4.5 — Unsigned integers, and why to avoid them

Unsigned integers

In the previous lesson (4.4 -- Signed integers), we covered signed integers, which are a set of types that can hold positive and negative whole numbers, including 0.

C++ also supports unsigned integers. Unsigned integers are integers that can only hold non-negative whole numbers.

Defining unsigned integers

To define an unsigned integer, we use the unsigned keyword. By convention, this is placed before the type:

unsigned short us;
unsigned int ui;
unsigned long ul;
unsigned long long ull;

Unsigned integer range

A 1-byte unsigned integer has a range of 0 to 255. Compare this to the 1-byte signed integer range of -128 to 127. Both can store 256 different values, but signed integers use half of their range for negative numbers, whereas unsigned integers can store positive numbers that are twice as large.

Here’s a table showing the range for unsigned integers:

An n-bit unsigned variable has a range of 0 to (2ⁿ)-1.

When no negative numbers are required, unsigned integers are well-suited for networking and systems with little memory, because unsigned integers can store more positive numbers without taking up extra memory.

Remembering the terms signed and unsigned

New programmers sometimes get signed and unsigned mixed up. The following is a simple way to remember the difference: in order to differentiate negative numbers from positive ones, we use a negative sign. If a sign is not provided, we assume a number is positive. Consequently, an integer with a sign (a signed integer) can tell the difference between positive and negative. An integer without a sign (an unsigned integer) assumes all values are positive.

Unsigned integer overflow

What happens if we try to store the number 280 (which requires 9 bits to represent) in a 1-byte (8-bit) unsigned integer? The answer is overflow.

If an unsigned value is out of range, it is divided by one greater than the largest number of the type, and only the remainder kept.

The number 280 is too big to fit in our 1-byte range of 0 to 255. 1 greater than the largest number of the type is 256. Therefore, we divide 280 by 256, getting 1 remainder 24. The remainder of 24 is what is stored.

Here’s another way to think about the same thing. Any number bigger than the largest number representable by the type simply “wraps around” (sometimes called “modulo wrapping”). 255 is in range of a 1-byte integer, so 255 is fine. 256, however, is outside the range, so it wraps around to the value 0. 257 wraps around to the value 1. 280 wraps around to the value 24.

Let’s take a look at this using 2-byte shorts:

#include <iostream>

int main()
{
    unsigned short x{ 65535 }; // largest 16-bit unsigned value possible
    std::cout << "x was: " << x << '\n';

    x = 65536; // 65536 is out of our range, so we get modulo wrap-around
    std::cout << "x is now: " << x << '\n';

    x = 65537; // 65537 is out of our range, so we get modulo wrap-around
    std::cout << "x is now: " << x << '\n';

    return 0;
}

What do you think the result of this program will be?

(Note: If you try to compile the above program, your compiler should issue warnings about overflow or truncation -- you’ll need to disable “treat warnings as errors” to run the program)

x was: 65535
x is now: 0
x is now: 1

It’s possible to wrap around the other direction as well. 0 is representable in a 2-byte unsigned integer, so that’s fine. -1 is not representable, so it wraps around to the top of the range, producing the value 65535. -2 wraps around to 65534. And so forth.

#include <iostream>

int main()
{
    unsigned short x{ 0 }; // smallest 2-byte unsigned value possible
    std::cout << "x was: " << x << '\n';

    x = -1; // -1 is out of our range, so we get modulo wrap-around
    std::cout << "x is now: " << x << '\n';

    x = -2; // -2 is out of our range, so we get modulo wrap-around
    std::cout << "x is now: " << x << '\n';

    return 0;
}

x was: 0
x is now: 65535
x is now: 65534

The above code triggers a warning in some compilers, because the compiler detects that the integer literal is out-of-range for the given type. If you want to compile the code anyway, temporarily disable “Treat warnings as errors”.

The controversy over unsigned numbers

Many developers (and some large development houses, such as Google) believe that developers should generally avoid unsigned integers.

This is largely because of two behaviors that can cause problems.

First, with signed values, it takes a little work to accidentally overflow the top or bottom of the range because those values are far from 0. With unsigned numbers, it is much easier to overflow the bottom of the range, because the bottom of the range is 0, which is close to where the majority of our values are.

Consider the subtraction of two unsigned numbers, such as 2 and 3:

#include <iostream>

// int is 4 bytes
int main()
{
	unsigned int x{ 2 };
	unsigned int y{ 3 };

	std::cout << x - y << '\n'; // 2 - 3 = 4294967295

	return 0;
}

You and I know that 2 - 3 is -1, but -1 can’t be represented as an unsigned integer, so we get overflow and the following result:

4294967295

Another common unwanted wrap-around happens when an unsigned integer is repeatedly decremented by 1 (using the -- operator) and becomes less than 0. You’ll see an example of this when loops are introduced.

Second, and more insidiously, unexpected behavior can result when you mix signed and unsigned integers. In C++, if a mathematical operation (e.g. arithmetic or comparison) has one signed integer and one unsigned integer, the signed integer will usually be converted to an unsigned integer. And the result will thus be unsigned. For example:

#include <iostream>

// int is 4 bytes
int main()
{
	unsigned int u{ 2 };
	signed int s{ 3 };

	std::cout << u - s << '\n'; // 2 - 3 = 4294967295

	return 0;
}

In this case, if u were signed, the correct result would be produced. But because u is unsigned (which is easy to miss), s gets converted to unsigned, and the result (-1) is unsigned, so we get overflow and an unexpected answer.

Here’s another example:

#include <iostream>

// int is 4 bytes
int main()
{
    signed int s { -1 };
    unsigned int u { 1 };

    if (s < u) // -1 is implicitly converted to 4294967295, and 4294967295 < 1 is false
        std::cout << "-1 is less than 1\n";
    else
        std::cout << "1 is less than -1\n"; // this statement executes

    return 0;
}

This program is well formed, compiles, and is logically consistent to the eye. But it prints the wrong answer. And while your compiler should warn you about a signed/unsigned mismatch in this case, your compiler will also generate identical warnings for other cases that do not suffer from this problem (e.g. when both numbers are positive), making it hard to detect when there is an actual problem.

Additionally, there are other problematic cases that are challenging to detect. Consider the following:

#include <iostream>

// int is 4 bytes
void doSomething(unsigned int x)
{
    // Run some code x times

    std::cout << "x is " << x << '\n';
}

int main()
{
    doSomething(-1);

    return 0;
}

The author of doSomething() was expecting someone to call this function with only positive numbers. But the caller is passing in -1 -- clearly a mistake, but one made regardless. What happens in this case?

The signed argument of -1 gets implicitly converted to an unsigned parameter. -1 isn’t in the range of an unsigned number, so it wraps around to 4294967295. Then your program goes ballistic.

Even more problematically, it can be hard to prevent this from happening. Unless you’ve configured your compiler to be aggressive about producing signed/unsigned conversion warnings (and you should), your compiler probably won’t even complain about this.

All of these problems are commonly encountered, produce unexpected behavior, and are hard to find, even using automated tools designed to detect problem cases.

Given the above, the somewhat controversial best practice that we’ll advocate for is to avoid unsigned types except in specific circumstances.