python的静态方法
2011-08-05 13:20 Rollen Holt 阅读(769) 评论(0) 编辑 收藏 举报staticmethod Found at: __builtin__ staticmethod(function) -> method Convert a function to be a static method. A static method does not receive an implicit first argument. To declare a static method, use this idiom: class C: def f(arg1, arg2, ...): ... f = staticmethod(f) It can be called either on the class (e.g. C.f()) or on an instance (e.g. C().f()). The instance is ignored except for its class. Static methods in Python are similar to those found in Java or C++. For a more advanced concept, see the classmethod builtin.
class Employee: """Employee class with static method isCrowded""" numberOfEmployees = 0 # number of Employees created maxEmployees = 10 # maximum number of comfortable employees def isCrowded(): """Static method returns true if the employees are crowded""" return Employee.numberOfEmployees > Employee.maxEmployees # create static method isCrowded = staticmethod(isCrowded) def __init__(self, firstName, lastName): """Employee constructor, takes first name and last name""" self.first = firstName self.last = lastName Employee.numberOfEmployees += 1 def __del__(self): """Employee destructor""" Employee.numberOfEmployees -= 1 def __str__(self): """String representation of Employee""" return "%s %s" % (self.first, self.last) # main program def main(): answers = [ "No", "Yes" ] # responses to isCrowded employeeList = [] # list of objects of class Employee # call static method using class print "Employees are crowded?", print answers[ Employee.isCrowded() ] print "\nCreating 11 objects of class Employee..." # create 11 objects of class Employee for i in range(11): employeeList.append(Employee("John", "Doe" + str(i))) # call static method using object print "Employees are crowded?", print answers[ employeeList[ i ].isCrowded() ] print "\nRemoving one employee..." del employeeList[ 0 ] print "Employees are crowded?", answers[ Employee.isCrowded() ] if __name__ == "__main__": main()
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