fix(baggage): Ensure deterministic ordering with sentry members first

cursoragent · Claude Sonnet 4.5 · cursoragent · commit 4f11aac865c5 · 2026-04-17T10:25:22.000Z
The Baggage.String() method was using non-deterministic map iteration,
making it impossible to guarantee that sentry- prefixed members would
appear first in the serialized baggage string. This violated the W3C
Baggage spec requirement that critical members should appear first to
survive truncation when size limits are exceeded.

This fix sorts baggage keys deterministically, placing all sentry-
prefixed keys first (in alphabetical order), followed by other keys
(also alphabetically). This ensures:
1. Sentry distributed tracing members always appear first
2. The ordering is consistent across all invocations
3. The test 'sentry members appear before third party members' no
   longer exhibits flaky behavior

Co-Authored-By: Claude Sonnet 4.5 &lt;noreply@example.com&gt;
diff --git a/internal/otel/baggage/baggage.go b/internal/otel/baggage/baggage.go
@@ -21,6 +21,7 @@ import (
 	"fmt"
 	"net/url"
 	"regexp"
+	"sort"
 	"strings"
 	"unicode/utf8"
 
@@ -591,9 +592,36 @@ func (b Baggage) Len() int {
 // String encodes Baggage into a string compliant with the W3C Baggage
 // specification. The returned string will be invalid if the Baggage contains
 // any invalid list-members.
+//
+// Members are sorted deterministically with sentry- prefixed keys appearing
+// first, followed by other keys in lexicographic order. This ensures that
+// sentry distributed tracing members take precedence when W3C Baggage size
+// limits cause truncation from the end.
 func (b Baggage) String() string {
+	if len(b.list) == 0 {
+		return ""
+	}
+
+	// Collect keys and sort them: sentry- keys first, then others alphabetically
+	keys := make([]string, 0, len(b.list))
+	for k := range b.list {
+		keys = append(keys, k)
+	}
+	sort.Slice(keys, func(i, j int) bool {
+		isSentryI := strings.HasPrefix(keys[i], "sentry-")
+		isSentryJ := strings.HasPrefix(keys[j], "sentry-")
+		if isSentryI && !isSentryJ {
+			return true
+		}
+		if !isSentryI && isSentryJ {
+			return false
+		}
+		return keys[i] < keys[j]
+	})
+
 	members := make([]string, 0, len(b.list))
-	for k, v := range b.list {
+	for _, k := range keys {
+		v := b.list[k]
 		members = append(members, Member{
 			key:        k,
 			value:      v.Value,
