面试题库

30 天从前端工程师到 Agent 工程师

D2 工具调用原理：JSON Schema、tool_use 循环；不用框架手写 Agent Loop

• 怎么防止 Agent 循环停不下来？只加一个最大步数够吗？How do you keep an agent loop from running forever — is a max-step counter enough?

  国内高频海外高频深入#agent-loop#reliability#cost

  分析过程 · 先想清楚再作答

  1. 后半句是明摆着的陷阱。只答「加一个计数器」是及格线，面试官真正想听的是你知道计数器拦不住什么。
  2. 先解释它为什么会停不下来：停止原因一直是 tool_calls，通常是因为工具返回的东西没帮模型前进——结果为空、字段答非所问、错误文案没说清该怎么改，于是它换个参数一试再试。所以第一层其实不是护栏，是把工具的返回值和错误文案写得有信息量。
  3. 再给硬护栏，三条互补：步数上限最直接；token 与成本预算拦的是「步数不多但每步都很贵」；单轮的墙上时钟超时拦的是「一步就卡了两分钟」。只有步数上限的系统，照样会被一次超长上下文的调用打爆预算。
  4. 语义层面再加一条：检测重复调用。同一个工具、同一份参数连续出现两次以上，几乎可以断定它在原地打转，直接截断并把「你已经用完全相同的参数调过这个工具了，换个思路或者告诉用户你做不到」回传给模型，往往比等步数耗尽更快收敛。
  5. 触顶之后必须有交代：不能静默返回空字符串，要给用户一句能理解的话；同时把触顶记成一个指标，触顶率上升通常意味着某个工具的描述或返回值该改了，而不是把上限调大。
  6. 可以预期的追问：上限设多少？没有普适值。聊天类任务 5 到 10 步通常够，需要多轮检索的任务可以更高。正确做法是看线上的步数分布，取 p99 再留一点余量，而不是拍脑袋——上限设得越死，你的系统就越靠近固定流程那一端，越不像一个 Agent。

  How to reason about it · think before answering

  1. The second half is an open trap. 'Add a counter' is the passing grade; what they want is whether you know what a counter cannot catch.
  2. Explain why it runs away first: the finish reason stays tool_calls because the tool results are not moving the model forward — empty results, fields that do not answer the question, error text that never says what correct looks like. So the first line of defense is not a guard rail at all; it is writing tool results and error messages that carry information.
  3. Then three complementary hard limits: a step cap is the obvious one; a token and cost budget catches 'few steps, all of them expensive'; a per-step wall-clock timeout catches 'one call hung for two minutes'. A system with only a step cap can still blow its budget on a single enormous context.
  4. Add a semantic guard: detect repeats. The same tool with identical arguments twice in a row is almost always spinning. Cut it short and tell the model so — 'you already called this tool with exactly these arguments' — which usually converges faster than waiting for the counter to run out.
  5. Hitting the cap needs an honest ending: never return an empty string, give the user a sentence they can act on, and record cap hits as a metric. A rising cap-hit rate usually means a tool's description or return value needs fixing, not that the cap should be raised.
  6. Expect: what number do you pick? There is no universal one. Chat-style tasks usually fit in five to ten steps; retrieval-heavy tasks need more. Read the production distribution, take p99 plus headroom, and remember that the tighter the cap, the closer your system sits to a fixed workflow rather than an agent.

  答题要点

  • 根因通常是工具返回值或错误文案没信息量，模型无法前进只能反复重试，先把这层写好
  • 三条硬护栏互补：最大步数、token 与成本预算、单步墙上时钟超时，只有步数上限并不够
  • 语义护栏：同一工具加同一份参数连续重复调用即判定原地打转，截断并把这个事实回传给模型
  • 触顶要给用户一句交代，不能静默返回空；同时把触顶率当指标，上升说明工具该改而不是把上限调大
  • 上限值按线上步数分布取 p99 加余量；上限越死越接近固定流程，越不像 Agent

  Key points

  • The root cause is usually uninformative tool results or error text, so fix that layer before adding guards
  • Three complementary hard limits: max steps, a token and cost budget, and a per-step wall-clock timeout
  • Add a semantic guard: identical tool plus identical arguments twice in a row means it is spinning — cut it and tell the model
  • Give the user an honest message when the cap is hit, and track the cap-hit rate as a signal that a tool needs fixing
  • Size the cap from the production step distribution, not intuition; a tighter cap makes the system a workflow rather than an agent

D4 模型接入与系统提示词：多 provider 抽象与 fallback、覆盖默认人设（dg P03/P04/M04）

• 设计模型 fallback 策略时要权衡哪些因素？What trade-offs shape a model fallback strategy?

