Examples of problems with integers

Hello! A few days back we talked about problems with floating point numbers.

This got me thinking – but what about integers? Of course integers have all kinds of problems too – anytime you represent a number in a small fixed amount of space (like 8/16/32/64 bits), you’re going to run into problems.

So I asked on Mastodon again for examples of integer problems and got all kinds of great responses again. Here’s a table of contents.

Like last time, I’ve written some example programs to demonstrate these problems. I’ve tried to use a variety of languages in the examples (Go, Javascript, Java, and C) to show that these problems don’t just show up in super low level C programs – integers are everywhere!

Also I’ve probably made some mistakes in here, I learned several things while writing this.

example 1: the small database primary key

One of the most classic (and most painful!) integer problems is:

1. You create a database table where the primary key is a 32-bit unsigned integer, thinking “4 billion rows should be enough for anyone!”
2. You are massively successful and eventually, your table gets close to 4 billion rows
3. oh no!
4. You need to do a database migration to switch your primary key to be a 64-bit integer instead

If the primary key actually reaches its maximum value I’m not sure exactly what happens, I’d imagine you wouldn’t be able to create any new database rows and it would be a very bad day for your massively successful service.

example 2: integer overflow/underflow

Here’s a Go program:

package main

import "fmt"

func main() {
var x uint32 = 5
var length uint32 = 0
if x < length-1 {
fmt.Printf("%d is less than %d\n", x, length-1)
}
}


This slightly mysteriously prints out:

5 is less than 4294967295


That true, but it’s not what you might have expected.

what’s going on?

0 - 1 is equal to the 4 bytes 0xFFFFFFFF.

There are 2 ways to interpret those 4 bytes:

1. As a signed integer (-1)
2. As an unsigned integer (4294967295)

Go here is treating length - 1 as a unsigned integer, because we defined x and length as uint32s (the “u” is for “unsigned”). So it’s testing if 5 is less than 4294967295, which it is!

what do we do about it?

I’m not actually sure if there’s any way to automatically detect integer overflow errors in Go. (though it looks like there’s a github issue from 2019 with some discussion)

Some brief notes about other languages:

• Lots of languages (Python, Java, Ruby) don’t have unsigned integers at all, so this specific problem doesn’t come up
• In C, you can compile with clang -fsanitize=unsigned-integer-overflow. Then if your code has an overflow/underflow like this, the program will crash.
• Similarly in Rust, if you compile your program in debug mode it’ll crash if there’s an integer overflow. But in release mode it won’t crash, it’ll just happily decide that 0 - 1 = 4294967295.

The reason Rust doesn’t check for overflows if you compile your program in release mode (and the reason C and Go don’t check) is that – these checks are expensive! Integer arithmetic is a very big part of many computations, and making sure that every single addition isn’t overflowing makes it slower.

aside: how do computers represent negative integers?

I mentioned in the last section that 0xFFFFFFFF can mean either -1 or 4294967295. You might be thinking – what??? Why would 0xFFFFFFFF mean -1?

So let’s talk about how computers represent negative integers for a second.

I’m going to simplify and talk about 8-bit integers instead of 32-bit integers, because there are less of them and it works basically the same way.

You can represent 256 different numbers with an 8-bit integer: 0 to 255

00000000 -> 0
00000001 -> 1
00000010 -> 2
...
11111111 -> 255


But what if you want to represent negative integers? We still only have 8 bits! So we need to reassign some of these and treat them as negative numbers instead.

Here’s the way most modern computers do it:

1. Every number that’s 128 or more becomes a negative number instead
2. How to know which negative number it is: take the positive integer you’d expect it to be, and then subtract 256

So 255 becomes -1, 128 becomes -128, and 200 becomes -56.

Here are some maps of bits to numbers:

00000000 -> 0
00000001 -> 1
00000010 -> 2
01111111 -> 127
10000000 -> -128 (previously 128)
10000001 -> -127 (previously 129)
10000010 -> -126 (previously 130)
...
11111111 -> -1 (previously 255)


This gives us 256 numbers, from -128 to 127.

And 11111111 (or 0xFF, or 255) is -1.

For 32 bit integers, it’s the same story, except it’s “every number larger than 2^31 becomes negative” and “subtract 2^32”. And similarly for other integer sizes.

