在 PYTHON 中使用 SOCKET 的线程中不能引发异常

Question

我尝试从 Python 线程引发异常。如果线程内部有一个套接字接收，我就不会这样做。

这里是客户端的代码：

from concurrent import futures
import traceback, time, socket

def cpu_process(thread_idx: int):
    socket_file = 'socket0'

    with socket.socket(socket.AF_UNIX, socket.SOCK_STREAM) as gpu_client:
        gpu_client.connect(socket_file)
        data_len_bin = int(123).to_bytes(4, 'big')
        gpu_client.sendall(data_len_bin)
        response_bytes = gpu_client.recv(1024) # If you comment this line, everything will run as expected.

    time.sleep(4)
    return thread_idx

def moderate():
    future_results = []
    try:
        with futures.ThreadPoolExecutor(max_workers=4) as executor:
            for thread_idx in range(8):
                future_results.append(executor.submit(cpu_process, thread_idx))

            for future in future_results:
                try:
                    response = future.result(timeout=3)
                except Exception as e:
                    print(f"Inner exception traceback:\n {traceback.format_exc()}")
                    raise Exception('CPU Process Exception').with_traceback(e.__traceback__)
    
    except Exception as e:
        print("Exterior exception")
        return 'Fail'
    return 'Success'

if __name__ == "__main__":
    moderate() # "Exterior exception"

我有一个监听端口但没有返回任何内容的服务器。如果我评论上面代码中的

response_bytes = gpu_client.recv(1024)

行，一切顺利，最后打印

Exterior exception

。同时，当前代码从不打印

Exterior exception

.

我的问题是：

1. 为什么即使我提出 TimeoutError

   gpu_client.recv(1024)

   也会阻止线程？
2. 克服障碍的最佳方法是什么？

非常感谢你。