~ operator returns signed value

Discussion in 'C Programming' started by RezaRob, Sep 23, 2007.

  1. RezaRob

    RezaRob Guest

    gcc is clearly returning -1 when I do this
    ~ (unsigned short) 0

    which is not the case if "unsigned int" is used instead of "unsigned
    short".

    What does the C standard say about that? (reference to the standard
    is appreciated.)

    Thanks.

    Reza.
     
    RezaRob, Sep 23, 2007
    #1
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  2. RezaRob

    pete Guest

    If INT_MAX is greater than or equal to USHRT_MAX,
    then (~ (unsigned short) 0) means the same thing as
    (~ (int)(unsigned short) 0).



    ISO/IEC 9899: 1990
    6.2.1 Arithmetic operands
    6.2.1.1 Characters and integers
    A char, a short int, or an int bit-field,
    or their signed or unsigned varieties,
    or an enumeration type, may be used in an expression
    wherever an int or unsigned int may be used.
    If an int can represent all values of the original type,
    the value is converted to an int; otherwise,
    it is converted to an unsigned int.
    These are called the integral romotions.
    All other arithmetic types are unchanged by the integral promotions.
     
    pete, Sep 23, 2007
    #2
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  3. RezaRob

    pete Guest

    That should be "These are called the integral promotions."
     
    pete, Sep 23, 2007
    #3
  4. RezaRob

    RezaRob Guest

    Pete, I really appreciate this.

    Reza.
     
    RezaRob, Sep 24, 2007
    #4
  5. RezaRob

    pete Guest

    The following three expressions have the same type and value:
    (USHRT_MAX)
    ((unsigned short) ~0u)
    ((unsigned short) -1)
     
    pete, Sep 24, 2007
    #5
  6. RezaRob

    Martin Wells Guest

    Are you sure about the middle one?

    1: 0u is of type unsigned int.

    2: ~0u is of type unsigned int and is equal to UINT_MAX.

    3: (unsigned short)~0u will be UINT_MAX % USHRT_MAX, or at least I
    think it will, so it can't be USHRT_MAX.

    but of course I'm open to correction!

    Martin
     
    Martin Wells, Sep 25, 2007
    #6
  7. RezaRob

    user923005 Guest

    Sure, there's a cast in it to unsigned short.
    ;-)
    [snip]
     
    user923005, Sep 25, 2007
    #7
  8. Martin Wells said:
    Oh good. :) Re your Point 3, a single counter-example should suffice.
    Consider a system with UINT_MAX == USHRT_MAX (e.g. typical 1990s MS-DOS
    systems, or an SIL64 system such as at least one Cray. Clearly, on such a
    system, UINT_MAX % USHRT_MAX will be 0 (for the same reason that x % x is
    0), and yet (unsigned short)~0u will be equivalent to ~0u, which is most
    emphatically NOT 0.
     
    Richard Heathfield, Sep 25, 2007
    #8
  9. RezaRob

    Martin Wells Guest

    Richard:

    Sorry I've red that a few times but I still don't understand what
    you're saying.

    Was pete right or wrong?

    Martin
     
    Martin Wells, Sep 25, 2007
    #9
  10. Martin Wells said:
    You claimed that "(unsigned short)~0u will be UINT_MAX % USHRT_MAX". I
    disputed that claim, by pointing out a counter-example.
    Presumably. :)
     
    Richard Heathfield, Sep 25, 2007
    #10
  11. RezaRob

    Martin Wells Guest

    Richard:

    I'm still lost. I'll try think it through using MS-DOS as an example:

    1: 0u == the number 0
    2: ~0u == UINT_MAX == the number 65535
    3: (unsigned short)~0u == the number USHRT_MAX % 65535 == the
    number 65535 % 65535 == the number 0.

    This leads me to believe that pete was wrong... unless I'm missing
    something.

    Martin
     
    Martin Wells, Sep 25, 2007
    #11
  12. Martin Wells said:
    Please explain why you think this is the case. I can see no justification
    for your assumption that (unsigned short)~0u is equal to USHRT_MAX % 65535
    (and yes, I know we're talking 16-bit systems. You originally wrote that
    it's equal to UINT_MAX % USHRT_MAX, which is also incorrect.

    You might reasonably claim that, on a 16-bit system, (unsigned short)~0u is
    equal to UINT_MAX % (USHRT_MAX + 1), but that would just be equal to
    UINT_MAX (and indeed USHRT_MAX).
    You appear to be missing a + 1. :)
     
    Richard Heathfield, Sep 25, 2007
    #12
  13. RezaRob

    pete Guest

    This part of your post: "UINT_MAX % USHRT_MAX"
    is wrong.
     
    pete, Sep 25, 2007
    #13
  14. RezaRob

    Martin Wells Guest

    pete:

    I'm gonna give the middle one another go. . .

    Before I begin though, I'm gonna pretend I'm working with the
    following machine:

    CHAR_BIT == 11
    sizeof(short) == 2 (all 22 bits are value bits)
    sizeof(int) == 3 (31 bits are value bits, 1 is padding)

    Therefore, we have:

    UINT_MAX = 2147483647
    USHRT_MAX = 4194303

    1: 0u

    Expression Type: unsigned int
    Expression Value: 0

    2: ~0u

    Expression Type: unsigned int
    Expression Value: 2147483647 (i.e. UINT_MAX)

    3: (unsigned short)~0u

    == UINT_MAX % (USHRT_MAX + 1)
    == 2147483647 % 4194304
    == 4194303

    Oh, so that worked out OK. I suppose what I was tryna get my head
    around from the start was how you could so blindy make the assumption

    Given: a is a positive integer
    b is a positive integer
    a is greater than or equal to b
    That:

    ( pow(2,a) - 1 ) % pow(2,b)

    is equal to:

    pow(2,b) - 1

    Martin
     
    Martin Wells, Sep 26, 2007
    #14
  15. RezaRob

    Martin Wells Guest


    Shuda written "2 are padding" instead of "1 is".

    Martin
     
    Martin Wells, Sep 26, 2007
    #15
  16. RezaRob

    pete Guest

    Actually, I got one of those wrong.
    The type of USHRT_MAX isn't unsigned short.
     
    pete, Sep 26, 2007
    #16
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