Sizes of pointers

Discussion in 'C Programming' started by James Harris \(es\), Jul 30, 2013.

  1. The latest draft is

    One of the few differences between the draft and the published standard
    is actually relevant here. The draft adds another exception to the rule
    that an array expression is converted to a pointer, namely when it's the
    operand of the new _Alignof operator. The published standard removes
    this, because _Alignof (for unclear reasons) can only be applied to a
    parenthesized type name, not to an expression
    Keith Thompson, Aug 9, 2013
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  2. (snip on modulo and pointers)
    As well as I understand it, it should satisfy the alignment
    requirements, but isn't necessarily optimal. A processor might
    slow down for some alignment, and the is no requirement that C
    avoid such slow alignments.

    I suppose that goes under quality of implementation, but some
    might be disapointed with the results.

    -- glen
    glen herrmannsfeldt, Aug 9, 2013
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  3. James Harris \(es\)

    James Kuyper Guest

    The C standard imposes no such requirement; but market forces favor
    vendors who create C implementations that impose alignment requirements
    that reasonably reflect the impact of mis-aligned access. Impose a
    stricter alignment than necessary, and the generated code wastes memory;
    impose too lenient of an alignment, and the generated code wastes time
    on mis-aligned access. A vendor who doesn't pay attention to it's
    customer's preferences with regard to time vs. memory space trade-offs
    will lose business to one who does. That includes giving them
    optimization options to adjust those preferences.
    James Kuyper, Aug 9, 2013
  4. (snip)
    I was thinking about this some more. If a machine with sign
    magnitude representation had signed addressing, you would not be
    able to pretend that they were unsigned. The magnitude decreases
    when you increment a negative address.

    I don't remember any, but it would be possible in the early
    years of computing, maybe in the decimal addressing days.

    -- glen
    glen herrmannsfeldt, Aug 9, 2013
  5. (snip on alignment requirements)
    And hardware can progress faster than software can keep up.

    For the 8087, only two byte alignment mattered for floating point,
    for the 80486, four byte alignment was faster, and, on the pentium
    and later eight byte alignment for double was significantly faster.

    Yet C implementations were doing four byte alignment well into
    the pentium and later days. (Hopefully fixed by now.)

    -- glen
    glen herrmannsfeldt, Aug 9, 2013
  6. James Harris \(es\)

    James Kuyper Guest

    If imposing stricter alignment requirements to achieve better processing
    speeds would have won an implementor enough new customers to cover the
    development costs, some implementor would have done so. A situation like
    that can exist only if users are sufficiently apathetic about the issue
    that they don't impose effective pressure on the implementors. And if
    the users are that apathetic about the issue, does it really matter?
    James Kuyper, Aug 9, 2013
  7. James Harris \(es\)

    Eric Sosman Guest


    Addresses on the Honeywell 8200[*] weren't signed, but all
    the registers were -- including the program counter. So if you
    set the PC's sign bit, the "PC = PC + 1" that advanced from one
    instruction to the next actually *decremented* the instruction
    address. The game, of course, was to concoct an instruction
    sequence that would run forward for a while, set the sign bit,
    run backwards through the same set of instructions, and somehow
    regain sanity before crashing.

    [*] Circa 1967. Dim memory: an H-800 48-bit word-oriented
    compute unit (maybe more than one?) lashed to an H-200 character-
    oriented machine to do the I/O. Three-address arithmetic, with
    a lot of built-in masking to pack multiple fields in those wide
    words and to deal with varying byte widths. More registers (it
    seemed) than the checkout lines of the world's biggest WalMart.

    Eric Sosman, Aug 9, 2013
  8. [...]

    You've been saying, if I understand you correctly, that pointers have
    or should have the same mathematical properties as unsigned integers.

    There's a lot more to mathematics than integers, or even numbers.

    C pointers can be described by a mathematical model that doesn't
    define the results of certain operations (just as a model for real
    numbers doesn't define division by zero or sqrt(-1)).

    C pointers can be thought of as mathematical entities that share
    some, but not all, characteristics with integers. Mathematics
    abounds with such entities. Look into group theory if you're not
    already familiar with it.

    The strong distinction C makes between integers and pointers is
    not the result of mathematical ignorance, as you've suggested.
    It's the result of an understanding of which characteristics are
    appropriate for an abstract model of pointers that can be implemented
    on a wide variety of physical hardware.
    Keith Thompson, Aug 9, 2013
  9. James Harris \(es\)

    Ike Naar Guest

    Any of {0,1,2,3,4,5,6,7} if mem is char aligned
    Any of {0,2,4,6} if mem is short and char aligned
    Any of {0,4} if mem is int,short,char aligned
    Ike Naar, Aug 10, 2013
  10. James Harris \(es\)

    James Kuyper Guest

    I've got Rosario1903's nonsense killfiled, so I'm not sure what he's
    said about what "address" is.

