Fix inverted notify/notify_all in BQueue.Interrupt

[winklemad]

Problem

BQueue.Interrupt(notifyAll) calls the wrong Condition method for each branch:

def Interrupt(self, notifyAll: bool = False):
    with self.__condition:
        if notifyAll:
            self.__condition.notify()        # wakes only ONE waiter
        else:
            self.__condition.notify_all()    # wakes ALL waiters

The two branches are swapped relative to the parameter's meaning: notifyAll=True wakes only a single blocked Get() waiter, while the default notifyAll=False wakes all of them. So a caller that relies on Interrupt(notifyAll=True) to release every consumer blocked in Get() only releases one; the rest keep waiting until their timeout (or forever, if called with timeout=None).

The current in-tree caller (core/channel.py Close()) uses the default and has a single consumer thread, so it happens to work — but the method's contract is still inverted.

Reproduce

import threading, time
from unitree_sdk2py.utils.bqueue import BQueue

q = BQueue(maxLen=10)
woke = []
def consumer(i):
    q.Get(timeout=5.0)   # blocks: queue is empty
    woke.append(i)

for i in range(2):
    threading.Thread(target=consumer, args=(i,), daemon=True).start()
time.sleep(0.3)          # both threads are now blocked in Get()

q.Interrupt(notifyAll=True)   # intent: wake BOTH consumers
time.sleep(0.3)
print(sorted(woke))      # [0]  -> only one woke; expected [0, 1]

Fix

Swap the two branches so notifyAll=True calls notify_all() and notifyAll=False calls notify(). After the fix the snippet prints [0, 1], and the single-consumer Interrupt() path used by channel.py is unchanged.