C++:

int i = 5;
i ^= printf("The initial value of i is %d\n", i)^
printf("i=i+1; // this increments i\n")^
printf("Trigger very obscure FPU bug %c",(int)((float)8.5953287712*(double)8.5953287713-'?'))/10;
printf("i has now been incremented by 1 : %d\n", i);

Output:

The initial value of i is 5  
i=i+1; // this increments i  
Trigger very obscure FPU bug  
i has now been incremented by 1 : 6

I didn’t test other values but they’re probably OK.

C++ templates my beloved

#pragma pack(1)
template <size_t i> struct increment;
template <> struct increment<0> {
	char _plusone;
	constexpr static size_t value = 1;
};

template <size_t i> struct increment<i> {
	char _plusone;  // this increments i
    constexpr static size_t value = sizeof(increment<i - 1>);
};

template <unsigned i>
using increment_v<i> = increment<i>::value;

C# .NET using reflection, integer underflow, and a touch of LINQ. Should work for all integer types.

// this increments i
private static T Increment<T>(T i)
{
    var valType = typeof(T);
    var maxField = valType.GetField("MaxValue");
    var minField = valType.GetField("MinValue");
    if (maxField != null)
    {
        T maxValue = (T)maxField.GetValue(i);
        T minValue = (T)minField.GetValue(i);

        var methods = valType.GetTypeInfo().DeclaredMethods;
        var subMethod = methods.Where(m => m.Name.EndsWith("op_Subtraction")).First();
               
        T interim = (T)subMethod.Invoke(
            null,
            [i, maxValue]);

        return (T)subMethod.Invoke(
            null, 
            [interim, minValue]);
    }
    throw new ArgumentException("Not incrementable.");
}

Conditional adder:

if x==1:
    return 2
else if x==2:
    return 3
...

C:

int increment(int i) {
    return (int) (1[(void*) i])

However, if you wanna go blazingly fast you gotta implement O(n) algorithms in rust. Additionally you want safety in case of integer overflows.

use std::error::Error;

#[derive(Debug, Error)]
struct IntegerOverflowError;

struct Incrementor {
    lookup_table: HashMap<i32, i33>
}

impl Incrementor {
    fn new() -> Self {
        let mut lut = HashMap::new();
        for i in 0..i32::MAX {
            lut.insert(i, i+1)
        }
        Incrementor { lookup_table: lut }
    }

    fn increment(&self, i: i32) -> Result<i32, IntegerOverflowError> {
        self.lookup_table.get(i)
            .map(|i| *i)
            .ok_or(IntegerOverflowError)
}

On mobile so I don’t even know if they compile though.

writing code on my phone on the train with three minutes of time, let’s go and think of the worst possible solution!

int i = 1337;

for (byte x = 0; x < 32; x++) {
    if (i & (1 << x) == 1) {
        i &= 0xFFFFFFFF ^ ( // dangit I'm out of time
    } else {
        i |= 1 << x;
        break;
    }
}

Instead of a loop, you’d use 32 of these copy-pasted if statements and a goto :p

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

Trying to avoid using any arithmetic operators, and sticking just to binary (extending beyond 16 bit unsigned ints is left as an exercise for the interested reader):

#!/usr/bin/perl

# This increments $i

my $i=1;
print "Start: $i ";

if (($i & 0b1111111111111111) == 0b1111111111111111) {die "Overflow";}
if (($i & 0b0000000000000001) == 0b0000000000000000) {$i=(($i & 0b1111111111111110) | 0b0000000000000001);}
else
{
        if (($i & 0b0111111111111111) == 0b0111111111111111) {$i=(($i & 0b0000000000000000) | 0b1000000000000000);}
        if (($i & 0b0011111111111111) == 0b0011111111111111) {$i=(($i & 0b1000000000000000) | 0b0100000000000000);}
        if (($i & 0b0001111111111111) == 0b0001111111111111) {$i=(($i & 0b1100000000000000) | 0b0010000000000000);}
        if (($i & 0b0000111111111111) == 0b0000111111111111) {$i=(($i & 0b1110000000000000) | 0b0001000000000000);}
        if (($i & 0b0000011111111111) == 0b0000011111111111) {$i=(($i & 0b1111000000000000) | 0b0000100000000000);}
        if (($i & 0b0000001111111111) == 0b0000001111111111) {$i=(($i & 0b1111100000000000) | 0b0000010000000000);}
        if (($i & 0b0000000111111111) == 0b0000000111111111) {$i=(($i & 0b1111110000000000) | 0b0000001000000000);}
        if (($i & 0b0000000011111111) == 0b0000000011111111) {$i=(($i & 0b1111111000000000) | 0b0000000100000000);}
        if (($i & 0b0000000001111111) == 0b0000000001111111) {$i=(($i & 0b1111111100000000) | 0b0000000010000000);}
        if (($i & 0b0000000000111111) == 0b0000000000111111) {$i=(($i & 0b1111111110000000) | 0b0000000001000000);}
        if (($i & 0b0000000000011111) == 0b0000000000011111) {$i=(($i & 0b1111111111000000) | 0b0000000000100000);}
        if (($i & 0b0000000000001111) == 0b0000000000001111) {$i=(($i & 0b1111111111100000) | 0b0000000000010000);}
        if (($i & 0b0000000000000111) == 0b0000000000000111) {$i=(($i & 0b1111111111110000) | 0b0000000000001000);}
        if (($i & 0b0000000000000011) == 0b0000000000000011) {$i=(($i & 0b1111111111111000) | 0b0000000000000100);}
        if (($i & 0b0000000000000001) == 0b0000000000000001) {$i=(($i & 0b1111111111111100) | 0b0000000000000010);}
}
print "End: $i\n";

Where’s NumIcrementorFactory?

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

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;
}

// this increments i: 
i = i+i---i;

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.

I like to shake the bytes around a little

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