btf: account for split BTF in loadRawSpec

The calculation for the first type id assumes that base.firstTypeID
is zero. This is only true if base wasn't parsed from split BTF
itself. Refuse split BTF as base but use base.lastTypeID as a
defensive measure.

Signed-off-by: Lorenz Bauer <[email protected]>

btf/btf.go

@@ -233,12 +233,21 @@ func loadRawSpec(btf io.ReaderAt, bo binary.ByteOrder, base *Spec) (*Spec, error
 		baseStrings *stringTable
 		baseTypes   []Type
 		firstTypeID TypeID
+		err         error
 	)
 
 	if base != nil {
+		if base.firstTypeID != 0 {
+			return nil, fmt.Errorf("can't use split BTF as base")
+		}
+
 		baseStrings = base.strings
 		baseTypes = base.types
-		firstTypeID = TypeID(len(baseTypes))
+
+		firstTypeID, err = base.nextTypeID()
+		if err != nil {
+			return nil, err
+		}
 	}
 
 	rawTypes, rawStrings, err := parseBTF(btf, bo, baseStrings)
@@ -257,7 +266,7 @@ func loadRawSpec(btf io.ReaderAt, bo binary.ByteOrder, base *Spec) (*Spec, error
 		namedTypes:  typesByName,
 		typeIDs:     typeIDs,
 		types:       types,
-		firstTypeID: TypeID(len(baseTypes)),
+		firstTypeID: firstTypeID,
 		strings:     rawStrings,
 		byteOrder:   bo,
 	}, nil
@@ -598,9 +607,9 @@ func (s *Spec) Add(typ Type) (TypeID, error) {
 		return id, nil
 	}
 
-	id := s.firstTypeID + TypeID(len(s.types))
-	if id < s.firstTypeID {
-		return 0, fmt.Errorf("type ID overflow")
+	id, err := s.nextTypeID()
+	if err != nil {
+		return 0, err
 	}
 
 	s.typeIDs[typ] = id
@@ -613,6 +622,16 @@ func (s *Spec) Add(typ Type) (TypeID, error) {
 	return id, nil
 }
 
+// nextTypeID returns the next unallocated type ID or an error if there are no
+// more type IDs.
+func (s *Spec) nextTypeID() (TypeID, error) {
+	id := s.firstTypeID + TypeID(len(s.types))
+	if id < s.firstTypeID {
+		return 0, fmt.Errorf("no more type IDs")
+	}
+	return id, nil
+}
+
 // TypeByID returns the BTF Type with the given type ID.
 //
 // Returns an error wrapping ErrNotFound if a Type with the given ID