Python的return self和return一个新的对象区别

目的:设计一个有理数相加。如3/5 + 7/15 = 80/75

return self

输入:

class Rational0:
    def __init__(self, num, den=1):
        self.num = num
        self.den = den

    def plus(self, another):
        den = self.den * another.den
        num = (self.num * another.den + self.den * another.num)
        return self

    def print(self):
        print(str(self.num) + "/" + str(self.den))

输出:

>>> r1 = Rational0(3, 5)
>>> r1.print()
3/5
>>> r1
<__main__.Rational0 object at 0x000001D1D91A5F60>
>>> r2 = r1.plus(Rational0(7,15))     # 新对象r2
>>> r2
<__main__.Rational0 object at 0x000001D1D91A5F60>     # 地址相同
>>> r1.plus(Rational0(7,15)).print()   #或者r2.print()
3/5
>>> r2.num
3
>>> r2.den
5

从地址中可以看出Rational0(3, 5)r1.plus(Rational0(7,15))返回的对象是同一个r1地址,原因在于return self

return self做出修改

class Rational0:
    def __init__(self, num, den=1):
        self.num = num
        self.den = den

    def plus(self, another):
        den = self.den * another.den 
        num = (self.num * another.den + self.den * another.num)
        self.den = den
        self.num = num
        return self
        #return Rational0(num, den)

    def print(self):
        print(str(self.num) + "/" + str(self.den))

r1 = Rational0(3, 5)
r2 = r1.plus(Rational0(7,15))
r2.print()

输出:

>>> r1 = Rational0(3, 5)
>>> r1.print()
3/5
>>> r1
<__main__.Rational0 object at 0x00000239CADF5E80>
>>> r2 = r1.plus(Rational0(7,15))
>>> r2.print()
80/75
>>> r2
<__main__.Rational0 object at 0x00000239CADF5E80>
>>> r2.num
80
>>> r2.den
75

分析:


return新的对象

输入:

class Rational0:
    def __init__(self, num, den=1):
        self.num = num
        self.den = den

    def plus(self, another):
        den = self.den * another.den
        num = (self.num * another.den + self.den * another.num)
        return Rational0(num, den)

    def print(self):
        print(str(self.num) + "/" + str(self.den))

输出:

>>> r1 = Rational0(3, 5)
>>> r1.print()
3/5
>>> r1
<__main__.Rational0 object at 0x000001F857C35DD8>
>>> r2 = r1.plus(Rational0(7,15))     # 新对象r2
<__main__.Rational0 object at 0x000001F857C48940>  #地址不同
>>>r2.print()
80/75
>>> r2.num
80
>>> r2.den
75

从地址中可以看出Rational0(3, 5)r1.plus(Rational0(7,15))返回的对象不是同一个,原因在于return Rational0(num, den)

posted on 2018-09-08 15:31  星辰之衍  阅读(1680)  评论(0编辑  收藏  举报

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