    If "address" is a pointer, as seems consistent with the what I've seen
    in responses to his messages, the only thing guaranteed for this code is
    a diagnostic message about the constraint violation (6.5.5p2).

    If "address" is the result of converting a pointer to uintptr_t, the
    only thing guaranteed about the value of address%8 is that it is in the
    range 0-7, and that's true regardless of how the pointer value is aligned.
    James Kuyper, Aug 10, 2013
  11. A pointer is a specific C variable which can be dereferenced, and on any
    sane C implementation holds a memory address.
    An address is any representation which can be converted ultimately to
    impulses along the bus to read the memory. So it's meaningful to take
    modulus or do other operations on a address which are forbidden in
    pointers. It's also meaningful to talk about a "negative" address, though
    ultimately, like all computer values, it's just set bits and unset bits
    and any construction we put upon them is human.
    Malcolm McLean, Aug 10, 2013
  12. If you're going to make this kind of distinction (which is a
    perfectly valid one), I suggest not using terms that are synonyms
    in the C standard. For example, the unary "&" operator yields
    the *address* of its operand, which is a *pointer* value. Perhaps
    "machine address" would be a better term for what you're calling
    a memory address.

    On the Cray T90 and related vector systems, a void* or char* value
    consists of a 64-bit machine-level pointer to a 64-bit word, with
    a 3-bit offset stored in the otherwise unused high-order 3 bits.
    This is done entirely in software; there's no hardware support for
    addressing anything smaller than a 64-bit word. Converting from
    a pointer type to an integer type simply copies the bits. Given:

    char arr[2];

    both (unsigned)&arr[0] % 8 and (unsigned)&arr[1] % 8 would almost
    certainly have the same value, which could be any value from 0 to 7.
    (I'd use uintptr_t, but the C compiler didn't support C99.)
    Keith Thompson, Aug 10, 2013
  13. And how is what you prefer relevant if it's not supported either
    by the C standard or by all real-world implementations?

    Feel free to go off and invent your own language that works the
    way you want it it.

    Pointers. Are. Not. Integers.
    Keith Thompson, Aug 10, 2013
  14. I suppose, but maybe the standard shouldn't use some terms which
    might not be right. Since a "pointer" might not be just an
    "address", maybe there should be different wording for &.

    -- glen
    glen herrmannsfeldt, Aug 10, 2013
  15. How can a "pointer" be anything other than an "address" (with
    both terms used the way the C standard uses them)? I suppose you
    could say that a null pointer isn't an address; is that what you're
    referring to?

    A C pointer / address might be something other than a machine-level
    address, but there are plenty of cases where the standard uses terms
    in more specific ways than they might be used in other contexts
    ("string", "byte", "object", etc.).
    Keith Thompson, Aug 11, 2013
  16. We could implement a weird and wonderful system where pointers were random
    words in a dictionary, then you looked them up in a hash table to resolve
    to disk writes for dynamic memory or variable names for automatic memory.

    So a pointer wouldn't be an address, in the sense I used the term (something
    which can map to electrical impulses on a bus attached to memory).
    Malcolm McLean, Aug 11, 2013
  17. Not so different from JVM object references. You aren't allowed
    to look at the bits. (There are no operations to do it.)
    If the reference is an array, the offset is supplied when it
    is dereferenced.

    -- glen
    glen herrmannsfeldt, Aug 11, 2013
  18. You need to understand that a Java object reference is a pointer or you
    can't really understand how Java works. But it's hard for programmers
    who learn Java as a first language, because they can't manipulate the

    Also in C we can easily do memory management. If we've got flash pointers
    and ram pointers, we can read and write to both, but writing to flash is
    expensive, so we won't want the compiler producing code to erase the flash
    on a simple pointer dereference. But we can provide a function flash_write,
    then we can test that the destination pointer is in the flash area of
    memory, and correctly aligned. We can do pretty much everything in C,
    except maybe a few assembly instructions to actually start the erase cycle.

    In Java you can't. You've got to patch the JVM.
    Malcolm McLean, Aug 11, 2013
  19. But it would be a address in the sense used by the C standard. If the
    unary "&" operator yields a random word in a dictionary, then that's
    what an address is.
    Keith Thompson, Aug 11, 2013
  20. James Harris \(es\)

    Joe Pfeiffer Guest

    The mapping would be odd, but it would be a mapping (I assume you don't
    really mean "variable names", since if you did there would need to be
    another mapping).
    Joe Pfeiffer, Aug 12, 2013
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