1、a=a+2，表示一個新的對象，新的對象名字還是a，但是指向的內存地址已經變了

>>> a=2 >>> id(a) 140406287260016 >>> a=a+2 >>> a 4 >>> id(a) 140406287259968

所以對于tuple對象（不可變對象），也是可以這樣操作的

>>> tuple1=(1,2) >>> id(tuple1) 4521580448 >>> tuple1=tuple1+(3,) >>> tuple1 (1, 2, 3) >>> id(tuple1) 4521658880

2、a+=2對于有些對象的操作是表示原來的對象，對有些對象的操作是生成了一個新對象

不可變對象tuple1，操作完后，內存地址已經發生變化，生成一個新的對象
>>> tuple1=(1,2) >>> type(tuple1) <type 'tuple'> >>> tuple1+=(3,) >>> id(tuple1) 4521658880 >>> tuple1+=(4,5) >>> id(tuple1) 4520649072

而list對象，可變對象，+=操作、append操作、extend操作，都是在原對象上操作

>>> list1=[1,2] >>> id(list1) 4521614656 >>> list1+=[3] >>> id(list1) 4521614656 >>> list1.append(4) >>> id(list1) 4521614656 >>> list1.extend(5) Traceback (most recent call last):File "<stdin>", line 1, in <module> TypeError: 'int' object is not iterable >>> list1.extend([5]) >>> id(list1) 4521614656 >>>

3、

x = [1,2,3] print "before func(), global! x = ",x,"id(x) = ",id(x) def func(): global x print "in func(), local! original x = ",x,"id(x) = ",id(x) x = x + [1] print "in func(), local! now x = ",x,"id(x) = ",id(x) func() print "after func(), global! x = ",x,"id(x) = ",id(x) 結果： [python] view plain copy before func(), global! x = [1, 2, 3] id(x) = 47781768 in func(), local! original x = [1, 2, 3] id(x) = 47781768 in func(), local! now x = [1, 2, 3, 1] id(x) = 47795720 after func(), global! x = [1, 2, 3, 1] id(x) = 47795720

global就保證了，即使我的變量x在函數中指向對象變了，外部的x也會指向新的對象

?

x = [1,2,3] print "before func(), global! x = ",x,"id(x) = ",id(x) def func(x): print "in func(), local! original x = ",x,"id(x) = ",id(x) x = x + [1] print "in func(), local! now x = ",x,"id(x) = ",id(x) func(x) print "after func(), global! x = ",x,"id(x) = ",id(x) 結果： before func(), global! x = [1, 2, 3] id(x) = 46339976 in func(), local! original x = [1, 2, 3] id(x) = 46339976 in func(), local! now x = [1, 2, 3, 1] id(x) = 46390664 after func(), global! x = [1, 2, 3] id(x) = 46339976

x = x + [1]，是新建了一個對象，id(x) = ?46390664

?

利用id(x)，查看下x += [1]對象是怎么變化的吧： x = [1,2,3] print "before func(), global! x = ",x,"id(x) = ",id(x) def func(x): print "in func(), local! original x = ",x,"id(x) = ",id(x) x += [1] print "in func(), local! now x = ",x,"id(x) = ",id(x) func(x) print "after func(), global! x = ",x,"id(x) = ",id(x) 結果： before func(), global! x = [1, 2, 3] id(x) = 46536584 in func(), local! original x = [1, 2, 3] id(x) = 46536584 in func(), local! now x = [1, 2, 3, 1] id(x) = 46536584 after func(), global! x = [1, 2, 3, 1] id(x) = 46536584
id(x)全程一樣，x += [1]，python直接就在原對象上操作

?

?

參考：

1、http://blog.csdn.net/emaste_r/article/details/47395055

轉載于:https://www.cnblogs.com/shengulong/p/7906582.html

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