Isn’t beginner++ gonna leave it unchanged?
range(val) iterates from 0 to val-1, so the final i+1 is val

// this increments i
var i = new AtomicInteger(0);
i.increment(); 

Your intermediate increment looks like serious JavaScript code I’ve seen.

74181

(A + 1)
A0:A3 = (Input Register)
S0:S3 = Low
Mode = Low
CaryN = High
Q1:Q4 = (Output)

https://en.wikipedia.org/wiki/74181

Why not wait for a random bit flip to increment it?

int i = 0;
while (i != i + 1);
//i is now incremented

I like to shake the bytes around a little

i = ( i << 1 + 2 ) >> 1

Java has AtomicInteger, which is probably one of the more complicated, but also robust, ways of setting an integer.

That’s a tricky problem, I think you might be able to create a script that increments it recursively.

I’m sure this project that computes Fibonacci recursively spawning several docker containers can be tweaked to do just that.

https://github.com/dgageot/fiboid

I can’t think of a more efficient way to do this.

$i = 0;
$s = "foobar";
$i+=$s=~/(oo)/; # This increments $i
say $i;

Something like that should do it:

i = ~((~i + 1) + ~0) + 1

Use bitwise operations to simulate an adder. Bonus points if you only use NOR

Let f(x) = 1/((x-1)^(2)). Given an integer n, compute the nth derivative of f as f^((n))(x) = (-1)^{(n)(n+1)!/((x-1)}(n+2)), which lets us write f as the Taylor series about x=0 whose nth coefficient is f^((n))(0)/n! = (-1)^(-2)(n+1)!/n! = n+1. We now compute the nth coefficient with a simple recursion. To show this process works, we make an inductive argument: the 0th coefficient is f(0) = 1, and the nth coefficient is (f(x) - (1 + 2x + 3x^(2) + … + nx^(n-1)))/x(n) evaluated at x=0. Note that each coefficient appearing in the previous expression is an integer between 0 and n, so by inductive hypothesis we can represent it by incrementing 0 repeatedly. Unfortunately, the expression we’ve written isn’t well-defined at x=0 since we can’t divide by 0, but as we’d expect, the limit as x->0 is defined and equal to n+1 (exercise: prove this). To compute the limit, we can evaluate at a sufficiently small value of x and argue by monotonicity or squeezing that n+1 is the nearest integer. (exercise: determine an upper bound for |x| that makes this argument work and fill in the details). Finally, evaluate our expression at the appropriate value of x for each k from 1 to n, using each result to compute the next, until we are able to write each coefficient. Evaluate one more time and conclude by rounding to the value of n+1. This increments n.

Your CPU has big registers, so why not use them!

#include <x86intrin.h>
#include <stdio.h>

static int increment_one(int input)
{
    int __attribute__((aligned(32))) result[8]; 
    __m256i v = _mm256_set_epi32(0, 0, 0, 0, 0, 0, 1, input);
    v = (__m256i)_mm256_hadd_ps((__m256)v, (__m256)v);
    _mm256_store_si256((__m256i *)result, v);
    return *result;
}

int main(void)
{
    int input = 19;
    printf("Input: %d, Incremented output: %d\n", input, increment_one(input));
    return 0;
}

First, imagine a number in JavaScript. (Bit of a nail biter here, huh?)

let i = 5

Then, we will construct an incrementor. This is really simple: here is the method.

1. Make a bracket-string-centric version of eval().

[]["filter"]["constructor"]("return i+1")()

2. Reconstruct stringy eval() by using +[] as 0, +!+[] as 1, and implicit conversions as ways to create strings. For example, ‘false’ is (![]+[]), so ‘f’ is (![]+[])[+[]].

[][
  (![] + [])[+[]] + // f
  ([![]] + [][[]])[+!+[] + [+[]]] + // i
  (![] + [])[!+[] + !+[]] + // l
  (!![] + [])[+[]] + // t
  (!![] + [])[!+[] + !+[] + !+[]] + // e
  (!![] + [])[+!+[]] // r
][
  ([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+ // c
  (!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+ // o
  ([][[]]+[])[+!+[]]+ // n
  (![]+[])[!+[]+!+[]+!+[]]+ // s
  (!![]+[])[+[]]+ // t
  (!![]+[])[+!+[]]+ // r
  ([][[]]+[])[+[]]+ // u
  ([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+ // c
  (!![]+[])[+[]]+ // t
  (!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+ // o
  (!![]+[])[+!+[]] // r
]("return i+1")()

3. Draw the rest of the fucking owl. Final code:

let i = 5; // haha yay

[][
  (![] + [])[+[]] + // f
  ([![]] + [][[]])[+!+[] + [+[]]] + // i
  (![] + [])[!+[] + !+[]] + // l
  (!![] + [])[+[]] + // t
  (!![] + [])[!+[] + !+[] + !+[]] + // e
  (!![] + [])[+!+[]] // r
][
  ([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+ // c
  (!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+ // o
  ([][[]]+[])[+!+[]]+ // n
  (![]+[])[!+[]+!+[]+!+[]]+ // s
  (!![]+[])[+[]]+ // t
  (!![]+[])[+!+[]]+ // r
  ([][[]]+[])[+[]]+ // u
  ([][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]]+[])[!+[]+!+[]+!+[]]+ // c
  (!![]+[])[+[]]+ // t
  (!![]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+[]]]+ // o
  (!![]+[])[+!+[]] // r
](
  (!![]+[])[+!+[]]+ // r
  (!![]+[])[!+[]+!+[]+!+[]]+ // e
  (!![]+[])[+[]]+ // t
  ([][[]]+[])[+[]]+ // u
  (!![]+[])[+!+[]]+ // r
  ([][[]]+[])[+!+[]]+ // n
  (+[![]]+[][(![]+[])[+[]]+(![]+[])[!+[]+!+[]]+(![]+[])[+!+[]]+(!![]+[])[+[]]])[+!+[]+[+!+[]]]+ // ' '
  ([![]]+[][[]])[+!+[]+[+[]]]+ // i
  (+(+!+[]+(!+[]+[])[!+[]+!+[]+!+[]]+[+!+[]]+[+[]]+[+[]])+[])[!+[]+!+[]]+ // +
  +!+[] // 1
)()
// no virus i swear. execute arbitrary code in your browser console.

Anyway, that’s just everyday JS work. It’s like step 5 after resizing the button, but a bit before centering the div.

based on this. _{some translation methods done differently.}

int toIncrement = ...;
int result;
do {
  result = randomInt();
} while (result != (toIncrement + 1));
print(result);