Python对Int值的错误识别 [英] Python misidentification of int values

问题描述

一段时间以来，我一直在编写一个依赖于其他函数来返回答案的函数.完成后，它表示我在某处错误地使用了"*".我经历了所有事情，看不到问题.我终于发现这是因为先前定义了一个简单的阶乘函数，如下所示:

I have been for some time writing a function which relies on many other to return an answer. When I finished it, it stated I had wrongly used the "*" somewhere. I went through everything and could not see a problem. I finally found that it was because of a simple factorial function earlier defined, as shown below:

if n == 0:
    return 1
elif n < 0 or (type(float)):
    print "%s is an invalid input; Positive integer value is required" % (n)
else:
    return (n)*fact(n-1)

但是，如果我输入n = 6，它将返回6是一个浮点数，从而退出该函数的其余部分，我希望有人能够回答为什么会发生这种情况以及如何解决它.

However if I enter n = 6, it returns that 6 is a floating number, and thus exits the rest of the function, and I hope someone can answer why this is happening and how to get around it.

非常感谢

推荐答案

elif n < 0 or (type(float)):

即使n >= 0，(type(float))部分也将返回float的type，并且该值是非空值或Truthy值.因此，该条件将始终得到满足.您需要的是isinstance

Even when n >= 0, (type(float)) part will return the type of float and that is non-empty or Truthy value. So, the condition will be met always. What you need is, isinstance

elif n < 0 or isinstance(n, float):

但我建议

elif not isinstance(n, int) or n < 0:

现在，如果n不是int，则将引发错误.

Now, if n is anything other than an int the error will be thrown.

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