+不支持的操作数类型:“函数”和“整数”错误

时间:2018-11-06 21:03:58

标签: python function for-loop

我的代码:

def start_input():
    start = int(input("\nAt what number shall we start, master? "))
    return start

def finish_input():
    end = int(input("\nwhen shall i finish, master? "))
    return end

def step_input():
    rise = int(input("\nby what ammount shall your numbers rise, master? "))
    return rise

def universal_step():
    rise = 3
    return rise

def the_counting():
    print("your desired count: ")
    for i in range ( start_input, finish_input +1, step_input): #can be also changed for automated step
        return print(i, finish_input ="  ")

def main():
    start_input()
    finish_input()
    step_input() #This can be changed for the universal_step function for no input if wanted
    the_counting()

main()

input("\n\nPress the enter key to exit.")

因此无需将代码放入以前可以完全起作用的函数中,现在我得到的只是“不支持的操作数类型,用于+:'function'和'int'error”,这在def计数中功能。我是python的新手,不知道为什么和发生了什么。感谢您的帮助:)

1 个答案:

答案 0 :(得分:3)

您在range中使用的所有东西都是函数,而不是变量;您必须致电(添加通话对象)以获取其价值,并进行以下更改:

for i in range ( start_input, finish_input +1, universal_step):

到(间隔PEP8):

for i in range(start_input(), finish_input() + 1, universal_step()):