python int（）函数 [英] python int( ) function

问题描述

如果小数（例如49.9）被发送到 next 变量，则下面的代码显示错误。你能告诉我为什么吗？为什么 int（）将其转换为整数？

The code below shows error if a decimal (eg. 49.9) is sent to next variable. Can you please tell me why? Why does int() converts it into an integer?

next=raw_input("> ")
how_much = int(next)
if how_much < 50:
    print"Nice, you're not greedy, you win"
    exit(0)
else:
    dead("You greedy bastard!")

如果我不使用 int（）或 float（）并且只需使用：

If I dont use int() or float() and just use:

how_much=next

然后它移动到else即使我输入 49.8 。

then it moves to "else" even if I give the input as 49.8.

推荐答案

正如其他答案所提到的， int 操作将如果字符串输入不可转换为int（例如float或字符），则会崩溃。您可以做的是使用一个小帮助方法来尝试为您解释字符串：

As the other answers have mentioned, the int operation will crash if the string input is not convertible to an int (such as a float or characters). What you can do is use a little helper method to try and interpret the string for you:

def interpret_string(s):
    if not isinstance(s, basestring):
        return str(s)
    if s.isdigit():
        return int(s)
    try:
        return float(s)
    except ValueError:
        return s

所以它将获取一个字符串并尝试将其转换为int，然后浮动，否则返回字符串。这更像是查看可转换类型的一般示例。如果您的值从该函数仍然是一个字符串返回将是一个错误，然后您需要向用户报告并请求新输入。

So it will take a string and try to convert it to int, then float, and otherwise return string. This is more just a general example of looking at the convertible types. It would be an error for your value to come back out of that function still being a string, which you would then want to report to the user and ask for new input.

如果它既不是浮点也不是int，也许是一个返回无的变体：

Maybe a variation that returns None if its neither float nor int:

def interpret_string(s):
    if not isinstance(s, basestring):
        return None
    if s.isdigit():
        return int(s)
    try:
        return float(s)
    except ValueError:
        return None

val=raw_input("> ")
how_much=interpret_string(val)
if how_much is None:
    # ask for more input? Error?

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