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.

Conditional adder:

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

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

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

This could even be made generic to any integer type with a little more effort

C# .NET using reflection, integer underflow, and a touch of LINQ. Should work for all integer types. (edit: also works with char values)

// 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.");
}

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";

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

My favourite one is:

i -=- 1

I like to shake the bytes around a little

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

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

Typing on mobile please excuse.

i = 0
while i != 1:
  pass

# i is now 1

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;
}

Create a python file that only contains this function

def increase_by_one(i):
    # this increments i
    f=open(__file__).read()
    st=f[28:-92][0]
    return i+f.count(st)

Then you can import this function and it will raise an index error if the comment is not there, coming close to the most literal way

Any code which does not contain the comment “this increments i:” will produce a compile error and fail to run.

could be interpreted in python

Something like that should do it:

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

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.