'int' 对象在 python 中不可调用答案

【问题标题】：'int' object is not callable in python'int' 对象在 python 中不可调用
【发布时间】：2014-11-06 16:06:19
【问题描述】：

我得到了这个，我期待它在打印 x.withdraw() 时打印 410。

Kyle 12345 500
Traceback (most recent call last):
    File "bank.py", line 21, in <module>
        print x.withdraw()
TypeError: 'int' object is not callable

这是我的代码：

class Bank:
    def __init__(self, name, id, balance, withdraw):
        self.name = name
        self.id = id
        self.balance = balance
        self.withdraw = withdraw
    def print_info(self):
        return "%s %d %d" % (self.name, self.id, self.balance)
    def withdraw(self):
        if self.withdraw > self.balance:
            return "ERROR: Not enough funds for this transfer"
        elif self.withdraw < self.balance and self.withdraw >= 0:
            self.balance = self.balace - self.withdraw
            return self.balance
        else:
            return "Not a legitimate amount of funds"

x = Bank("Kyle", 12345, 500, 90)
print x.print_info()
print x.withdraw()

我需要修复类本身的问题还是我的方法调用有问题？

【问题讨论】：

标签： python class methods

【解决方案1】：

您在实例上设置了同名属性：

self.withdraw = withdraw

您现在尝试调用的是该属性，而不是方法。 Python 不区分方法和属性，它们不存在于单独的命名空间中。

为属性使用不同的名称； withdrawn（撤回的过去时）作为一个更好的属性名称浮现在脑海中：

class Bank:
    def __init__(self, name, id, balance, withdrawn):
        self.name = name
        self.id = id
        self.balance = balance
        self.withdrawn = withdrawn
    def print_info(self):
        return "%s %d %d" % (self.name, self.id, self.balance)
    def withdraw(self):
        if self.withdrawn > self.balance:
            return "ERROR: Not enough funds for this transfer"
        elif self.withdrawn < self.balance and self.withdrawn >= 0:
            self.balance = self.balance - self.withdrawn
            return self.balance
        else:
            return "Not a legitimate amount of funds"

（我还更正了一个错字；您在一个您打算使用balance 的位置使用了balace）。

【讨论】：