Money Oriented Programmer
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Some people are under the misconception that Unicode is simply a 16-bit code where each character takes 16 bits and therefore there are 65,536 possible characters. This is not, actually, correct. It is the single most common myth about Unicode, so if you thought that, don’t feel bad.

Disclaimer:

Any contents (excluding intellectual properties from third parties) I posted in StackOverflow and StackExchange sites can be claimed as yours without prior written permission from me. No attributions are needed. Any kind of risks of using them are your own responsibilities. This statement never expires.

If you can’t explain it to a six-year-old, you don’t understand it yourself.

Anything that need to be highly performant will eventually be written in C++. (Me)

By the way, my blood type is Microsoft .Net and I love typing semicolons. I am just a single point on the complex plane. I am not the strongest and smartest species in this universe. I love mathematics, physics, and programming. I am the only one who knows the last digit of π. I am a sort of person who don't want to be bothered with reading lengthy documentations even for things that potentially cause catastrophic effects. I considered it as my learning style. I can tell you more but then I have to kill you.

  • Euler's identity is often cited as an example of deep mathematical beauty. It links five fundamental mathematical constants: 0, 1, π, e, and i.

  • C++ can also be cited as an example of deep programming beauty. It can link all four types of brackets: [],<>,(), and {}.

    auto foo = []<typename T>(T a, T b) { return a + b; };
    

TypeScript:

let foo = (x: number) => (y: number) => x * y;
console.log(foo(2)(3));

C#:

Func<int, Func<int, int>> foo = (x) => (y) => x * y;
Console.WriteLine(foo(2)(3));

C++:

auto foo = [](int x)
{
    return [=](int y)
    {
        return x * y;
    };
};
cout << foo(2)(3) << endl;

Java:

Function<Integer, Function<Integer, Integer>> foo = (x) -> (y) -> x * y;
System.out.println(foo.apply(2).apply(3));

Python:

foo = lambda x: lambda y: x * y
print(foo(2)(3))

Finding e

const fact = (x: number) => [...Array(x)].reduce((i, j, k) => i * (k + 1), 1);
const e: number = [...Array(20)].reduce((i, j, k) => i + (k + 1) / fact(k), 0) / 2;

Useful links:

https://ccrma.stanford.edu/~jos/mdft/mdft.html

https://learnopengl.com/

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