我只使用一个异常处理程序，它将捕获KeyboardInterrupt并存储异常。然后，在迭代完成的那一刻，如果有一个异常待处理，我将中断循环并重新引发该异常（以使正常的异常处理有机会发生）。
这有效（已在Python 2.7上测试）：
x = 1
print "Script started."
stored_exception=None
while True:
try:
print "Processing file #",x,"started...",
# do something time-cosnuming
time.sleep(1)
print " finished."
if stored_exception:
break
x += 1
except KeyboardInterrupt:
stored_exception=sys.exc_info()
print "Bye"
print "x=",x
if stored_exception:
raise stored_exception[0], stored_exception[1], stored_exception[2]
sys.exit()
由于已在评论中发现，此答案对于原始海报不令人满意，这是基于线程的解决方案：
import time
import sys
import threading
print "Script started."
class MyProcessingThread(threading.Thread):
def __init__(self):
threading.Thread.__init__(self)
def run(self):
print "Processing file #",x,"started...",
# do something time-cosnuming
time.sleep(1)
print " finished."
for x in range(1,4):
task = MyProcessingThread()
task.start()
try:
task.join()
except KeyboardInterrupt:
break
print "Bye"
print "x=",x
sys.exit()
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