That’s how we end up with 0xFFFFFFFF meaning -1.

there are multiple ways to represent negative integers

The way we just talked about of representing negative integers (“it’s the equivalent positive integer, but you subtract 2^n”) is called two’s complement, and it’s the most common on modern computers. There are several other ways though, the wikipedia article has a list.

weird thing: the absolute value of -128 is negative

This Go program has a pretty simple abs() function that computes the absolute value of an integer:

package main

import (
"fmt"
)

func abs(x int8) int8 {
if x < 0 {
return -x
}
return x
}

func main() {
fmt.Println(abs(-127))
fmt.Println(abs(-128))
}


This prints out:

127
-128


This is because the signed 8-bit integers go from -128 to 127 – there is no +128! Some programs might crash when you try to do this (it’s an overflow), but Go doesn’t.

Now that we’ve talked about signed integers a bunch, let’s dig into another example of how they can cause problems.

example 3: decoding a binary format in Java

Let’s say you’re parsing a binary format in Java, and you want to get the first 4 bits of the byte 0x90. The correct answer is 9.

public class Main {
public static void main(String[] args) {
byte b = (byte) 0x90;
System.out.println(b >> 4);
}
}


This prints out “-7”. That’s not right!

what’s going on?

There are two things we need to know about Java to make sense of this:

1. Java doesn’t have unsigned integers.
2. Java can’t right shift bytes, it can only shift integers. So anytime you shift a byte, it has to be promoted into an integer.

Let’s break down what those two facts mean for our little calculation b >> 4:

• In bits, 0x90 is 10010000. This starts with a 1, which means that it’s more than 128, which means it’s a negative number
• Java sees the >> and decides to promote 0x90 to an integer, so that it can shift it
• The way you convert a negative byte to an 32-bit integer is to add a bunch of 1s at the beginning. So now our 32-bit integer is 0xFFFFFF90 (F being 15, or 1111)
• Now we right shift (b >> 4). By default, Java does a signed shift, which means that it adds 0s to the beginning if it’s positive, and 1s to the beginning if it’s negative. (>>> is an unsigned shift in Java)
• We end up with 0xFFFFFFF9 (having cut off the last 4 bits and added more 1s at the beginning)
• As a signed integer, that’s -7!

what can you do about it?

I don’t the actual idiomatic way to do this in Java is, but the way I’d naively approach fixing this is to put in a bit mask before doing the right shift. So instead of:

b >> 4


we’d write

(b & 0xFF) >> 4


b & 0xFF seems redundant (b is already a byte!), but it’s actually not because b is being promoted to an integer.

Now instead of 0x90 -> 0xFFFFFF90 -> 0xFFFFFFF9, we end up calculating 0x90 -> 0xFFFFFF90 -> 0x00000090 -> 0x00000009, which is the result we wanted: 9.

And when we actually try it, it prints out “9”.

Also, if we were using a language with unsigned integers, the natural way to deal with this would be to treat the value as an unsigned integer in the first place. But that’s not possible in Java.

example 4: misinterpreting an IP address or string as an integer

I don’t know if this is technically a “problem with integers” but it’s funny so I’ll mention it: Rachel by the bay has a bunch of great examples of things that are not integers being interpreted as integers. For example, “HTTP” is 0x48545450 and 2130706433 is 127.0.0.1.

She points out that you can actually ping any integer, and it’ll convert that integer into an IP address, for example:

$ping 2130706433 PING 2130706433 (127.0.0.1): 56 data bytes$ ping 132848123841239999988888888888234234234234234234
PING 132848123841239999988888888888234234234234234234 (251.164.101.122): 56 data bytes


(I’m not actually sure how ping is parsing that second integer or why ping accepts these giant larger-than-2^64-integers as valid inputs, but it’s a fun weird thing)

example 5: security problems because of integer overflow

Another integer overflow example: here’s a search for CVEs involving integer overflows. There are a lot! I’m not a security person, but here’s one random example: this json parsing library bug

My understanding of that json parsing bug is roughly:

• you load a JSON file that’s 3GB or something, or 3,000,000,000
• due to an integer overflow, the code allocates close to 0 bytes of memory instead of ~3GB amount of memory
• but the JSON file is still 3GB, so it gets copied into the tiny buffer with almost 0 bytes of memory
• this overwrites all kinds of other memory that it’s not supposed to

The CVE says “This vulnerability mostly impacts process availability”, which I think means “the program crashes”, but sometimes this kind of thing is much worse and can result in arbitrary code execution.

