AskaEth
15181e7063
根因（agent-193，文件:行）：loop 的 tool_history 按调用逐条产出 RoleTool 消息， 而 anthropic_encode 把每条各自编码成一条独立 user 消息 ⇒ 上游只认紧邻 assistant 的那一条，第 2..N 个 tool_use id 被判「没有 tool_result」⇒ `messages.1: tool_use ids were found without tool_result blocks immediately after`。 单调用回合只有 1 条结果消息，所以长期潜伏；KB1 让 knowledge 真正成功后它才被暴露。 修法：编码器在**线格式**层把相邻的 RoleTool 合并进同一条 user 消息（块序 = 调用序）， 遇到任何非工具结果回合即复位（不同 assistant 回合的结果绝不混入同一条）。 未碰 openai_encode（OpenAI 要求每个 tool_call_id 一条独立 tool 消息）。 测试 3 个新文件：编码器线格式（非流式 + 流式）/ loop 端到端（真适配器打假上游）/ 回填侧契约；含配对断言（每个 tool_use 在下一条消息里有同 id 的 tool_result）。 变异：还原旧编码 ⇒ 4 条红，端到端逐字复现真机 400。 是否本轮引入：否（编码与回填两文件在本轮零改动，属基线潜伏缺陷）。
214 lines
7.0 KiB
Go
214 lines
7.0 KiB
Go
package loop
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import (
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"context"
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"encoding/json"
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"fmt"
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"io"
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"net/http"
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"net/http/httptest"
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"reflect"
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"sync"
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"testing"
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"git.yeij.top/AskaEth/Cyrene_v2/internal/config"
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"git.yeij.top/AskaEth/Cyrene_v2/internal/runtime/model"
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)
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// 本文件是并行工具调用的**端到端**回归：用真适配器（AnthropicChatAdapter）
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// 打一个按 Anthropic 契约校验请求的假上游，跑完整回合。真机 P0 的原文
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// 400 ——
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//
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// messages.1: `tool_use` ids were found without `tool_result` blocks
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// immediately after: call_01_1RBBKboRTJCl2S80t0FJ3155.
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// Each `tool_use` block must have a corresponding `tool_result` block in the next message.
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//
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// 在这里被逐字复刻：上游只放行「assistant 回合里的每个 tool_use 都在紧随其后的
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// 那一条消息里被同 id 的 tool_result 应答」的请求。因此修复一旦回退，本用例会
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// 以同一个 400 失败 —— 而不是靠对内部结构做断言。
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// strictAnthropicUpstream 是一个按 Anthropic 配对契约校验的假上游。
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// 它按顺序逐次回放 responses（超出则重复最后一条）。
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type strictAnthropicUpstream struct {
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mu sync.Mutex
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bodies []map[string]any
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responses []string
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srv *httptest.Server
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}
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func newStrictAnthropicUpstream(t *testing.T, responses ...string) *strictAnthropicUpstream {
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t.Helper()
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up := &strictAnthropicUpstream{responses: responses}
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up.srv = httptest.NewServer(http.HandlerFunc(func(w http.ResponseWriter, r *http.Request) {
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var body map[string]any
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_ = json.NewDecoder(r.Body).Decode(&body)
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up.mu.Lock()
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up.bodies = append(up.bodies, body)
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idx := len(up.bodies) - 1
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if idx >= len(up.responses) {
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idx = len(up.responses) - 1
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}
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payload := up.responses[idx]
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up.mu.Unlock()
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if bad := unpairedToolUse(body); bad != "" {
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w.Header().Set("Content-Type", "application/json")
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w.WriteHeader(http.StatusBadRequest)
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_, _ = fmt.Fprintf(w, `{"error":{"message":%q,"type":"invalid_request_error"}}`, bad)
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return
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}
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w.Header().Set("Content-Type", "application/json")
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w.WriteHeader(http.StatusOK)
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_, _ = io.WriteString(w, payload)
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}))
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t.Cleanup(up.srv.Close)
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return up
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}
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// requests 返回假上游收到的请求体快照数量。
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func (up *strictAnthropicUpstream) requests() int {
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up.mu.Lock()
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defer up.mu.Unlock()
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return len(up.bodies)
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}
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// bodyAt 返回第 i 个请求体。
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func (up *strictAnthropicUpstream) bodyAt(i int) map[string]any {
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up.mu.Lock()
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defer up.mu.Unlock()
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return up.bodies[i]
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}
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// unpairedToolUse 复刻上游的校验：对每条 assistant 消息，取出它声明的全部
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// tool_use id，要求紧随其后的那一条消息里出现**同 id、同数量、同顺序**的
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// tool_result 块；违反时返回与上游同形的错误文案（空串表示通过）。
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func unpairedToolUse(body map[string]any) string {
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raw, _ := body["messages"].([]any)
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for i, m := range raw {
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msg, _ := m.(map[string]any)
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uses := wireBlockIDs(msg, "tool_use")
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if len(uses) == 0 {
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continue
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}
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if i+1 >= len(raw) {
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return fmt.Sprintf("messages.%d:`tool_use` ids were found without `tool_result` blocks immediately after: %s",
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i, joinIDs(uses))
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}
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next, _ := raw[i+1].(map[string]any)
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results := wireBlockIDs(next, "tool_result")
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if !reflect.DeepEqual(uses, results) {
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return fmt.Sprintf("messages.%d:`tool_use` ids were found without `tool_result` blocks immediately after: %s",
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i, joinIDs(missingIDs(uses, results)))
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}
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}
