What is the output and why

Question

a = "hello_world"
b = "hello_" + "world"
print(a is b)

Answer 1

Google for:

Python interning

it will explain what's happening.

also, try this:

a = "hello_world"
b = "hello_" + "world"
c = "hello_"
c += "world"

print(a) #hello_world
print(b) #hello_world
print(c) #hello_world

print(id(a)) # some id
print(id(b)) # same as id(a)
print(id(c)) # different id

print(a is b) #True
print(a is c) #False
print(b is c) #False

Basically, concatenating the strings before assigning to variables as in a and b is the same as assigning the same string to both.

This results in the interning optimization being applied, so a and b gets the same address, because pointing to the same memory location is more efficient.

c is intitialized with a different string, so even if it eventually has the value of 'hello_world', it's id is different.

https://www.geeksforgeeks.org/python/object-interning-in-python/

Answer 2

What do you think and why don't you try?