1
0
mirror of https://github.com/satwikkansal/wtfpython synced 2024-12-22 21:00:09 +01:00

Add new example

Closes https://github.com/satwikkansal/wtfPython/issues/9
This commit is contained in:
Satwik Kansal 2017-09-01 03:49:20 +05:30
parent f76511c15b
commit 8ae68f11b6

53
README.md vendored
View File

@ -1396,6 +1396,59 @@ tuple()
---
### Let's see if you can guess this?
Originally, suggested by @PiaFraus in [this](https://github.com/satwikkansal/wtfPython/issues/9) issue.
**Output:**
```py
>>> a, b = a[b] = {}, 5
>>> a
{5: ({...}, 5)}
```
#### 💡 Explanation:
* According to [Python language reference](https://docs.python.org/2/reference/simple_stmts.html#assignment-statements), assignment statements have the form
```
(target_list "=")+ (expression_list | yield_expression)
```
and
> An assignment statement evaluates the expression list (remember that this can be a single expression or a comma-separated list, the latter yielding a tuple) and assigns the single resulting object to each of the target lists, from left to right.
* The `+` in `(target_list "=")+` means there can be **one or more** target lists. In this case, target lists are `a, b` and `a[b]` (note the expression list is exactly one, which in our case is `{}, 5`).
* After the expression list is evaluated, it's value is unpacked to the target lists from **left to right**. So, in our case, first the `{}, 5` tuple is unpacked to `a, b` and we now have `a = {}` and `b = 5`.
* `a` is now assigned to `{}` which is a mutable object.
* The second target list is `a[b]` (you may expect this to throw an error because both `a` and `b` have not been defined in the statements before. But remember, we just assigned `a` to `{}` and `b` to `5`).
* Now, we are setting the key `5` in the dictionary to the tuple `({}, 5)` creating a circular reference (the `{...}` in the output refers to the same object that `a` is already referencing). Another simpler example of circular reference could be
```py
>>> some_list = some_list[0] = [0]
>>> some_list
[[...]]
>>> some_list[0]
[[...]]
>>> some_list is some_list[0]
[[...]]
```
Similar is the case in our example (`a[b][0]` is the same object as `a`)
* So to sum it up, you can break the example down to
```py
a, b = {}, 5
a[b] = a, b
```
And the circular reference can be justified by the fact that `a[b][0]` is the same object as `a`
```py
>>> a[b][0] is a
True
```
---
### Minor Ones
* `join()` is a string operation instead of list operation. (sort of counter-intuitive at first usage)