>>> names = [1,2,3]
>>> names = [1,2,[3,4]]
>>> n = names.copy()
>>> n is names
False
is the memory address stored in the comparison variable? In that case, shouldn"t the memory addresses stored in n and names be the same?
this knowledge point has not been understood.
ask for help.
is compares whether two variables point to the same object, copy () and generates a new object, so is returns False
(1) is compares whether two references point to the same object (reference comparison).
ab:
2 ncopy
"b = a" n = names.copy()
:
:
:
:
1a,b
2c,c[0],c[1]a,b
3"e = c.copy()":,c[0],c[1],c[0],c[1]
4ee[0],e[1])a,b.ece,cec.copy()isFalse.:
probably explain this, small students are not talented, speech skills may not be very difficult, I hope I can help you.
>>> a = 1
>>> names = [a]
>>> n = names.copy()
>>> n is names
False
>>> n[0] is names[0]
Truen and names be the same?
first of all, make it clear that you want to compare n and names, or the internal elements of the two.
names opens up memory space for n because it is a list (mutable), copy operation, so here n and names are compared to False with is.
and the copy case of another immutable:
>>> import copy
>>> a = (1, 2)
>>> copy.copy(a) is a
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