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 <[email protected]> Reviewed-by: Keith Randall <[email protected]>
xphuhu · Feb 20, 2020 · 24fa159 · 24fa159
1 parent 49f8d45
commit 24fa159
Showing 1 changed file with 7 additions and 1 deletion.
diff --git 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)