How does lambda in python 3 handle the argument reference?

Question

I'm learning flask, and now I'm reading the flask code.
I come into a block that I can not understand completely.

def implements_to_string(cls):
    cls.__unicode__ = cls.__str__
    cls.__str__ = lambda x: x.__unicode__().encode('utf-8')
    return cls

@implements_to_string
class Test(object):
    def __init__ (self):
        pass

test = Test()
print(test.__str__)
print(test.__str__())

The first print shows the lambda method as:

<bound method Test.<lambda> of <__main__.Test object at 0x7f98d70d1210>>

The second:

<__main__.Test object at 0x7fcc4394d210>

So when does the x in the lambda in func implements_to_string become the cls object?
Is it just an inner mechanism I just need to remember now?
Or is there something else behind need to know?

The x parameter when calling the lambda function is not the Test class but the instance test. Python invokes instancemethods with the instance (self) as first argument. — resi
– resi, Commented May 13, 2016 at 9:36
@resi Thank you very much. I under stand now that the lambda function under a class environment will take the first param as self — John
– John, Commented May 13, 2016 at 10:02

AKS · Accepted Answer · 2016-05-13 10:06:41Z

1

From the documentation:

Small anonymous functions can be created with the lambda keyword. This function returns the sum of its two arguments: lambda a, b: a+b. Lambda functions can be used wherever function objects are required. They are syntactically restricted to a single expression. Semantically, they are just syntactic sugar for a normal function definition.

Your implementation of implements_to_string is similar to what I have here:

def implements_to_string(cls):
    cls.__unicode__ = cls.__str__

    def lambda_func(self):
        return self.__unicode__().encode('utf-8')

    cls.__str__ = lambda_func
    return cls

So when does the x in the lambda in func implements_to_string become the cls object?

When you use print(test.__str__) you are printing the method itself and its representation is printed.

But when you use print(test.__str__()) you are first executing the function and printing what is returned by the method.

edited May 13, 2016 at 10:06

answered May 13, 2016 at 9:38

AKS

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3 Comments

Martin Valgur Over a year ago

You may want to use the conventional self variable name instead of x in the included code snippet to make it even clearer what is going on.

John Over a year ago

I figure out now that the first param in lambda under class circumstance will be treated as self as in normal function. Thank you @AKS and @Martin.

AKS Over a year ago

I just used x to keep it closer to what the user have in the snippet.

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