Jens Gustedt's Blog

September 21, 2010

Anatomy of integer types in C

Filed under: C99, integers — Jens Gustedt @ 04:06

Integer types in C may present subtle traps (sic!) that many people are not aware of when doing seemingly simple things like ~0 or

(1 << (sizeof(int)*CHAR_BIT - 1))

Most times, on almost all processors these will produce the desired effects, but sometimes such a code will fail, crash, spit a lot of warnings. I will try to analyze this a bit, to show what may go wrong, here, and how you can get around a lot of possible problems.

This is a discussion that is a bit involved, you should know a bit of things about integer types and how to use them.

The simple case: unsigned char

The first and only case that is guaranteed to be simple are unsigned char. On most commodity architectures this will correspond to uint8_t, but to be most general let us assume here that it has CHAR_BIT >= 8 bits. The C99 standard enforces that this type takes values from 0 to UCHAR_MAX = M - 1 where M, the modulus, is given exactly by the formula 2CHAR_BIT.

All computation for unsigned char is computation modulo M: e.g for CHAR_BIT == 8 the result of 128 + 127 when we assign it to an unsigned char is 255, 128 + 128 is 0 and the result of 128 + 129 is 1.

There are no hidden bits in an unsigned char. The ith bit (counting from 0) is simply set by something like:

(((unsigned char)1) << i)

Simple and clean.

Other unsigned types: the first traps appear

In most aspects other unsigned types, say T, that are not _Bool are like unsigned char. They have a modulus MT (generally different from the one for unsigned char), MT = 2w, , the maximum value is MT-1, computation is done modulo MT, and the bits can be accessed in a similar way as before. Good.

(_Bool is special since converting any value that is not 0 to _Bool results in 1.)

The only thing is, that MT has not much to do with the size of the type: sizeof(T)*CHAR_BIT is only an upper bound for the width w, but it mustn’t forcibly be that value. There can be bits that don’t contribute to the value of the variable, so-called padding bits. The system may e.g use such padding bits for

  1. nothing, i.e never access these bits
  2. garbage, access them occasionally, e.g in case of “overflow”, but never clear them
  3. error correction, make sure that one bit error in the variable can be repaired
  4. the sign bit of the corresponding signed type

The first two cases are relatively harmless, although for 2 we already may have a hard time to distinguish it from 3 or 4. Either of the two cases appears on every C99 platform: _Bool is unsigned (it has no negative values) and it only has 1 value bit. Since it must at least have a width of char, it has at least 7 padding bits. A platform may use the padding bits or not, just storing 1 for a true value or UCHAR_MAX.

The case 3 is a hard one already, since it may lead to a so-called trap representation, i.e an invalid bit pattern that is stored in an object. If we suppose that we manipulate the bit pattern directly, placing a wrong bit in the padding bits can mix up the error correction and signal an error to the application. On a POSIX system, accessing such a mixed up object might signal an arithmetic exception (SIGFPE) that leads to the termination of the program.

The case 4 is really nasty. Below when discussing signed types we will see more on that.

Determine the width of an unsigned type

If we suppose that for our unsigned type T we want to find the width. From what we know this is now easy: (T)-1, minus one transformed into a value of type T will always be M-1. Since M is forcibly a power of two, we just have to check for the values w that are less or equal to sizeof(T)*CHAR_BIT, easy. In the following we will assume that we have a macro P99_TWIDTH that does that work for us.

Signed integer types: at a first glance

Signed integer types only add a little tiny bit to all that, namely the sign bit. This tells when a value is positive (sign bit 0) or negative (sign bit 1). Now there are several possibilities on how to interpret the bit pattern of a signed integer as a whole. C allows the compiler implementor to choose between three different representations:

sign and magnitude:
this is probably the most natural one for people not too involved in bit manipulation. The value bits determine the magnitude of the value and the sign bit, well, the sign.
two’s complement:
the sign bit has the symbolical value -2N
one’s complement:
the sign bit has the symbolical value -2N-1

Thus for the later two, we take the value that is given by the value bits and then subtract a large value if the sign bit is set, that makes it negative.

So why are there three different such representations? Because each has a specific problem. For the first and the third we find two representations with value 0. For sign and magnitude this is obvious; the value 0 can have the sign bit switched on or off. For one’s complement the largest value for the value bits is -2N-1, so subtracting that same value leads to 0. The encodings of 0 with sign bit on in these representations are called negative zeros. Whether or not these are trap representations (crash your application, raise a signal, do something weird…) is left to the compiler implementor.

