在 SMACH 并发容器的不同状态下使用相同的数据
Using same data in different states of SMACH Concurrent-container
假设我们有一个并发 SMACH 容器 sm_con,其中包括两个状态机 SM1 和 SM2。我需要找到一种方法让 SM1 持续更新一些数据,并让 SM2 访问(并最终修改)相同的数据。我考虑通过将 sm_con 的用户数据传递给 SM1 和 SM2 来解决这个问题输入和输出键希望如果 SM1 修改数据它会自动覆盖 sm_cons 用户数据(有点像使用c++ 中的指针)但这不起作用。
相应的代码如下所示:
import smach
import smach_ros
import rospy
class st1(smach.State):
def __init__(self, outcomes=['successful', 'preempted']):
smach.State.__init__(self, outcomes, input_keys=['in_test'], output_keys=['out_test'])
def execute(self, userdata):
if self.preempt_requested():
self.service_preempt()
return 'preempted'
rospy.logerr('test 1: '+str(userdata.in_test))
userdata.out_test=userdata.in_test+1
return 'successful'
class st2(smach.State):
def __init__(self, outcomes=['successful', 'preempted']):
smach.State.__init__(self, outcomes, input_keys=['in_test'])
def execute(self, userdata):
#time.sleep(2)
if self.preempt_requested():
self.service_preempt()
return 'preempted'
rospy.logerr('test 2: ' + str(userdata.in_test))
return 'successful'
if __name__=="__main__":
rospy.init_node('test_state_machine')
sm_con = smach.Concurrence(outcomes=['success'],
default_outcome='success'
)
sm_con.userdata.testdata = 0
with sm_con:
sm_1 = smach.StateMachine(outcomes=['success', 'preempted'], input_keys=['testdata'], output_keys=['testdata'])
with sm_1:
smach.StateMachine.add('ST1', st1(),
remapping={'in_test': 'testdata', 'out_test': 'testdata'},
transitions={'successful': 'ST1'})
sm_2 = smach.StateMachine(outcomes=['success', 'preempted'], input_keys=['testdata'])
with sm_2:
smach.StateMachine.add('ST2', st2(),
remapping={'in_test':'testdata'},
transitions={'successful': 'ST2'})
smach.Concurrence.add('SM1', sm_1)
smach.Concurrence.add('SM2', sm_2)
# Execute SMACH plan
outcome = sm_con.execute()
print('exit-outcome:' + outcome)
# Wait for ctrl-c to stop the application
rospy.spin()
运行 这段代码,输出 'test 1: ...' 表明在 SM1 中用户数据得到递增,而输出 'test 2: ...' 显示 SM2 不访问递增的数据,因为输出保持不变 0.
如何修改SM1中的一些数据并在SM2中访问修改后的数据?
我找到了一个解决方法,使用here所述的可变对象。
应用上面的代码,它看起来像下面这样:
import smach
import smach_ros
import rospy
class st1(smach.State):
def __init__(self, outcomes=['successful', 'preempted']):
smach.State.__init__(self, outcomes, input_keys=['in_test'])
def execute(self, userdata):
if self.preempt_requested():
self.service_preempt()
return 'preempted'
rospy.logerr('test 1: '+str(userdata.in_test))
userdata.in_test[0]=userdata.in_test[0]+1
return 'successful'
class st2(smach.State):
def __init__(self, outcomes=['successful', 'preempted']):
smach.State.__init__(self, outcomes, input_keys=['in_test'])
def execute(self, userdata):
#time.sleep(2)
if self.preempt_requested():
self.service_preempt()
return 'preempted'
rospy.logerr('test 2: ' + str(userdata.in_test[0]))
return 'successful'
if __name__=="__main__":
rospy.init_node('test_state_machine')
sm_con = smach.Concurrence(outcomes=['success'],
default_outcome='success'
)
sm_con.userdata.testdata = [0]
with sm_con:
sm_1 = smach.StateMachine(outcomes=['success', 'preempted'], input_keys=['testdata'])
with sm_1:
smach.StateMachine.add('ST1', st1(),
remapping={'in_test': 'testdata'},
transitions={'successful': 'ST1'})
sm_2 = smach.StateMachine(outcomes=['success', 'preempted'], input_keys=['testdata'])
with sm_2:
smach.StateMachine.add('ST2', st2(),
remapping={'in_test':'testdata'},
transitions={'successful': 'ST2'})
smach.Concurrence.add('SM1', sm_1)
smach.Concurrence.add('SM2', sm_2)
# Execute SMACH plan
outcome = sm_con.execute()
print('exit-outcome:' + outcome)
# Wait for ctrl-c to stop the application
rospy.spin()
