cmd/compile: add a flag to print the source line for generated rules

When working on rulegen, I often find myself
searching the rules files to find the source of
generated code. Add a flag to make that easier.

The flag needs to be off by default,
so that adding a single rule doesn't cause a massive diff.

Change-Id: I5a6f09129dc6fceef7c9cd1ad7eee24f3880ba91
Reviewed-on: https://go-review.googlesource.com/c/go/+/213700
Run-TryBot: Josh Bleecher Snyder <josharian@gmail.com>
Reviewed-by: Keith Randall <khr@golang.org>

diff --git a/src/cmd/compile/internal/ssa/gen/rulegen.go b/src/cmd/compile/internal/ssa/gen/rulegen.go
index d5ad96c890cc295decb9ea11600e20c5f4dcc625..fa0b7c8a6b7be6da4ec858a69c865bc47c17da64 100644 (file)
--- a/src/cmd/compile/internal/ssa/gen/rulegen.go
+++ b/src/cmd/compile/internal/ssa/gen/rulegen.go
@@ -54,7 +54,10 @@ import (
 
 // If multiple rules match, the first one in file order is selected.
 
-var genLog = flag.Bool("log", false, "generate code that logs; for debugging only")
+var (
+       genLog  = flag.Bool("log", false, "generate code that logs; for debugging only")
+       addLine = flag.Bool("line", false, "add line number comment to generated rules; for debugging only")
+)
 
 type Rule struct {
        rule string
@@ -598,6 +601,9 @@ func fprint(w io.Writer, n Node) {
                        fprint(w, n)
                }
        case *RuleRewrite:
+               if *addLine {
+                       fmt.Fprintf(w, "// %s\n", n.loc)
+               }
                fmt.Fprintf(w, "// match: %s\n", n.match)
                if n.cond != "" {
                        fmt.Fprintf(w, "// cond: %s\n", n.cond)