Python 3.x 函数名称未定义错误

我想用python编写一个函数，告知我给出的数字是否为素数，目前为止我的代码是：

def enter_input():
    n=input("Enter the number to be checked\n")
    return n

def check():
    o=int(enter_input())
    r=o
    print(type(o))
    print(type(r))
    bool=True
    for i in range(2,r,1):
        if(o%i==0):
            return False
    return True

def display():
    print("Entered Number Is {0}".format(p))
    print("The Number is a Prime:{0}".format(check()))


enter_input()
check()
display()

---

但是我在输入数字后得到了这个错误：

运行时：-

Enter the number to be checked
23
Traceback (most recent call last):
    File "Chech_Prime.py", line 8, in check
    o=int(enter_input())
    NameError: name 'enter_input' is not defined

回复关闭

---

我哪里出错了？这是我第一次尝试函数调用，我调用enter_input（）函数的地方似乎找不到它。感谢您的帮助

您写了一些有点奇怪的东西，因此我将解释为什么这应该做您正在检查的事情：

这是您的代码，我将在两行之间进行注释：

def enter_input():
    n=input("Enter the number to be checked\n")
    return n 

def check(): # you create the check function, and then a display function, and is not necesary, because you want to do both thinks in the same function
    o=int(enter_input()) # you call this function a lot of times, and you only want to input the value once, so, is better to call the input method inside the function
    r=o # is better to name variables more descriptive, I mean, check() as a function, what checks? maybe you want to call it is_prime(), more descriptive, o and p... doesn't say anything, but input sounds better
    print(type(o))
    print(type(r)) #this auxiliar variable is not needed
    bool=True
    for i in range(2,r,1):
        if(o%i==0):
            return False
    return True

def display():
    print("Entered Number Is {0}".format(p))
    print("The Number is a Prime:{0}".format(check()))# here you again call the function, and again it ask to you to add a input... not good


enter_input()
check() # and again.. three times, that's so confusing even to follow when you are adding the inputs
display()

我的方法是：

def print_is_prime():
    value_to_eval=int(input("Enter the number to be checked\n"))

    print("the type of the value is " + str(value_to_eval.__class__.__name__))

    is_prime =True #initial value is correct
    for i in range(2,value_to_eval,1):
        if(value_to_eval%i==0):
            is_prime = False #here, you detect is not prime
            break # so with break you can avoid to execute innecesary all the loop

    is_prime = True and is_prime # with and operator, true is the neutral, so the value will be always the other value, in this case, is_prime, this like 0 + x, the value will be always x

    print("Entered Number Is {0}".format(value_to_eval))
    print("The Number is a Prime:{0}".format(is_prime))

print_is_prime()

---

你写了一些有点奇怪的东西，所以我将解释为什么这应该做你正在观察的事情：

这是您的代码，我将在两行之间进行注释：

def enter_input():
    n=input("Enter the number to be checked\n")
    return n 

def check(): # you create the check function, and then a display function, and is not necesary, because you want to do both thinks in the same function
    o=int(enter_input()) # you call this function a lot of times, and you only want to input the value once, so, is better to call the input method inside the function
    r=o # is better to name variables more descriptive, I mean, check() as a function, what checks? maybe you want to call it is_prime(), more descriptive, o and p... doesn't say anything, but input sounds better
    print(type(o))
    print(type(r)) #this auxiliar variable is not needed
    bool=True
    for i in range(2,r,1):
        if(o%i==0):
            return False
    return True

def display():
    print("Entered Number Is {0}".format(p))
    print("The Number is a Prime:{0}".format(check()))# here you again call the function, and again it ask to you to add a input... not good


enter_input()
check() # and again.. three times, that's so confusing even to follow when you are adding the inputs
display()

我的方法是：

def print_is_prime():
    value_to_eval=int(input("Enter the number to be checked\n"))

    print("the type of the value is " + str(value_to_eval.__class__.__name__))

    is_prime =True #initial value is correct
    for i in range(2,value_to_eval,1):
        if(value_to_eval%i==0):
            is_prime = False #here, you detect is not prime
            break # so with break you can avoid to execute innecesary all the loop

    is_prime = True and is_prime # with and operator, true is the neutral, so the value will be always the other value, in this case, is_prime, this like 0 + x, the value will be always x

    print("Entered Number Is {0}".format(value_to_eval))
    print("The Number is a Prime:{0}".format(is_prime))

print_is_prime()

---

“p”变量是在哪里定义的？很抱歉是“p”，它应该是“o”。但是问题仍然存在。好吧，我修改了一点你的代码试试看，在中途我会给你写一个解释让我知道这是否对你有效，我会很快更新，与你的代码进行比较，告诉你哪里犯了小错误，为什么我做了我做的事我给你添加了解释为什么你的代码不起作用，还有我为什么做我的。再见！“p”变量是在哪里定义的？很抱歉是“p”，它应该是“o”。但是问题仍然存在。好吧，我修改了一点你的代码试试看，在中途我会给你写一个解释让我知道这是否对你有效，我会很快更新，与你的代码进行比较，告诉你哪里犯了小错误，为什么我做了我做的事我给你添加了解释为什么你的代码不起作用，还有我为什么做我的。再见！谢谢Damian，修改后的代码成功了。我试图学习函数以及它们相互调用的方式，所以我制作了3个不同的函数，而不是一个。我会检查新的代码并解决问题。@Ritesh\u BM我一直很乐意帮忙，别忘了接受(✓) 如果你喜欢，请投票：）谢谢Damian，修改后的代码起作用了。我试图学习函数以及它们相互调用的方式，所以我制作了3个不同的函数，而不是一个。我会检查新的代码并解决问题。@Ritesh\u BM我一直很乐意帮忙，别忘了接受(✓) 如果你愿意，可以投票：）