Python 避免孤儿进程

Question

我正在使用 python 进行基准测试。这可能会花费大量时间，我想设置一个(全局)超时。我使用以下脚本(总结):

class TimeoutException(Exception):
    pass
def timeout_handler(signum, frame):
    raise TimeoutException()

# Halt problem after half an hour
signal.alarm(1800)
try:
    while solution is None:
        guess = guess()
        try:
            with open(solutionfname, 'wb') as solutionf:
                solverprocess = subprocess.Popen(["solver", problemfname], stdout=solutionf)
                solverprocess.wait()
        finally:
            # `solverprocess.poll() == None` instead of try didn't work either
            try:
                solverprocess.kill()
            except:
                # Solver process was already dead
                pass
except TimeoutException:
    pass
# Cancel alarm if it's still active
signal.alarm(0)

然而，它有时会不断产生孤儿进程，但我无法可靠地重现这种情况。有谁知道防止这种情况的正确方法是什么？

Answer 1

您只需在终止进程后等待即可。