Python 定义函数重复循环
简单的修复,但我如何才能使这停止循环?结果是,它只是重复地请求值。还有一种方法,我可以检查所有的值,以确保在20到40之间Python 定义函数重复循环,python,Python,简单的修复,但我如何才能使这停止循环?结果是,它只是重复地请求值。还有一种方法,我可以检查所有的值,以确保在20到40之间 def main(): print("Welcome to the autostacker") print(getDim()) def getDim(): height = int(input("Enter the height: ")) width = int(input("Enter the width: ")) len
def main():
print("Welcome to the autostacker")
print(getDim())
def getDim():
height = int(input("Enter the height: "))
width = int(input("Enter the width: "))
length = int(input("Enter the width: "))
return getDim()
main()
我想,getDim中的return语句是循环的罪魁祸首。您可以检查这些值是否在如下所示的范围内
def main():
print("Welcome to the autostacker")
print(getDim())
def getDim():
height = int(input("Enter the height: "))
width = int(input("Enter the width: "))
length = int(input("Enter the width: "))
if 20 <= height <=40 and 20 <= width <= 40 and 20 <= length <= 40:
print "all the values lie in the range"
# DO WHAT YOU WANT
else:
print " values are not in range"
# DO SOMETHING ELSE
return height, width,length
main()
差不多
def main():
print("Welcome to the autostacker")
h, w, l = getDim()
if 20 <= h <= 40:
if 20 <= w <= 40:
if 20 <= l <= 40:
print "height is %d width is %d length is %d" % (h, w, l)
else:
print 'length out of range'
else:
print 'width out of range'
else:
print 'height out of range'
def getDim():
height = int(input("Enter the height: "))
width = int(input("Enter the width: "))
length = int(input("Enter the length: "))
return height, width, length
main()
确保不要在getDim中调用getDim,除非您有某种条件最终使其停止调用本身。将return getDim更改为返回高度、重量、长度如何显示正在输入的字段?您当前拥有的是一个永无止境的递归循环。[20