My impression is that there are a large variety of different flavours of security vulnerabilities caused by integer overflows.

example 6: the case of the mystery byte order

One person said that they’re do scientific computing and sometimes they need to read files which contain data with an unknown byte order.

Let’s invent a small example of this: say you’re reading a file which contains 4 bytes - 00, 00, 12, and 81 (in that order), that you happen to know represent a 4-byte integer. There are 2 ways to interpret that integer:

1. 0x00001281 (which translates to 4737). This order is called “big endian”
2. 0x81120000 (which translates to 2165440512). This order is called “little endian”.

Which one is it? Well, maybe the file contains some metadata that specifies the endianness. Or maybe you happen to know what machine it was generated on and what byte order that machine uses. Or maybe you just read a bunch of values, try both orders, and figure out which makes more sense. Maybe 2165440512 is too big to make sense in the context of whatever your data is supposed to mean, or maybe 4737 is too small.

A couple more notes on this:

• this isn’t just a problem with integers, floating point numbers have byte order too
• this also comes up when reading data from a network, but in that case the byte order isn’t a “mystery”, it’s just going to be big endian. But x86 machines (and many others) are little endian, so you have to swap the byte order of all your numbers.

example 7: modulo of negative numbers

This is more of a design decision about how different programming languages design their math libraries, but it’s still a little weird and lots of people mentioned it.

Let’s say you write -13 % 3 in your program, or 13 % -3. What’s the result?

It turns out that different programming languages do it differently, for example in Python -13 % 3 = 2 but in Javascript -13 % 3 = -1.

There’s a table in this blog post that describes a bunch of different programming languages’ choices.

example 8: compilers removing integer overflow checks

We’ve been hearing a lot about integer overflow and why it’s bad. So let’s imagine you try to be safe and include some checks in your programs – after each addition, you make sure that the calculation didn’t overflow. Like this:

#include <stdio.h>

#define INT_MAX 2147483647

int check_overflow(int n) {
n = n + 100;
if (n + 100 < 0)
return -1;
return 0;
}

int main() {
int result = check_overflow(INT_MAX);
printf("%d\n", result);
}


check_overflow here should return -1 (failure), because INT_MAX + 100 is more than the maximum integer size.

$gcc check_overflow.c -o check_overflow && ./check_overflow -1$ gcc -O3 check_overflow.c  -o check_overflow && ./check_overflow
0


That’s weird – when we compile with gcc, we get the answer we expected, but with gcc -O3, we get a different answer. Why?

what’s going on?

My understanding (which might be wrong) is:

1. Signed integer overflow in C is undefined behavior. I think that’s because different C implementations might be using different representations of signed integers (maybe they’re using one’s complement instead of two’s complement or something)
2. “undefined behaviour” in C means “the compiler is free to do literally whatever it wants after that point” (see this post With undefined behaviour, anything is possible by Raph Levine for a lot more)
3. Some compiler optimizations assume that undefined behaviour will never happen. They’re free to do this, because – if that undefined behaviour did happen, then they’re allowed to do whatever they want, so “run the code that I optimized assuming that this would never happen” is fine.
4. So this if (n + 100 < 0) check is irrelevant – if that did happen, it would be undefined behaviour, so there’s no need to execute the contents of that if statement.

So, that’s weird. I’m not going to write a “what can you do about it?” section here because I’m pretty out of my depth already.

I certainly would not have expected that though.

My impression is that “undefined behaviour” is really a C/C++ concept, and doesn’t exist in other languages in the same way except in the case of “your program called some C code in an incorrect way and that C code did something weird because of undefined behaviour”. Which of course happens all the time.

example 9: the && typo

This one was mentioned as a very upsetting bug. Let’s say you have two integers and you want to check that they’re both nonzero.

In Javascript, you might write:

if a && b {
/* some code */
}


But you could also make a typo and type:

if a & b {
/* some code */
}


This is still perfectly valid code, but it means something completely different – it’s a bitwise and instead of a boolean and. Let’s go into a Javascript console and look at bitwise vs boolean and for 9 and 4:

> 9 && 4
4
> 9 & 4
0
> 4 && 5
5
> 4 & 5
4


It’s easy to imagine this turning into a REALLY annoying bug since it would be intermittent – often x & y does turn out to be truthy if x && y is truthy.