c++cintegerlanguage-lawyerdata-representation

What do the C and C++ standards say about bit-level integer representation and manipulation?


I know the C and C++ standards don't dictate a particular representation for numbers (could be two's complement, sign-and-magnitude, etc.). But I don't know the standards well enough (and couldn't find if it's stated) to know if there are any particular restrictions/guarantees/reserved representations made when working with bits. Particularly:

  1. If all the bits in an integer type are zero, does the integer as whole represent zero?
  2. If any bit in an integer type is one, does the integer as a whole represent non-zero? (if this is a "yes" then some representations like sign-and-magnitude would be additionally restricted)
  3. Is there a guaranteed way to check if any bit is not set?
  4. Is there a guaranteed way to check if any bit is set? (#3 and #4 kind of depend on #1 and #2, because I know how to set, for example the 5th bit (see #5) in some variable x, and I'd like to check a variable y to see if it's 5th bit is 1, I would like to know if if (x & y) will work (because as I understand, this relies on the value of the representation and not whether nor not that bit is actually 1 or 0))
  5. Is there a guaranteed way to set the left-most and/or right-most bits? (At least a simpler way than taking a char c with all bits true (set by c = c | ~c) and doing c = c << (CHAR_BIT - 1) for setting the high-bit and c = c ^ (c << 1) for the low-bit, assuming I'm not making any assumptions I should't be, given these questions)
  6. If the answer to #1 is "no" how could one iterate over the bits in an integer type and check if each one was a 1 or a 0?

I guess my overall question is: are there any restrictions/guarantees/reserved representations made by the C and C++ standards regarding bits and integers, despite the fact that an integer's representation is not mandated (and if the C and C++ standards differ in this regard, what's their difference)?

I came up with these questions while doing my homework which required me to do some bit manipulating (note these aren't questions from my homework, these are much more "abstract").

Edit: As to what I refer to as "bits," I mean "value forming" bits and am not including "padding" bits.


Solution

  • (1) If all the bits in an integer type are zero, does the integer as whole represent zero?

    Yes, the bit pattern consisting of all zeroes always represents 0:

    The representations of integral types shall define values by use of a pure binary numeration system.49 [§3.9.1/7]

    49 A positional representation for integers that uses the binary digits 0 and 1, in which the values represented by successive bits are additive, begin with 1, and are multiplied by successive integral power of 2, except perhaps for the bit with the highest position.


    (2) If any bit in an integer type is one, does the integer as a whole represent non-zero? (if this is a "yes" then some representations like sign-and-magnitude would be additionally restricted)

    No. In fact, signed magnitude is specifically allowed:

    [ Example: this International Standard permits 2’s complement, 1’s complement and signed magnitude representations for integral types. —end example ] [§3.9.1/7]


    (3) Is there a guaranteed way to check if any bit is not set?

    I believe the answer to this is "no," if you consider signed types. It is equivalent to equality testing with a bit pattern of all ones, which is only possible if you have a way to produce a signed number with bit pattern of all ones. For an unsigned number this representation is guaranteed, but casting from unsigned to signed is undefined if the number is unrepresentable:

    If the destination type is signed, the value is unchanged if it can be represented in the destination type (and bit-field width); otherwise, the value is implementation-defined. [§4.7/3]


    (4) Is there a guaranteed way to check if any bit is set?

    I don't think so, because signed magnitude is allowed—0 would compare equal to −0. But it should be possible with unsigned numbers.


    (5) Is there a guaranteed way to set the left-most and/or right-most bits?

    Again, I believe the answer is "yes" for unsigned numbers, but "no" for signed numbers. Shifts are undefined for negative signed numbers:

    Otherwise, if E1 has a signed type and non-negative value, and E1 × 2E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined. [§5.8/2]