Is using an unsigned rather than signed int more likely to cause bugs? Why?

Question

In the Google C++ Style Guide, on the topic of "Unsigned Integers", it is suggested that

Because of historical accident, the C++ standard also uses unsigned integers to represent the size of containers - many members of the standards body believe this to be a mistake, but it is effectively impossible to fix at this point. The fact that unsigned arithmetic doesn't model the behavior of a simple integer, but is instead defined by the standard to model modular arithmetic (wrapping around on overflow/underflow), means that a significant class of bugs cannot be diagnosed by the compiler.

What is wrong with modular arithmetic? Isn't that the expected behaviour of an unsigned int?

What kind of bugs (a significant class) does the guide refer to? Overflowing bugs?

Do not use an unsigned type merely to assert that a variable is non-negative.

One reason that I can think of using signed int over unsigned int, is that if it does overflow (to negative), it is easier to detect.

Answer 1

Some of the answers here mention the surprising promotion rules between signed and unsigned values, but that seems more like a problem relating to mixing signed and unsigned values, and doesn't necessarily explain why signed variables would be preferred over unsigned outside of mixing scenarios.

In my experience, outside of mixed comparisons and promotion rules, there are two primary reasons why unsigned values are bug magnets as follows.

Unsigned values have a discontinuity at zero, the most common value in programming

Both unsigned and signed integers have a discontinuities at their minimum and maximum values, where they wrap around (unsigned) or cause undefined behavior (signed). For unsigned these points are at zero and UINT_MAX. For int they are at INT_MIN and INT_MAX. Typical values of INT_MIN and INT_MAX on system with 4-byte int values are -2^31 and 2^31-1, and on such a system UINT_MAX is typically 2^32-1.

The primary bug-inducing problem with unsigned that doesn't apply to int is that it has a discontinuity at zero. Zero, of course, is a very common value in programs, along with other small values like 1,2,3. It is common to add and subtract small values, especially 1, in various constructs, and if you subtract anything from an unsigned value and it happens to be zero, you just got a massive positive value and an almost certain bug.

Consider code iterates over all values in a vector by index except the last^0.5:

for (size_t i = 0; i < v.size() - 1; i++) { // do something } 

This works fine until one day you pass in an empty vector. Instead of doing zero iterations, you get v.size() - 1 == a giant number¹ and you'll do 4 billion iterations and almost have a buffer overflow vulnerability.

You need to write it like this:

for (size_t i = 0; i + 1 < v.size(); i++) { // do something } 

So it can be "fixed" in this case, but only by carefully thinking about the unsigned nature of size_t. Sometimes you can't apply the fix above because instead of a constant one you have some variable offset you want to apply, which may be positive or negative: so which "side" of the comparison you need to put it on depends on the signedness - now the code gets really messy.

There is a similar issue with code that tries to iterate down to and including zero. Something like while (index-- > 0) works fine, but the apparently equivalent while (--index >= 0) will never terminate for an unsigned value. Your compiler might warn you when the right hand side is literal zero, but certainly not if it is a value determined at runtime.

Counterpoint

Some might argue that signed values also have two discontinuities, so why pick on unsigned? The difference is that both discontinuities are very (maximally) far away from zero. I really consider this a separate problem of "overflow", both signed and unsigned values may overflow at very large values. In many cases overflow is impossible due to constraints on the possible range of the values, and overflow of many 64-bit values may be physically impossible). Even if possible, the chance of an overflow related bug is often minuscule compared to an "at zero" bug, and overflow occurs for unsigned values too. So unsigned combines the worst of both worlds: potentially overflow with very large magnitude values, and a discontinuity at zero. Signed only has the former.

Many will argue "you lose a bit" with unsigned. This is often true - but not always (if you need to represent differences between unsigned values you'll lose that bit anyways: so many 32-bit things are limited to 2 GiB anyways, or you'll have a weird grey area where say a file can be 4 GiB, but you can't use certain APIs on the second 2 GiB half).

Even in the cases where unsigned buys you a bit: it doesn't buy you much: if you had to support more than 2 billion "things", you'll probably soon have to support more than 4 billion.

Logically, unsigned values are a subset of signed values

Mathematically, unsigned values (non-negative integers) are a subset of signed integers (just called _integers).². Yet signed values naturally pop out of operations solely on unsigned values, such as subtraction. We might say that unsigned values aren't closed under subtraction. The same isn't true of signed values.

