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by plus 437 days ago
For those who are curious, `...` is a placeholder value in Python called Ellipsis. I don't believe it serves any real purpose other than being a placeholder. But it is an object and it implements `__eq__`, and is considered equal to itself. So `...==...` evaluates to `True`. When you prefix a `True` with `-`, it is interpreted as a prefix negation operator and implicitly converts the `True` to a `1`, so `-(...==...)` is equal to `-1`. Then, you add another prefix `-` to turn the `-1` back into `1`.

`--(...==...)--(...==...)` evaluates to `2` because the first block evaluates to 1, as previously mentioned, and then the next `-` is interpreted as an infix subtraction operator. The second `-(...==...)` evaluates to `-1`, so you get `1 - -1` or `2`.

When chaining multiple together, you can leave off the initial `--`, because booleans will be implicitly converted to integers if inserted into an arithmetic expression, e.g. `True - -1` -> `1 - -1` -> `2`.

> There should be one-- and preferably only one --obvious way to do it.

This article is obviously completely tongue-in-cheek, but I feel the need to point out that this sentence is not meant to be a complete inversion of the Perl philosophy of TIMTOWTDI. The word "obvious" is crucial here - there can be more than one way, but ideally only one of the ways is obvious.

4 comments

Numpy actively uses … to make slicing multidimensional arrays less verbose. There are also uses in FastAPI along the lines of «go with the default».
Expanding on this a little, I will be replacing all occurrences of 2 with two blobs fighting, with shields:

    >>> 0^((...==...)--++--(...==...))^0
    2
excellent explanation, to add to this since I was curious about the composition, '%c' is an integer presentation type that tells python to format numbers as their corresponding unicode characters[1] so

'%c' * (length_of_string_to_format) % (number, number, ..., length_of_string_to_format_numbers_later)

is the expression being evaluated here after you collapse all of the 1s + math formatting each number in the tuple as a unicode char for each '%c' escape in the string corresponding to its place in the tuple.

[1] https://docs.python.org/3/library/string.html#format-specifi...

>> There should be one-- and preferably only one --obvious way to do it.

Except for package management, of course. There, we need lots and lots of ways.

And apparently string formatting which should have an ever growing number of ways to handle it. :shrug: