我刚刚了解了回调的概念,我决定尝试实现自己的回调。我的努力是卓有成效的,我确实设法模拟了回调的功能。不幸的是,我注意到我的实现导致堆栈每周期增加2个函数调用,我假设,如果代码运行足够长的时间,最终会导致堆栈溢出
我想知道,如何实现这段代码来防止堆栈在每个循环中增长?或者这是实施的必然产物,在这种情况下,如何规避这个问题
import time
import inspect
def doSomething(x):
return x + 0.00000001
def continue_processing(runningTotal,termination_condition,callback,callback_args,timeout=5):
startTime = time.time()
while (time.time() - startTime < timeout and not(termination_condition(runningTotal))):
runningTotal = doSomething(runningTotal)
print(f"Returning control to calling function, running total is {runningTotal}")
return callback(runningTotal,*callback_args)
def process(runningTotal,n,beginTime):
if(runningTotal < n):
print(f"Continue processing, running total is {runningTotal}\nTime elapsed {time.time() - beginTime}\nCurrent stack size: {len(inspect.stack())}")
continue_processing(runningTotal,lambda x: x>n,process,(n,beginTime))
if __name__ == '__main__':
beginTime = time.time()
try:
process(0,1,beginTime)
except KeyboardInterrupt:
print("Program interrupted!")
exit(0)
print(f"Completed in {time.time() - beginTime}"
问题是回调是递归的,它(间接地)调用自己——这就是堆栈溢出的原因。下面是如何避免这种情况。注意,我还修改了您的代码,使之符合PEP 8 - Style Guide for Python Code准则,使其更具可读性。我强烈建议你阅读和遵循它,特别是如果你只是学习语言
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