如何跳过子进程超时

import subprocess
import signal
import select

def timeout_handler(signum, frame):
    # 超时处理函数，用于终止子进程
    raise TimeoutError("子进程超时")

def run_command_with_timeout(command, timeout):
    # 设置超时处理函数
    signal.signal(signal.SIGALRM, timeout_handler)
    # 设置超时时间
    signal.alarm(timeout)

    try:
        # 创建子进程并执行命令
        process = subprocess.Popen(command, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
        # 监听子进程的输出和错误流
        while process.poll() is None:
            # 使用select函数设置超时时间
            ready, _, _ = select.select([process.stdout, process.stderr], [], [], timeout)
            if not ready:
                # 超时时间到达，终止子进程
                process.terminate()
                raise TimeoutError("子进程超时")

        # 取消定时器
        signal.alarm(0)

        # 获取子进程的输出和错误信息
        stdout, stderr = process.communicate()
        return stdout.decode(), stderr.decode()

    except TimeoutError as e:
        return None, str(e)

# 示例调用
command = ["your_command", "arg1", "arg2"]
timeout = 10  # 超时时间为10秒
stdout, stderr = run_command_with_timeout(command, timeout)
if stderr:
    print("子进程执行出错:", stderr)
else:
    print("子进程执行结果:", stdout)