Parse fully-qualified associated types in generics without whitespace

This breaks code that looks like this:

    let x = foo as bar << 13;

Change such code to look like this:

    let x = (foo as bar) << 13;

Closes #17362.

[breaking-change]

1 files changed, 8 insertions, 8 deletions

diff --git a/src/libcore/hash/sip.rs b/src/libcore/hash/sip.rs
index ab6b0986c68..2501037107a 100644
--- a/src/libcore/hash/sip.rs
+++ b/src/libcore/hash/sip.rs
@@ -50,19 +50,19 @@ pub struct SipState {
 macro_rules! u8to64_le {
     ($buf:expr, $i:expr) =>
     ($buf[0+$i] as u64 |
-     $buf[1+$i] as u64 << 8 |
-     $buf[2+$i] as u64 << 16 |
-     $buf[3+$i] as u64 << 24 |
-     $buf[4+$i] as u64 << 32 |
-     $buf[5+$i] as u64 << 40 |
-     $buf[6+$i] as u64 << 48 |
-     $buf[7+$i] as u64 << 56);
+     ($buf[1+$i] as u64) << 8 |
+     ($buf[2+$i] as u64) << 16 |
+     ($buf[3+$i] as u64) << 24 |
+     ($buf[4+$i] as u64) << 32 |
+     ($buf[5+$i] as u64) << 40 |
+     ($buf[6+$i] as u64) << 48 |
+     ($buf[7+$i] as u64) << 56);
     ($buf:expr, $i:expr, $len:expr) =>
     ({
         let mut t = 0;
         let mut out = 0u64;
         while t < $len {
-            out |= $buf[t+$i] as u64 << t*8;
+            out |= ($buf[t+$i] as u64) << t*8;
             t += 1;
         }
         out