python - Python不抛出错误或异常，websocket on_message

Question

当我将太多参数传递给f()时，为什么python不抛出错误或异常？显然，由于该错误，未执行f()。我正在使用VS-Code和python 3.8。 (如果要复制，则必须安装websocket_client软件包)

import websocket
import json

connection = "wss://ws-feed.pro.coinbase.com"
subscription = json.dumps({
    "type": "subscribe",
    "channels": [
        {
            "name": "level2",
            "product_ids": ["BTC-USD"]
        }
    ]
})

def f(msg):
    print(msg)

def ws_message(ws, message):
    f("hi",True) #<--should throw a too many arguments error/exception
    

def ws_open(ws):
    ws.send(subscription)

def ws_thread(*args):
    ws = websocket.WebSocketApp(
        connection,  on_message=ws_message, on_open=ws_open)
    ws.run_forever()

ws_thread()

Answer 1

这是由于实现了here的故障安全机制所致。如果您想跟踪异常，则应为记录器设置适当的级别。为此，请执行以下操作:

import websocket
websocket._logging._logger.level = -99 # This will enable all levels of logging

# Rest of your code goes here

这样，运行代码后，您将看到:

  File "blablabla/.local/lib/python3.7/site-packages/websocket/_app.py", line 346, in _callback
    callback(self, *args)
  File "blablabla/test/test.py", line 22, in ws_message
    f("hi", True) #<--should throw a too many arguments error/exception

您的代码仍将运行，但是您将能够捕获异常