在Python中,`不是int`是否会返回false？

Question

我正在使用参数解析器允许将端口号传递给我的程序.然后我尝试验证该值,并且第一个测试之一是is int测试:

    parser = argparse.ArgumentParser(description='Provides XML-RPC API.')
    parser.add_argument('--port', dest='port', default=9666, type=int,
                       help='port to listen on, 9666 by default.');
    args = parser.parse_args()

    if args.port is not int:
        raise TypeError("Port must be integer (%s given), use -h argument to get help."%(type(args.port).__name__))
    if args.port>=65536 or args.port<=0:
        raise IndexError("Port must be lower than 65536, 65535 is the maximum port number. Also, port can only be positive, non-zero number. Use -h argument to get help.")
    if args.port<=1204:
        print("Notice: ports up to 1024 are reserved to well known services, using this port numbers often leads to conflicts. Use -h argument to get help.")

无论我做什么,我都会收到错误:

Port must be integer (int given), use -h argument to get help.

我的意思是,到底是什么？即使错误说它是int,所以发生了什么？

Answer 1

args.port is not intFalse如果args.port设置为int类型将返回:

args.port = int
args.port is int  # True

或者您重新绑定int到args.port已绑定的同一对象:

int = args.port
args.port is int  # True

这是因为is测试两个引用是否指向完全相同的对象.

也许您对如何测试对象类型感到困惑,因为您使用该isinstance()函数,或者测试输出type():

type(args.port) is not int  # False if args.port is a different type
not isinstance(args.port, int)  # False if args.port is a different type and not a subclass either

这也不会是这种情况,因为您将ArgumentParser选项配置为仅接受整数:

parser.add_argument('--port', dest='port', default=9666, type=int,
                   help='port to listen on, 9666 by default.')

该type=int参数很重要,因为这意味着任何不能被解释为整数的东西都将被解析器拒绝.你永远不会过去parser.parse_args():

>>> import argparse
>>> parser = argparse.ArgumentParser(description='Provides XML-RPC API.')
>>> parser.add_argument('--port', dest='port', default=9666, type=int,
...                     help='port to listen on, 9666 by default.')
_StoreAction(option_strings=['--port'], dest='port', nargs=None, const=None, default=9666, type=<type 'int'>, choices=None, help='port to listen on, 9666 by default.', metavar=None)
>>> parser.parse_args(['--port', 'not an integer'])
usage: [-h] [--port PORT]
: error: argument --port: invalid int value: 'not an integer'

那是因为'not an integer'无法通过int()功能转换; 一个ValueError上升和argparse转换是一个错误消息要告诉你:

>>> int('not an integer')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: 'not an integer'

您可以提供限制范围的自己的会话功能:

def portnumber(value):
    port = int(value)  # may raise ValueError
    if not 0 <= port <= 65536:
        raise ValueError('Not a valid port number')
    return port

然后使用它作为type:

parser.add_argument('--port', dest='port', default=9666, type=portnumber,
                   help='port to listen on, 9666 by default.')