Only call signal.Notify once during agent startup

Calling twice appears to have no adverse effects, however serves to
confuse as to what the semantics of such code may be! This seems like it
was probably introduced while resolving conflicts during the merge of
the fix for #2404.
pull/4024/head
James Nugent 7 years ago
parent 464bad03cb
commit 96f871862e

@ -385,7 +385,6 @@ func (c *cmd) run(args []string) int {
// wait for signal
signalCh := make(chan os.Signal, 10)
signal.Notify(signalCh, os.Interrupt, syscall.SIGTERM, syscall.SIGHUP)
signal.Notify(signalCh, os.Interrupt, syscall.SIGTERM, syscall.SIGHUP, syscall.SIGPIPE)
for {

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