User:RaiseAfloppaFan3925

My username is actually raiseAfloppaFan3925, however because of the User: prefix, I prefer RaiseAfloppaFan3925.

▼(^v^)▼ ▼(^v^)▼ 3.1415926535897932... ▼(^v^)▼ ▼(^v^)▼

I hate compiler development. I mean I don't exactly HATE it like how some people hate JavaScript, but I still don't like it. My programs just end up being

int main() {
    return -1 == -1;
}

and I can't even add other types. Even if local variables didn't exist I would still have some fun if I had types, but NO. I CAN'T. Only "signed 32-bit integer" is what I can have, and nothing else. I look at other compilers and look at their support for even floating-point types, while that one Scheme compiler gets away with dynamic typing, Vala gets away with GLib, and then here's my compiler. It can't even compete with other toy compilers since they at least have global and local variables, while mine is just a fancy expression evaluator.

I still want to make one though. Do you think the B language (C before C) would be what saves me from this stalemate?

my esolangs

golden era

flop era

Mango (not to be confused with Mango, but it did use to live on that page)
Redshift and Redshift-A
RSI1
Hailstone
I Ate 562 Metric Tons Of Air And Now I Am Floating Into The Atmosphere's Various Layers At Approximately 0.401 Meters Per Second. If I Had Not Ate That Much Air Then I Would Be Fine Right Now. Now I Have To Exhale All Of This Air. The Lesson Here Is To Not Inhale 562 Metric Tons Of Air.
Purely Object-Oriented Producer-based/Constructorless Language

renaissance

Bad Apple!!! should be an esolang

According to the following proof, Bad Apple!!! should be turned into an esolang.

if true = true and false = false and true = NOT false and false = NOT true

thoughts

pythOwO deserves a C runtime.

LOLCODE deserves a compiler either like Go or like a subset/joint set version.

According to this very mathematically incorrect proof, $1$ is equal to $-1$ .

incorrect mathematical proof

$1={\sqrt {1}}\cdot {\sqrt {1}}$

  ${\sqrt {1}}=-1$ , since all positive numbers have two square roots; one positive and one negative.

  $1=-1\cdot -1$ 
 
  ${\sqrt {1}}=1$  too, so the equation below is valid.
 
  $1={\sqrt {-1\cdot -1}}$ 
 
 There is a simplification rule  ${\sqrt {ab}}={\sqrt {a}}\cdot {\sqrt {b}}$ , which simplifies the above expression into:
 
  $1={\sqrt {-1}}\cdot {\sqrt {-1}}$ 
 
 Since  ${\sqrt {-1}}=i$ , the above can be simplified to:
 
  $1=i\cdot i$ 
 
 Since  ${\sqrt {-1}}=i$ , then  $i^{2}=-1$ .
 
  $1=-1$