Explain this

Question

a = [1,2,3,4]
print(a is a[:])
#output: False (but why?)

Answer 1

MD. Ferdous Ibne Abu Bakar

"is" checks if both variable refer to the same object but a and a[:] would have different objects even both are looking same.

Use == to check.

use id() function to check identify of two objects

print (id(a))
print (id(a[:]))

will give different id

Answer 2

Hi Ferdous!
If a is a name of the list then, a[:] is its shallow copy. It means when assigning, a (re)binds the name and a[:] slice-assigns, replacing what was previously in the list.
For example,
a = [1,2,3,4]
a1 = a
a2 = a[:]
a1[0] = 0
print(a) #output: [0,2,3,4]
print(a2) #output: [1,2,3,4] no changes

So, both a,a1 are in a same location(same memory address), while a2 is in a different location. That's why print(a is a[:]) returns False.