  国内高频海外高频进阶#model-routing#reliability#cost

  分析过程 · 先想清楚再作答

  1. 题眼在「权衡」两个字——面试官不要你背一个 for 循环，他要看你知不知道 fallback 是有代价的。
  2. 先拆出第一个关键判断：不是所有错误都该 fallback，而分类的依据不是状态码的首位数字，是「换一家有没有可能变好」。400 请求体不合法、403 被安全策略拦截，换谁都一样，重试只是把同一个 bug 再犯一遍、白花两倍的钱和时间；408 超时、429 限流、5xx 服务端故障是对方的问题，换一家大概率能成。最容易答错的是 402 余额不足和 404 模型被下线或改名——它们同属 4xx、长得像「你的问题」，其实换一家完全可能成功；401 则要看凭证怎么管，三家共用一把网关 key 时换了也没用，各有各的 key 时 A 被吊销切到 B 完全能救。分类是 fallback 的第一步，不是重试。
  3. 再说成本：切换意味着同一段 prompt 你付了两次钱，三家链路最坏是三倍成本。这条一定要主动说出来，它区分了「写过」和「上过线」。
  4. 然后是延迟：串行 fallback 的总耗时是各家超时值的累加。如果每家给 15 秒、三家串下来用户要等 45 秒，那还不如早点失败。所以超时值必须按 provider 分别设，且要设总预算上限。
  5. 最后是雪崩，这是最容易被追问的点：主 provider 限流时你把全部流量瞬间压到备用上，很可能把备用也压垮。所以要加熔断——连续失败 N 次就暂时摘掉该 provider，过一段时间放少量流量试探。
  6. 可以预期的追问：怎么知道该摘多久？答案是指数退避 + 半开状态试探，和数据库连接池的熔断是同一套思路。

  How to reason about it · think before answering

  1. The word 'trade-offs' is the hinge: they are not asking for a for-loop, they want to know you understand fallback has costs.
  2. First key judgment: not every error deserves a fallback, and the test is not the leading digit of the status code but whether another provider could plausibly succeed. A 400 (malformed body) or 403 (blocked by safety policy) fails everywhere, so retrying repeats your own bug at double the cost and latency; a 408, 429 or 5xx is theirs and usually succeeds elsewhere. The two that people get wrong are 402 (out of credit) and 404 (model retired or renamed): both are 4xx, both look like your fault, and both are fixed by switching. A 401 depends on how credentials are managed — one shared gateway key fails everywhere, but per-provider keys mean a revoked key on A is survivable on B. Classification comes before retry.
  3. Cost: switching means paying for the same prompt twice, up to 3x across a three-provider chain. Volunteering this separates people who shipped from people who only read about it.
  4. Latency: serial fallback accumulates timeouts. Three providers at 15s each means a 45s wait — worse than failing fast. Timeouts must be per-provider with an overall budget.
  5. Thundering herd is the most common follow-up: when the primary rate-limits, shifting all traffic at once can take down the backup too. Hence circuit breaking — drop a provider after N consecutive failures, then probe with a trickle.
  6. Expect: how long do you drop it for? Exponential backoff with a half-open probe — the same pattern as database connection pool breakers.

  答题要点

  • 先分类再重试，判据是「换一家有没有可能变好」而不是状态码首位：400/403 不该切，408/429/5xx 该切，402 余额不足和 404 模型下线同样该切，401 取决于三家是否共用同一把凭证
  • 成本：每次 fallback 都要重付一遍 prompt 的钱，链路越长最坏成本越高
  • 延迟：串行 fallback 的耗时是各超时值累加，必须按 provider 分设超时并设总预算
  • 雪崩防护：主 provider 故障时全量流量压向备用会把备用也压垮，需要熔断 + 指数退避 + 半开试探

  Key points

  • Classify before retrying, judging by whether another provider could plausibly succeed rather than the leading digit: 400/403 must not fail over; 408/429/5xx should; so should 402 (out of credit) and 404 (model retired); 401 depends on whether the providers share one key
  • Cost: every fallback re-pays for the same prompt, so worst-case cost scales with chain length
  • Latency: serial fallback sums the timeouts, so set per-provider timeouts plus an overall budget
  • Thundering herd: shifting full traffic to the backup can topple it too — use circuit breaking with exponential backoff and half-open probes
• 如何在成本和延迟之间给不同任务选择合适的模型？How do you pick the right model per task, balancing cost against latency?

