为什么这个子进程命令没有正确超时？

Question

所以我有这个简单的 POC

import subprocess

proc = subprocess.run('cmd', timeout=1, shell=True, capture_output=True)

在给定的时间间隔（本例中为 1 秒）之后，它不会引发 TimeoutError

subprocess.run()

中的超时标志在 Linux 中工作得很好，但在 Windows 中似乎会产生错误

例如

import subprocess

proc = subprocess.run('ping -n 10 127.0.0.1', shell=True, capture_output=True, timeout=1)

似乎不会引发超时错误直到命令完成后

Answer 1

评论太长...作为仅供参考

PS C:\Users\OWNER> type .\Documents\t.py
import subprocess
roc = subprocess.run('ping -n 10 127.0.0.1', shell=True, capture_output=True, timeout=1)

PS C:\Users\OWNER> python .\Documents\t.py
python : Traceback (most recent call last):
At line:1 char:1
+ python .\Documents\t.py
+ ~~~~~~~~~~~~~~~~~~~~~~~
    + CategoryInfo          : NotSpecified: (Traceback (most recent call last)::String) [], RemoteException
    + FullyQualifiedErrorId : NativeCommandError
 
  File "C:\Users\OWNER\Documents\t.py", line 2, in <module>
    roc = subprocess.run('ping -n 10 127.0.0.1', shell=True, capture_output=True, timeout=1)
          ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "C:\Program Files\WindowsApps\PythonSoftwareFoundation.Python.3.12_3.12.2032.0_x64__qbz5n2kfra8p0\Lib\subprocess.py", line 550, in run
    stdout, stderr = process.communicate(input, timeout=timeout)
                     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "C:\Program Files\WindowsApps\PythonSoftwareFoundation.Python.3.12_3.12.2032.0_x64__qbz5n2kfra8p0\Lib\subprocess.py", line 1209, in communicate
    stdout, stderr = self._communicate(input, endtime, timeout)
                     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
  File "C:\Program Files\WindowsApps\PythonSoftwareFoundation.Python.3.12_3.12.2032.0_x64__qbz5n2kfra8p0\Lib\subprocess.py", line 1630, in _communicate
    raise TimeoutExpired(self.args, orig_timeout)
subprocess.TimeoutExpired: Command 'ping -n 10 127.0.0.1' timed out after 1 seconds