为什么python打印[...]
问题描述:
有人可以解释我为什么我的Python代码打印这个[...]是什么意思? 谢谢为什么python打印[...]
我的代码:
def foo(x,y):
x[0]= y[1]
x,y=y,x
return x+y
z=[1,2]
z2=[z,z]
t=foo(z,z2)
print(z)
print(z2)
print(t)
答
这是因为z
名单在[0]
位置引用本身,当你这样做:
def foo(x,y):
x[0]= y[1]
x,y=y,x
return x+y
z=[1,2]
#Here you have created a list containing 1,2
z2=[z,z]
#here is not creating two lists in itself, but it is referencing z itself(sort of like pointer), you can verify this by:
In [21]: id(z2[0])
Out[21]: 57909496
In [22]: id(z2[1])
Out[22]: 57909496
#see here both the location have same objects
t=foo(z,z2)
#so when you call foo and do x[0]= y[1], what actually happens is
# z[0] = z2[0] , which is z itself
# which sort of creates a recursive list
print(z)
print(z2)
print(t)
#you can test this by
In [17]: z[0]
Out[17]: [[...], 2]
In [18]: z[0][0]
Out[18]: [[...], 2]
In [19]: z[0][0][0]
Out[19]: [[...], 2]
In [20]: z[0][0][0][0]
Out[20]: [[...], 2]
#you can carry on forever like this, but since it is referencing itself, it wont see end of it
这是因为列表的引用本身。试试这个:'x = [1,2,3]'然后'x [2] = x'然后'print(x)'''print(x [2])'''print(x [2] [2 ])','print(x [2] [2] [1])'' –