So when manipulating bits we must be careful not to run in such a trap. But this is exactly what we do when we complement all the bits of 0: ~0 can lead to a trap representation when we have to deal with one’s complement.

The other representation, two’s complement, does not have that problem, and this one is the chosen one for most modern platforms. Unfortunately it has another problem for the minimal allowed value. The minimal possible value is simply MIN0 = -2N, namely if the sign bit is on and all other bits are off. But this value has a bizarre property: the positive value -MIN0 has no representation in the type. We may choose among to evil

  1. We allow a value of MIN0, but then the prefix operation “-” can raise a range condition since -MIN0 is not representable.
  2. We forbid MIN0 i.e mark it as a trap representation, potentially signaling an error if we do some bit manipulations.

So if MIN and MAX are the smallest and greatest value of the type besides the usual arithmetic under- or overflow we have to avoid the following computations

  • ~0, may lead to trap for one’s complement.
  • ~MAX, may lead to trap for sign and magnitude.
  • -MIN, may lead to range error for two’s complement.

Fitting unsigned and signed types together

To (almost) each unsigned integer type there is a signed one of the same size, the only exception being _Bool for which it makes not much sense to have a signed counterpart. Even if the sizes of `unsigned int` and `signed int` are the same, their widths Nu and Ns are not necessarily so. The standard only imposes that the number of value bits (the so-called precision) of the signed type is no larger than of the unsigned type. Expressed in terms of the width: Ns ≤ Nu+1.

The most common way the mapping between the bits of the unsigned and signed is as follows:

unsigned value
signed sign value

That is the sign bit of the signed is the highest order bit of the unsigned. Easy.

Things get a little more nasty if there is common padding:

unsigned padding value
signed padding sign value

This can be handled, but you’d have to avoid the common error to take sizeof(T)*CHAR_BIT for the width of the type.

To have fewer bits in the signed than in the unsigned is also possible:

unsigned padding value
signed padding sign value

But such a thing is not very likely, technically it makes not much sense.

Comes the last one, the really nasty:

unsigned padding value
signed padding sign value

Here the sign bit is hidden in the padding bit(s) of the unsigned.

  • We have no legal way to manipulate the sign bit through operations on the unsigned.
  • If the representation is two’s complement and MIN0 is allowed, -MIN0 is not even representable as an unsigned value.
  • A simple function like the following abs might signal an error on MIN0.
    unsigned abs(int a);

Fortunately platforms with this representations of signed are exotic. Historically, the reason for implementing such a design had been processors that did only implement signed arithmetic. There the easiest way for a compiler implementor is just to take the sign bit as an `overflow’ indicator for the unsigned types. Whether or not on such platforms the implementors had at least the pity to drop MIN0 from view I don’t know. Also I hope that nobody will ever port a C99 compiler to such a beast.

Getting access to all the bits

Fortunately, for unsigned char and signed char the later scenario can’t happen since unsigned char is not allowed to have padding bits. A union such as

union overlay {
   T x;
   unsigned char X[sizeof(T)];

will always allow you to access each individual bit of x, regardless what type T represents. But beware, no other type than unsigned char should be used for such a trick: signed char may having padding bits, and char you don’t even know the signdness of that beast, don’t you.

Edit: Since C11 signed char can’t have padding bits, either, but it still could have a trap representation.


  1. nit: you have a typo in “the simple case: unsigned char” :: the modulus is not 2 ^ UCHAR_MAX

    Comment by pmg — October 6, 2010 @ 11:47

  2. Thanks for this interesting article. Just for the record there is another typo in the code section of the last paragraph: ‘struct overlay’ should read ‘union overlay’ as stated in the text.

    Comment by aa — January 26, 2011 @ 10:45

    • right, thanks for the notice, corrected

      Comment by Jens Gustedt — January 26, 2011 @ 11:01

  3. This is a really useful article. I have one observation though. Unless I missed something, the C Standard doesn’t specify the ‘location’ of the padding bits. Your pictures show the padding bits always come as the most significant bits. This is probably true for every machine ever made, but would C technically allow padding elsewhere, or even have those bits non-contiguous?

    Comment by anvcom — March 31, 2016 @ 09:37

    • Have in mind that the “significance” of the bits is not directly related to the memory layout. So if we are speaking of “significance” of the bits, yes the padding bits are always the high order bits. This is just by definition. The least significant bit is the one that represents the value 1. Now the memory layout can be quite different, for example on little Endian architectures.

      Comment by Jens Gustedt — March 31, 2016 @ 10:35

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