  国内高频海外高频进阶#model-routing#cost#latency

  分析过程 · 先想清楚再作答

  1. 这题考的是「你有没有真的在花自己的钱」。答「用最好的模型」是最差的答案，答「按需选择」太空，要给出可执行的分档维度。
  2. 先建立核心事实：不同模型的价格能差 50 倍以上，而你的任务里很大一部分根本不需要最强的模型。用旗舰模型做意图识别，等于开跑车去楼下取快递。
  3. 然后给出三个可操作的路由维度：任务类型（分类抽取走便宜模型，长文推理走强模型）、延迟要求（前台用户在等就走低延迟，后台批处理可以慢而便宜）、输入长度（超长上下文只有部分模型支持且价格陡增）。
  4. 结论要落到数字上才有说服力：1 万轮对话每轮 2000 token，全走旗舰约 300 元一天；把六成粗活改走小模型后降到 125 元左右，一年省六万多，用户感知不到差别。
  5. 还要主动说出实现上的取舍：先按任务类型静态分档，不要一上来就做「让模型判断该用哪个模型」的动态路由——那个方案本身又要多一次模型调用，延迟和成本可能得不偿失，等有真实数据再优化。
  6. 可以预期的追问：怎么验证降档没有损失质量？答案是准备 golden set，对同一批输入跑两档模型，用人工或 LLM-as-judge 比对准确率，把降档决策建立在数据上而不是感觉上。

  How to reason about it · think before answering

  1. This question tests whether you have ever spent your own money. 'Use the best model' is the worst answer; 'it depends' is too vague — give actionable routing dimensions.
  2. Establish the core fact: model pricing spans 50x or more, and much of your workload does not need the strongest model. Using a flagship for intent detection is driving a sports car to fetch a parcel downstairs.
  3. Give three routing dimensions: task type (classification and extraction go cheap, long-form reasoning goes strong), latency requirement (foreground users need low latency, background batches can be slow and cheap), and input length (only some models handle very long context, and pricing rises steeply).
  4. Quantify it: 10k conversations a day at 2000 tokens each costs roughly 300 CNY/day on a flagship; routing the 60% of grunt work to a small model drops it to about 125 CNY/day, saving 60k+ CNY a year with no perceptible quality change.
  5. Volunteer the implementation trade-off: start with static tiers by task type. Dynamic routing that asks a model which model to use adds another model call, and the latency and cost may not pay for themselves — optimize once you have real data.
  6. Expect: how do you verify the cheaper tier did not hurt quality? A golden set — run both tiers over the same inputs and compare with human or LLM-as-judge scoring, so the decision rests on data rather than vibes.

  答题要点

  • 不同模型价格能差 50 倍以上，用旗舰模型做意图识别是明显的浪费
  • 三个路由维度：任务类型（分类抽取 vs 推理生成）、延迟要求（前台 vs 后台）、输入长度（是否需要超长上下文）
  • 实现上给调用层加 tier 参数，按任务静态分档挑起始 provider，fallback 逻辑完全复用
  • 先静态分档再考虑动态路由，让模型判断该用哪个模型本身要多一次调用，可能得不偿失
  • 用 golden set 对比两档模型的准确率，把降档决策建立在数据上

  Key points

  • Model pricing spans 50x or more, so a flagship doing intent detection is obvious waste
  • Three routing dimensions: task type, latency requirement, and input length
  • Add a tier parameter to the call layer, pick the starting provider statically, and reuse the fallback chain
  • Prefer static tiers first — dynamic model-picks-model routing adds a call and may not pay off
  • Validate downgrades against a golden set rather than intuition

D6 消息、上下文工程与压缩、会话存储/恢复/分叉（dg M06/M08/M09/M10）

• 长对话里上下文放不下了，你会怎么压缩？什么时候触发、压掉什么、保留什么？When a long conversation outgrows the context window, how do you compress it — when do you trigger, what do you drop, and what do you keep?