Want to find the "delta" between two unsigned indexes into a file? Well you better do the subtraction in the right order, or else you'll get the wrong answer. Of course, you often need a runtime check to determine the right order! When dealing with unsigned values as numbers, you'll often find that (logically) signed values keep appearing anyways, so you might as well start of with signed.

Counterpoint

As mentioned in footnote (2) above, signed values in C++ aren't actually a subset of unsigned values of the same size, so unsigned values can represent the same number of results that signed values can.

True, but the range is less useful. Consider subtraction, and unsigned numbers with a range of 0 to 2N, and signed numbers with a range of -N to N. Arbitrary subtractions result in results in the range -2N to 2N in _both cases, and either type of integer can only represent half of it. Well it turns out that the region centered around zero of -N to N is usually way more useful (contains more actual results in real world code) than the range 0 to 2N. Consider any of typical distribution other than uniform (log, zipfian, normal, whatever) and consider subtracting randomly selected values from that distribution: way more values end up in [-N, N] than [0, 2N] (indeed, resulting distribution is always centered at zero).

64-bit closes the door on many of the reasons to use unsigned values as numbers

I think the arguments above were already compelling for 32-bit values, but the overflow cases, which affect both signed and unsigned at different thresholds, do occur for 32-bit values, since "2 billion" is a number that can exceeded by many abstract and physical quantities (billions of dollars, billions of nanoseconds, arrays with billions of elements). So if someone is convinced enough by the doubling of the positive range for unsigned values, they can make the case that overflow does matter and it slightly favors unsigned.

Outside of specialized domains 64-bit values largely remove this concern. Signed 64-bit values have an upper range of 9,223,372,036,854,775,807 - more than nine quintillion. That's a lot of nanoseconds (about 292 years worth), and a lot of money. It's also a larger array than any computer is likely to have RAM in a coherent address space for a long time. So maybe 9 quintillion is enough for everybody (for now)?

When to use unsigned values

Note that the style guide doesn't forbid or even necessarily discourage use of unsigned numbers. It concludes with:

Do not use an unsigned type merely to assert that a variable is non-negative.

Indeed, there are good uses for unsigned variables:

• When you want to treat an N-bit quantity not as an integer, but simply a "bag of bits". For example, as a bitmask or bitmap, or N boolean values or whatever. This use often goes hand-in-hand with the fixed width types like uint32_t and uint64_t since you often want to know the exact size of the variable. A hint that a particular variable deserves this treatment is that you only operate on it with with the bitwise operators such as ~, |, &, ^, >> and so on, and not with the arithmetic operations such as +, -, *, / etc.

  Unsigned is ideal here because the behavior of the bitwise operators is well-defined and standardized. Signed values have several problems, such as undefined and unspecified behavior when shifting, and an unspecified representation.

• When you actually want modular arithmetic. Sometimes you actually want 2^N modular arithmetic. In these cases "overflow" is a feature, not a bug. Unsigned values give you what you want here since they are defined to use modular arithmetic. Signed values cannot be (easily, efficiently) used at all since they have an unspecified representation and overflow is undefined.

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^0.5 After I wrote this I realized this is nearly identical to Jarod's example, which I hadn't seen - and for good reason, it's a good example!

¹ We're talking about size_t here so usually 2^32-1 on a 32-bit system or 2^64-1 on a 64-bit one.

² In C++ this isn't exactly the case because unsigned values contain more values at the upper end than the corresponding signed type, but the basic problem exists that manipulating unsigned values can result in (logically) signed values, but there is no corresponding issue with signed values (since signed values already include unsigned values).

Answer 2

As stated, mixing unsigned and signed might lead to unexpected behaviour (even if well defined).

Suppose you want to iterate over all elements of vector except for the last five, you might wrongly write:

for (int i = 0; i < v.size() - 5; ++i) { foo(v[i]); } // Incorrect // for (int i = 0; i + 5 < v.size(); ++i) { foo(v[i]); } // Correct 

Suppose v.size() < 5, then, as v.size() is unsigned, s.size() - 5 would be a very large number, and so i < v.size() - 5 would be true for a more expected range of value of i. And UB then happens quickly (out of bound access once i >= v.size())

If v.size() would have return signed value, then s.size() - 5 would have been negative, and in above case, condition would be false immediately.

On the other side, index should be between [0; v.size()[ so unsigned makes sense. Signed has also its own issue as UB with overflow or implementation-defined behaviour for right shift of a negative signed number, but less frequent source of bug for iteration.