What is the use of () in the Python code?

Question

for x in range(10):
   arr.append(lambda:x**2)
arr[4]()

Expected output: 16 or nothing because no print statement But output, when I run, is 81. Why so?

You're running this in an interactive interpreter. (That, and https://stackoverflow.com/questions/12423614/local-variables-in-nested-functions) — user2357112, Jul 13 '20 at 09:54
each element in an array is a function (or lambda). So arr[4] is a function and () is a function call — Benjamin, Jul 13 '20 at 09:54
Also, it sounds like you already know what all the parentheses do in this code, if you understood enough to expect 16 or nothing. — user2357112, Jul 13 '20 at 09:55

Blusky · Accepted Answer · 2020-07-13T10:13:03.333

4

In your code, arr.append(lambda:x**2), x is scoped outside the function.

When you finish the loop, x is set to 9 (0 to 9).

Then, x**2 is 81, so arr[4]() will re-evaluate x, so the result is 81.

Just for fun, you can attempt what you want with the following:

f_generator = lambda i: lambda: i**2
arr = [f_generator(i) for i in range(10)]
arr[4]()  # 16

edited Jul 13 '20 at 10:13

answered Jul 13 '20 at 09:54

Blusky

2

Actually `arr[y]()` will return 81 for every `y`. Good answer, +1 OP – Gamopo Jul 13 '20 at 09:56
But when I call arr[4]() and arr[3]() it should return 81 and 81 but it only returns 81 once. Why so? – Mannu_050 Jul 14 '20 at 10:24
In your version, `arr[4]()` and `arr[3]()` are both returning 81. – Blusky Jul 15 '20 at 07:39

score 0 · Answer 2 · answered Jul 13 '20 at 09:58

0

To get the effect you want, you should remove the lambda like this:

arr = []
for x in range(10):
   arr.append(x**2)
print(arr[4])

because otherwise you are storing a function, not a number. Notice that now you don't need ()

answered Jul 13 '20 at 09:58

Gamopo

score 0 · Answer 3 · answered Jul 13 '20 at 09:59

0

() is used for calling any object, it can be a method or class like,

class Name:
    ...

Name()

or,

name = Name()

answered Jul 13 '20 at 09:59

Nj Nafir

3 Answers3