emillon

Joined 16 years ago

Karma
2,853
Submissions
394
Member since
Jan 10
Paris-based λ-term writer.

    https://github.com/emillon/
    http://blog.emillon.org/
    @etiennemillon
    me @ <username> . org
View on Hacker News

Recent submissions

Comment · 10 years ago
When I used pwsafe and keepassx they had this option.
Comment · 10 years ago
You might want to filter local and unroutable URLs, for example entering 127.0.0.1 renders a MAMP Pro page.
Comment · 10 years ago
It's a common feature of password managers.
Comment · 10 years ago
There are no semantic problems with this, but typing will get in the way: you can express it fairly easily if all the functions have the same type (such as Int -> Int): actually it's just 'foldr ($)'. But it is difficult to type a list of functions such as each member's return value has the same type as the next one's parameter (symbolically, [an-1 -> an, ..., a1 -> a2, a0 -> a1]). It's easier to refer to the composition of such functions, which is why you would see it as 'h . g . f'.
Comment · 10 years ago
It's more like the pipe operator in ocaml (http://blog.shaynefletcher.org/2013/12/pipelining-with-opera...). The lisp version has the extra advantage that you don't have to repeat it between all the intermediate functions. ((->> 2 (* 100) str count) vs 2 |> (* 100) |> str |> count).
Comment · 10 years ago
Comment · 10 years ago
This kind of proofs usually are separated in two steps:

- correction: define a loop invariant. Assuming the loop terminates, it will hold when the program exits the loop.

- termination: define a loop variant, something that changes at every iteration. If you can find a variant that is a strictly decreasing sequence of natural numbers (or an increasing, bounded sequence), you've established that the loop terminates.

Once you've done these two steps you know that the program terminates and the invariant holds at the end. (in your case, you won't be able to find a good variant - depending on the floating point semantics you use of course).

Comment · 11 years ago
This is correct, I missed this!
Comment · 11 years ago
That's assuming that this particular string was already present somewhere in the binary. Since it is only present as a reference, you would not see the string in a binary patch.

Live from the official Hacker News API

A modern reader for Hacker News · SvelteKit + Skeleton