cmd/compile: simplify exportsym debug message

No need to disambiguate if we're exporting or reexporting, because
it's obvious from the output.

Change-Id: I59053d34dc6f8b29e20749c7b03c3cb4f4d641ff
Reviewed-on: https://go-review.googlesource.com/104236
Reviewed-by: Robert Griesemer <gri@golang.org>

diff --git a/src/cmd/compile/internal/gc/export.go b/src/cmd/compile/internal/gc/export.go
index 37b09844799d144de08c3d15582d6288cb9e4256..4ea999109be6fd4d38307671864e4c18a28ca396 100644 (file)
--- a/src/cmd/compile/internal/gc/export.go
+++ b/src/cmd/compile/internal/gc/export.go
@@ -36,11 +36,7 @@ func exportsym(n *Node) {
        n.Sym.SetOnExportList(true)
 
        if Debug['E'] != 0 {
-               if n.Sym.Pkg == localpkg {
-                       fmt.Printf("export symbol %v\n", n.Sym)
-               } else {
-                       fmt.Printf("reexport name %v\n", n.Sym)
-               }
+               fmt.Printf("export symbol %v\n", n.Sym)
        }
 
        exportlist = append(exportlist, n)