  国内高频海外高频进阶#context-engineering#compression#cost

  分析过程 · 先想清楚再作答

  1. 这题的区分度不在「用摘要」三个字上，几乎人人都答得出。区分度在你有没有说出触发时机和保留清单——只答「让模型总结一下前面的对话」的，面试官会判定你没在长对话上线过。
  2. 先把问题拆成三问再逐个答：什么时候压、压掉什么、保留什么。这个拆法本身就是加分项，因为它说明你把压缩当成一个策略而不是一个函数。
  3. 触发用阈值不用定时器，也不能等报错。给一个具体数字并解释它：历史占用到预算的七成就动手，因为摘要本身是一次模型调用，有延迟也可能失败，卡到九成再压，一旦摘要超时下一轮就直接撞窗口上限了——七成是留给自己的抢救时间。
  4. 压掉的是过程性内容：中间推理、已经被消费完的工具原始返回值、用户后来推翻的需求。它们的共同点是价值已经沉淀进后面的结论里。保留的是系统提示词（它不属于历史）、最近若干条原文、以及用户明确声明过的约束和事实——后者写错了模型会当场失忆。
  5. 再补一条别人不会说的：切口必须对齐到一轮的开头。切在 assistant 的 tool_calls 和对应的 tool 结果中间，下一次请求就有了悬空调用，多数厂商的 API 直接返回 400。这一条最能证明你真的调过。
  6. 可以预期的追问有两个。一是摘要该用哪个模型：用便宜的小模型就行，摘要是抽取任务不是推理任务，这也接上了 D4 的分层路由。二是摘要调用失败了怎么办：降级到不摘要的滑动窗口（直接丢最早的几轮），保证请求发得出去，别让压缩失败连带整轮对话失败。

  How to reason about it · think before answering

  1. Saying 'summarize it' earns nothing — everyone says that. The signal is whether you name a trigger point and a keep-list; without those you sound like someone who never ran a long conversation in production.
  2. Split it into three questions before answering: when to compress, what to drop, what to keep. The split itself scores, because it frames compression as a policy rather than a function.
  3. Trigger on a threshold, not a timer, and never on an error. Give a number and justify it: compress at roughly 70% of the history budget, because summarizing is itself a model call that can be slow or fail. Waiting until 90% means one timed-out summary call and the next turn slams into the window limit.
  4. Drop the process: intermediate reasoning, raw tool payloads already consumed, requirements the user later reversed — their value has already settled into later conclusions. Keep the system prompt (it is not history), the most recent turns verbatim, and any constraint or fact the user stated explicitly. Getting that last one wrong makes the model visibly forget.
  5. Add the detail others miss: the cut must land on a turn boundary. Slicing between an assistant tool_calls message and its matching tool result leaves a dangling call, and most providers reject that request with a 400. This is the line that proves hands-on experience.
  6. Two follow-ups to expect. Which model summarizes? A cheap small one — summarization is extraction, not reasoning, which ties back to tiered routing. And what if the summary call fails? Degrade to a plain sliding window that drops the oldest turns, so a failed compression never fails the whole turn.

  答题要点

  • 阈值触发：历史占用到预算七成就压，因为摘要本身是一次会失败、有延迟的模型调用，必须留抢救余量
  • 压过程、留结论：丢中间推理和已消费的工具原始返回，保留系统提示词、最近若干条原文、用户明确声明的约束与事实
  • 切口必须对齐到一轮开头，切在 tool_calls 与 tool 结果之间会让下一次请求返回 400
  • 压缩是有损且不可逆的：原始历史另存一份只追加，发给模型的是压缩版，需要回溯或分叉时读原始版
  • 摘要用便宜的小模型；摘要失败要能降级成滑动窗口，别让压缩失败连累整轮对话

  Key points

  • Threshold-triggered at about 70% of the history budget, because the summary call itself is a slow, fallible model call that needs headroom
  • Drop process, keep conclusions: discard intermediate reasoning and consumed raw tool payloads; keep the system prompt, the recent turns verbatim, and explicit user constraints and facts
  • Align the cut to a turn boundary — slicing between tool_calls and its tool result makes the next request fail with a 400
  • Compression is lossy and irreversible: keep an append-only original, send the compressed version, and read the original when you need to backtrack or fork
  • Summarize with a cheap small model, and degrade to a sliding window if the summary call fails so compression failure never fails the turn