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return ""
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}
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// wireBlockIDs 按出现顺序取出 assistant 的 tool_use id / 结果的 tool_result id。
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func wireBlockIDs(msg map[string]any, blockType string) []string {
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blocks, _ := msg["content"].([]any)
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ids := make([]string, 0, len(blocks))
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for _, b := range blocks {
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blk, _ := b.(map[string]any)
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if blk["type"] != blockType {
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continue
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}
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key := "id"
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if blockType == "tool_result" {
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key = "tool_use_id"
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}
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if s, ok := blk[key].(string); ok {
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ids = append(ids, s)
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}
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}
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return ids
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}
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// joinIDs 渲染 id 列表，供错误文案使用。
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func joinIDs(ids []string) string {
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out := ""
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for i, id := range ids {
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if i > 0 {
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out += ", "
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}
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out += id
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}
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return out
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}
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// missingIDs 返回 uses 里没有在 results 中被一一应答的 id。
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func missingIDs(uses, results []string) []string {
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answered := make(map[string]int, len(results))
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for _, r := range results {
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answered[r]++
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}
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missing := make([]string, 0, len(uses))
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for _, u := range uses {
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if answered[u] > 0 {
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answered[u]--
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continue
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}
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missing = append(missing, u)
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}
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return missing
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}
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// parallelToolUseResponse 是首步模型响应：**同一步并行发出两个** tool_use
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// （真机那次是两个 knowledge 调用）。id 采用 OpenAI 风格的
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// call_01_*，与真机日志一致。
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func parallelToolUseResponse() string {
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return `{
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"id":"msg_1","model":"claude-test","role":"assistant",
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"content":[
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{"type":"tool_use","id":"call_01_a","name":"knowledge","input":{"query":"Cyrene"}},
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{"type":"tool_use","id":"call_01_b","name":"datetime","input":{"timezone":"Asia/Shanghai"}}
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],
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"stop_reason":"tool_use",
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"usage":{"input_tokens":10,"output_tokens":8}
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}`
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}
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const parallelFollowUpResponse = `{
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"id":"msg_2","model":"claude-test","role":"assistant",
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"content":[{"type":"text","text":"两条记录都在。"}],
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"stop_reason":"end_turn",
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"usage":{"input_tokens":20,"output_tokens":5}
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}`
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// TestRun_ParallelToolCalls_SecondStepAccepted 端到端复现真机 P0 的场景：
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// 同一步并行两个 tool_use，回灌结果后必须能完成第二步模型调用并给出最终答复。
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// 修复回退时上游按原样回 400，本用例即以那条错误失败。
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func TestRun_ParallelToolCalls_SecondStepAccepted(t *testing.T) {
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t.Parallel()
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up := newStrictAnthropicUpstream(t, parallelToolUseResponse(), parallelFollowUpResponse)
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ad := model.NewAnthropicChat(config.LLMConfig{
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AnthropicBaseURL: up.srv.URL, AnthropicModel: "claude-test",
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})
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td := dispatcherWith(t,
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&mockTool{name: "knowledge", result: map[string]any{"count": 10}},
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&mockTool{name: "datetime", result: map[string]any{"now": "2026-10-04 22:17"}},
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)
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l := New(Options{Agent: newTestAgent(t, ad, td), MaxRounds: 8})
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res, err := l.Run(context.Background(), "sess", []model.Message{
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model.TextMessage(model.RoleUser, "请调用 knowledge 工具检索关于「Cyrene」的记录"),
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})
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if err != nil {
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t.Fatalf("Run: %v（旧行为下这里是上游 400: tool_use ids were found without tool_result blocks）", err)
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}
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if res.Text != "两条记录都在。" {
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t.Fatalf("Text = %q, want 两条记录都在。", res.Text)
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}
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if res.ToolCallsExecuted != 2 {
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t.Fatalf("ToolCallsExecuted = %d, want 2（两个并行调用都要执行）", res.ToolCallsExecuted)
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}
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if up.requests() != 2 {
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t.Fatalf("上游请求数 = %d, want 2（第二步必须真的发出去）", up.requests())
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}
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// 第二步请求的配对形态也要成立（不只是「没报错」）。
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if bad := unpairedToolUse(up.bodyAt(1)); bad != "" {
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t.Fatalf("第二步请求仍违反配对契约: %s", bad)
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}
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}
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