布尔检查在功能中不起作用 [英] Boolean check not working in function

问题描述

此代码将成为程序的一部分，该程序将检查数字是否为质数.我知道它不是特别优雅，但我想让它仅出于体验而工作.我认为该函数失败了，因为if/elif上的逻辑是错误的，当我运行此代码时，它似乎直接进入了else子句.这是语法问题，还是不允许我在if子句中进行逻辑检查?

This code will be part of a program that will check if a number is prime or not. I know it's not particularly elegant, but I want to get it working simply for experience. I think that the function is failing because the logic on the if/elif is wrong, when I run this code, it seems to just go straight to the else clause. Is this a syntax problem, or am I not allowed to do logic checks in if clauses?

list = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]

def find_prime(list, n):
    if n in list == False:
        list.append(n)
        print "I'ts in there now."
    elif n in list == True:
        print "It's in there already."
    else:
        print "Error"

find_prime(list, 3)
find_prime(list, 51)

推荐答案

1. list是变量的坏名称.它掩盖了内置的list.

1. list is a bad name for a variable. It masks the built-in list.

if n in list == True:不会执行您要等待的操作:1 in [0, 1] == True返回False(因为，正如@Duncan指出的那样，1 in [0,1] == True是1 in [0,1] and [0,1] == True的简写).使用if n in li:和if n not in li:

if n in list == True: doesn't do what you await: 1 in [0, 1] == True returns False (because, as @Duncan notes, 1 in [0,1] == True is shorthand for 1 in [0,1] and [0,1] == True). Use if n in li: and if n not in li:

没有必要使用Error，因为元素在列表中或不在列表中.还有什么是编程错误.

No reason for an Error, since an element is in the list or it is not in the list. Anything else is programming error.

因此您的代码应如下所示:

So your code could look like this:

li = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47]

def find_prime(li, n):
    if n in li:
        print "It's in there already."
    else:
        li.append(n)
        print "It's in there now."

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