doc: allow buffered channel as semaphore without initialization

This rule not existing has been the source of many discussions
on golang-dev and on issues. We have stated publicly that it is
true, but we have never written it down. Write it down.

Fixes #6242.

LGTM=r, dan.kortschak, iant, dvyukov
R=golang-codereviews, r, dominik.honnef, dvyukov, dan.kortschak, iant, 0xjnml
CC=golang-codereviews
https://golang.org/cl/75130045

diff --git a/doc/effective_go.html b/doc/effective_go.html
index 6006724e5a940b1a4b83b3b6040bd772ac47da08..ae04899e190becc4b2e129480a4738d4dbe27f89 100644 (file)
--- a/doc/effective_go.html
+++ b/doc/effective_go.html
@@ -2942,26 +2942,19 @@ means waiting until some receiver has retrieved a value.
 <p>
 A buffered channel can be used like a semaphore, for instance to
 limit throughput.  In this example, incoming requests are passed
-to <code>handle</code>, which receives a value from the channel, processes
-the request, and then sends a value back to the channel
-to ready the "semaphore" for the next consumer.
+to <code>handle</code>, which sends a value into the channel, processes
+the request, and then receives a value from the channel
+to ready the &ldquo;semaphore&rdquo; for the next consumer.
 The capacity of the channel buffer limits the number of
-simultaneous calls to <code>process</code>,
-so during initialization we prime the channel by filling it to capacity.
+simultaneous calls to <code>process</code>.
 </p>
 <pre>
 var sem = make(chan int, MaxOutstanding)
 
 func handle(r *Request) {
-    &lt;-sem          // Wait for active queue to drain.
-    process(r)     // May take a long time.
-    sem &lt;- 1       // Done; enable next request to run.
-}
-
-func init() {
-    for i := 0; i &lt; MaxOutstanding; i++ {
-        sem &lt;- 1
-    }
+    sem &lt;- 1    // Wait for active queue to drain.
+    process(r)  // May take a long time.
+    &lt;-sem       // Done; enable next request to run.
 }
 
 func Serve(queue chan *Request) {
@@ -2973,10 +2966,9 @@ func Serve(queue chan *Request) {
 </pre>
 
 <p>
-Because data synchronization occurs on a receive from a channel
-(that is, the send "happens before" the receive; see
-<a href="/ref/mem">The Go Memory Model</a>),
-acquisition of the semaphore must be on a channel receive, not a send.
+Once <code>MaxOutstanding</code> handlers are executing <code>process</code>,
+any more will block trying to send into the filled channel buffer,
+until one of the existing handlers finishes and receives from the buffer.
 </p>
 
 <p>
@@ -2993,10 +2985,10 @@ Here's an obvious solution, but beware it has a bug we'll fix subsequently:
 <pre>
 func Serve(queue chan *Request) {
     for req := range queue {
-        &lt;-sem
+        sem &lt;- 1
         go func() {
             process(req) // Buggy; see explanation below.
-            sem &lt;- 1
+            &lt;-sem
         }()
     }
 }</pre>
@@ -3014,10 +3006,10 @@ to the closure in the goroutine:
 <pre>
 func Serve(queue chan *Request) {
     for req := range queue {
-        &lt;-sem
+        sem &lt;- 1
         go func(req *Request) {
             process(req)
-            sem &lt;- 1
+            &lt;-sem
         }(req)
     }
 }</pre>
@@ -3032,11 +3024,11 @@ name, as in this example:
 <pre>
 func Serve(queue chan *Request) {
     for req := range queue {
-        &lt;-sem
         req := req // Create new instance of req for the goroutine.
+        sem &lt;- 1
         go func() {
             process(req)
-            sem &lt;- 1
+            &lt;-sem
         }()
     }
 }</pre>

diff --git a/doc/go_mem.html b/doc/go_mem.html
index 3e769daecaba846ca628f4ba17a705180dd88fcc..69e7c8ce75817bdf816f91ff0ef4f13c35ce83b2 100644 (file)
--- a/doc/go_mem.html
+++ b/doc/go_mem.html
@@ -274,6 +274,41 @@ then the program would not be guaranteed to print
 crash, or do something else.)
 </p>
 
+<p class="rule">
+The <i>k</i>th send on a channel with capacity <i>C</i> happens before the <i>k</i>+<i>C</i>th receive from that channel completes.
+</p>
+
+<p>
+This rule generalizes the previous rule to buffered channels.
+It allows a counting semaphore to be modeled by a buffered channel:
+the number of items in the channel corresponds to the semaphore count,
+the capacity of the channel corresponds to the semaphore maximum,
+sending an item acquires the semaphore, and receiving an item releases
+the semaphore.
+This is a common idiom for rate-limiting work.
+</p>
+
+<p>
+This program starts a goroutine for every entry in the work list, but the
+goroutines coordinate using the <code>limit</code> channel to ensure
+that at most three are running work functions at a time.
+</p>
+
+<pre>
+var limit = make(chan int, 3)
+
+func main() {
+       for _, w := range work {
+               go func() {
+                       limit <- 1
+                       w()
+                       <-limit
+               }()
+       }
+       select{}
+}
+</pre>
+
 <h3>Locks</h3>
 
 <p>