Identify the error in this FastAPI code snippet: ```python from fastapi import FastAPI, HTTPException app = FastAPI() @app.get("/users/{user_id}") async def get_user(user_id: int): try: if user_id < 0: raise HTTPException(status_code=400, detail="Negative ID") except HTTPException: return {"error": "Invalid user ID"} ```

medium📝 Debug Q6 of 15

FastAPI - Error Handling

ARaising HTTPException inside try block is invalid syntax.

BCatching HTTPException and returning dict bypasses proper HTTP error response.

CThe endpoint should not use async def.

DThe route path is missing a trailing slash.

Step-by-Step Solution

Solution:

Step 1: Analyze exception handling logic
The code catches HTTPException and returns a plain dict instead of raising it.
Step 2: Understand FastAPI error response mechanism
Returning a dict bypasses FastAPI's automatic HTTP error response formatting, losing status code 400.
Final Answer:
Catching HTTPException and returning dict bypasses proper HTTP error response. -> Option B
Quick Check:
Catch then return dict = no proper HTTP error [OK]

Quick Trick: Raise HTTPException, don't catch and return dict [OK]

Common Mistakes:

MISTAKES

Thinking raising HTTPException inside try is invalid
Believing async def is wrong here
Confusing route path syntax

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