由于这只是一种解决方法,请参阅我相应的 issue-post here 了解更多信息。
假设我们有一个并发 SMACH 容器 sm_con,其中包括两个状态机 SM1 和 SM2。我需要找到一种方法让 SM1 持续更新一些数据,并让 SM2 访问(并最终修改)相同的数据。我考虑通过将 sm_con 的用户数据传递给 SM1 和 SM2 来解决这个问题输入和输出键希望如果 SM1 修改数据它会自动覆盖 sm_cons 用户数据(有点像使用c++ 中的指针)但这不起作用。
相应的代码如下所示:
import smach
import smach_ros
import rospy
class st1(smach.State):
def __init__(self, outcomes=['successful', 'preempted']):
smach.State.__init__(self, outcomes, input_keys=['in_test'], output_keys=['out_test'])
def execute(self, userdata):
if self.preempt_requested():
self.service_preempt()
return 'preempted'
rospy.logerr('test 1: '+str(userdata.in_test))
userdata.out_test=userdata.in_test+1
return 'successful'
class st2(smach.State):
def __init__(self, outcomes=['successful', 'preempted']):
smach.State.__init__(self, outcomes, input_keys=['in_test'])
def execute(self, userdata):
#time.sleep(2)
if self.preempt_requested():
self.service_preempt()
return 'preempted'
rospy.logerr('test 2: ' + str(userdata.in_test))
return 'successful'
if __name__=="__main__":
rospy.init_node('test_state_machine')
sm_con = smach.Concurrence(outcomes=['success'],
default_outcome='success'
)
sm_con.userdata.testdata = 0
with sm_con:
sm_1 = smach.StateMachine(outcomes=['success', 'preempted'], input_keys=['testdata'], output_keys=['testdata'])
with sm_1:
smach.StateMachine.add('ST1', st1(),
remapping={'in_test': 'testdata', 'out_test': 'testdata'},
transitions={'successful': 'ST1'})
sm_2 = smach.StateMachine(outcomes=['success', 'preempted'], input_keys=['testdata'])
with sm_2:
smach.StateMachine.add('ST2', st2(),
remapping={'in_test':'testdata'},
transitions={'successful': 'ST2'})
smach.Concurrence.add('SM1', sm_1)
smach.Concurrence.add('SM2', sm_2)
# Execute SMACH plan
outcome = sm_con.execute()
print('exit-outcome:' + outcome)
# Wait for ctrl-c to stop the application
rospy.spin()
运行 这段代码,输出 'test 1: ...' 表明在 SM1 中用户数据得到递增,而输出 'test 2: ...' 显示 SM2 不访问递增的数据,因为输出保持不变 0.
如何修改SM1中的一些数据并在SM2中访问修改后的数据?
我找到了一个解决方法,使用here所述的可变对象。
应用上面的代码,它看起来像下面这样:
import smach
import smach_ros
import rospy
class st1(smach.State):
def __init__(self, outcomes=['successful', 'preempted']):
smach.State.__init__(self, outcomes, input_keys=['in_test'])
def execute(self, userdata):
if self.preempt_requested():
self.service_preempt()
return 'preempted'
rospy.logerr('test 1: '+str(userdata.in_test))
userdata.in_test[0]=userdata.in_test[0]+1
return 'successful'
class st2(smach.State):
def __init__(self, outcomes=['successful', 'preempted']):
smach.State.__init__(self, outcomes, input_keys=['in_test'])
def execute(self, userdata):
#time.sleep(2)
if self.preempt_requested():
self.service_preempt()
return 'preempted'
rospy.logerr('test 2: ' + str(userdata.in_test[0]))
return 'successful'
if __name__=="__main__":
rospy.init_node('test_state_machine')
sm_con = smach.Concurrence(outcomes=['success'],
default_outcome='success'
)
sm_con.userdata.testdata = [0]
with sm_con:
sm_1 = smach.StateMachine(outcomes=['success', 'preempted'], input_keys=['testdata'])
with sm_1:
smach.StateMachine.add('ST1', st1(),
remapping={'in_test': 'testdata'},
transitions={'successful': 'ST1'})
sm_2 = smach.StateMachine(outcomes=['success', 'preempted'], input_keys=['testdata'])
with sm_2:
smach.StateMachine.add('ST2', st2(),
remapping={'in_test':'testdata'},
transitions={'successful': 'ST2'})
smach.Concurrence.add('SM1', sm_1)
smach.Concurrence.add('SM2', sm_2)
# Execute SMACH plan
outcome = sm_con.execute()
print('exit-outcome:' + outcome)
# Wait for ctrl-c to stop the application
rospy.spin()
由于这只是一种解决方法,请参阅我相应的 issue-post here 了解更多信息。