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cmd/compile: remove confusing comment, fix comment for symExport 

The symExport flag tells whether a symbol is in the export list
already or not (and it's also used to avoid being added to that
list). Exporting is based on that export list - no need to check
again.

Change-Id: I6056f97aa5c24a19376957da29199135c8da35f9
Reviewed-on: https://go-review.googlesource.com/39033
Reviewed-by: Dave Cheney <dave@cheney.net>
This commit is contained in:
Robert Griesemer 2017-03-30 16:40:52 -07:00
parent 673a8fdfe6
commit 42aa608f8a
2 changed files with 1 additions and 11 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

10
src/cmd/compile/internal/gc/bexport.go
View File

@ -265,11 +265,6 @@ func export(out *bufio.Writer, trace bool) int {
Fatalf("exporter: unexpected symbol: %v", sym)
}
// TODO(gri) Should we do this check?
// if !sym.Export() {
// continue
// }
if sym.Def == nil {
Fatalf("exporter: unknown export symbol: %v", sym)
}
@ -332,11 +327,6 @@ func export(out *bufio.Writer, trace bool) int {
Fatalf("exporter: unexpected symbol: %v", sym)
}
// TODO(gri) Should we do this check?
// if !sym.Export() {
// continue
// }
if sym.Def == nil {
Fatalf("exporter: unknown export symbol: %v", sym)
}

2
src/cmd/compile/internal/gc/go.go
View File

@ -61,7 +61,7 @@ type Sym struct {
}
const (
symExport = 1 << iota // to be exported
symExport = 1 << iota // added to exportlist (no need to add again)
symPackage
symExported // already written out by export
symUniq