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7 天 MCP：把工具接进任何 Agent

D5 写一个 MCP client：在自己的 Agent 循环里发现并调用工具、多 server 聚合与命名冲突

• 线上一个 MCP 服务端超时了。你的 Agent 循环应该怎么反应？One of your MCP servers times out in production. How should your agent loop react?

  国内高频海外高频进阶#client#reliability

  分析过程 · 先想清楚再作答

  1. 这题看的是工程直觉：能不能把「一个依赖挂了」和「这一轮对话失败」分开。答「重试三次」是把问题往后推了一步，面试官会立刻追问重试期间用户在等什么。
  2. 先分阶段。超时发生在两个完全不同的时刻：发现阶段（server/discover 或 tools/list）和调用阶段（tools/call）。两个阶段的正确反应不一样，混着答就会露怯。
  3. 发现阶段：逐个服务端 try/catch，失败的记进一张掉线表并继续下一个。整张工具表少几个工具，但循环照常起得来。记的必须是原因而不是一个布尔值，因为事后你要能回答少了什么、为什么少。
  4. 调用阶段：把失败翻译成一条 isError 为真的工具结果喂回模型，不要抛。理由是 MCP 本来就用 isError 表达「工具执行失败但协议是成功的」，模型看得见这句话就有机会换个工具或换个参数；抛出去只会把整轮对话打断，而且用户什么解释都得不到。
  5. 接着补三件配套的事。一是**每条请求都必须有超时**，stdio 上服务端不回你就永远不回；二是**幂等性决定能不能重试**，工具注解里的 idempotentHint 是提示不是保证，写操作的重试要靠客户端自己的去重键；三是**掉线要让用户看得见**，把掉线的服务端标在界面上或写进系统提示，否则模型会表现得像那个能力从来不存在，一本正经地说查不到。
  6. 结论：一个服务端超时，最坏的后果应该是少几个工具加一条明确的说明，而不是这一轮对话失败。
  7. 可预期的追问：要不要熔断？连续失败到阈值就把这个服务端标记为不可用一段时间，避免每一轮都白等一次超时；恢复用探活或下一次会话重连。再追问会问到超时值怎么定——按工具而不是按服务端定，一个跑三十秒的分析工具和一个查缓存的工具不该共用一个阈值。

  How to reason about it · think before answering

  1. This probes engineering instinct: can you separate 'one dependency is down' from 'this turn fails'. Answering 'retry three times' just moves the problem, and the interviewer will ask what the user is staring at meanwhile.
  2. Split by phase first. A timeout happens at two very different moments: discovery (server/discover or tools/list) and invocation (tools/call). The correct reaction differs, and blurring them shows you have not built this.
  3. Discovery: wrap each server in its own try/catch, record the failure with its reason in a down list, and continue to the next server. The tool table loses a few entries but the loop still starts. Record the reason, not a boolean, because afterwards you must be able to say what is missing and why.
  4. Invocation: translate the failure into a tool result with isError true and feed it back to the model rather than throwing. MCP already uses isError for 'the tool failed but the protocol succeeded', so the model can switch tools or arguments; throwing kills the turn and leaves the user with no explanation.
  5. Then three supporting points. Every request needs a timeout, because on stdio a silent server is silent forever. Idempotency decides whether a retry is safe, and the idempotentHint annotation is a hint, not a guarantee, so writes need a client-side dedup key. And outages must be visible, surfaced in the UI or in the system prompt, or the model will behave as if the capability never existed and confidently report nothing found.
  6. Conclusion: the worst outcome of one server timing out should be a few missing tools plus an explicit note, never a failed turn.
  7. Likely follow-ups: should you add a circuit breaker? Yes, after consecutive failures mark the server unusable for a while so you stop paying a timeout every turn, with recovery by health check or reconnect on the next session. And how do you set the timeout? Per tool rather than per server, since a thirty-second analysis tool and a cache lookup should not share a threshold.

  答题要点

  • 分阶段：发现阶段逐个服务端 try/catch 记进掉线表并继续，调用阶段一律不抛
  • 调用失败翻译成 isError 为真的工具结果喂回模型，让它换工具或换参数
  • 每条请求必须设超时；能不能重试取决于幂等性，注解只是提示不是保证
  • 掉线必须对用户和模型可见，否则会变成静默降级，模型会假装那个能力不存在

  Key points

  • Split by phase: per-server try/catch during discovery with a recorded reason, and never throw during invocation
  • Translate call failures into isError tool results so the model can switch tools or arguments
  • Every request needs a timeout; retry safety depends on idempotency, and the annotation is a hint, not a guarantee
  • Outages must be visible to user and model, otherwise silent degradation makes the model deny the capability ever existed