~ | -> | Cons | 2 | ~ | -> | Cons | 3 | ~ | -> | Nil          |
-        +--------------+    +--------------+    +--------------+    +--------------+
+            List box            List box            List box          List box
+        +--------------+    +--------------+    +--------------+    +----------+
+list -> | Cons | 1 | ~ | -> | Cons | 2 | ~ | -> | Cons | 3 | ~ | -> | Nil      |
+        +--------------+    +--------------+    +--------------+    +----------+
 ~~~
 
 > *Note:* the above diagram shows the logical contents of the enum. The actual
@@ -1197,7 +1198,8 @@ fn eq(xs: &List, ys: &List) -> bool {
         // If we have reached the end of both lists, they are equal.
         (&Nil, &Nil) => true,
         // If the current element in both lists is equal, keep going.
-        (&Cons(x, ~ref next_xs), &Cons(y, ~ref next_ys)) if x == y => eq(next_xs, next_ys),
+        (&Cons(x, ~ref next_xs), &Cons(y, ~ref next_ys))
+                if x == y => eq(next_xs, next_ys),
         // If the current elements are not equal, the lists are not equal.
         _ => false
     }
@@ -1256,7 +1258,7 @@ Using the generic `List<T>` works much like before, thanks to type inference:
 #     Cons(value, ~xs)
 # }
 let mut xs = Nil; // Unknown type! This is a `List<T>`, but `T` can be anything.
-xs = prepend(xs, 10); // The compiler infers the type of `xs` as `List<int>` from this.
+xs = prepend(xs, 10); // Here the compiler infers `xs`'s type as `List<int>`.
 xs = prepend(xs, 15);
 xs = prepend(xs, 20);
 ~~~
@@ -1303,7 +1305,8 @@ fn eq<T: Eq>(xs: &List<T>, ys: &List<T>) -> bool {
         // If we have reached the end of both lists, they are equal.
         (&Nil, &Nil) => true,
         // If the current element in both lists is equal, keep going.
-        (&Cons(ref x, ~ref next_xs), &Cons(ref y, ~ref next_ys)) if x == y => eq(next_xs, next_ys),
+        (&Cons(ref x, ~ref next_xs), &Cons(ref y, ~ref next_ys))
+                if x == y => eq(next_xs, next_ys),
         // If the current elements are not equal, the lists are not equal.
         _ => false
     }
@@ -1331,7 +1334,8 @@ impl<T: Eq> Eq for List<T> {
             // If we have reached the end of both lists, they are equal.
             (&Nil, &Nil) => true,
             // If the current element in both lists is equal, keep going.
-            (&Cons(ref x, ~ref next_xs), &Cons(ref y, ~ref next_ys)) if x == y => next_xs == next_ys,
+            (&Cons(ref x, ~ref next_xs), &Cons(ref y, ~ref next_ys))
+                    if x == y => next_xs == next_ys,
             // If the current elements are not equal, the lists are not equal.
             _ => false
         }