Re: Struct assignment

Discussion in 'C Programming' started by Eric Sosman, Jan 19, 2013.

  1. Eric Sosman

    Eric Sosman Guest

    On 1/19/2013 7:09 AM, Russell Shaw wrote:
    > Hi,
    > In gcc-4.7 C99, i get an error (in a function scope):
    >
    > struct {
    > int a;
    > } sa;
    >
    > struct {
    > int a;
    > } sb;
    >
    > sb = sa;
    >
    > error: incompatible types when assigning to type 'struct <anonymous>'
    > from type 'struct <anonymous>'
    >
    >
    > ISO/IEC 9899:201x Committee Draft -- August 11, 2008 WG14/N1336
    >
    > 6.5.16.1 Simple assignment
    > Para 1
    > -- the left operand has a qualified or unqualified version of a
    > structure or union type compatible with the type of the right;
    >
    >
    > 6.2.7 Compatible type and composite type
    > Para 1
    > Two types have compatible type if their types are the same. Additional
    > rules for determining whether two types are compatible are described in
    > 6.7.2 for type specifiers, in 6.7.3 for type qualifiers, and in 6.7.5
    > for declarators.48) Moreover, two structure, union, or enumerated types
    > declared in separate translation units are compatible if their tags and
    > members satisfy the following requirements: If one is declared with a
    > tag, the other shall be declared with the same tag. If both are complete
    > types, then the following additional requirements apply: there shall be
    > a one-to-one correspondence between their members such that each pair of
    > corresponding members are declared with compatible types, and such that
    > if one member of a corresponding pair is declared with a name, the other
    > member is declared with the same name. For two structures, corresponding
    > members shall be declared in the same order. For two structures or
    > unions, corresponding bit-fields shall have the same widths. For two
    > enumerations, corresponding members shall have the same values.
    >
    >
    > The above mentions "separate translation units", but i don't see why the
    > error should happen.


    Um, er, because your `sa' and `sb' are not in separate
    translation units? And hence aren't compatible?

    The fact that two types look the same (as yours do) doesn't
    mean they actually are the same; it's just a coincidence of
    their representation. On many systems `int' and `long' have
    the same representation and even the same behavior, yet they
    are distinct types nonetheless. (Incompatible types, too,
    though implicit conversions often disguise the fact.)

    Have you ever run into the (in)famous "struct declaration
    in parameter list" problem (FAQ Question 2.5)? That's very much
    like what's happening here: Having written a type declaration
    without a name (tag) or alias (typedef), you have no way to
    designate that same type again.

    --
    Eric Sosman
    d
    Eric Sosman, Jan 19, 2013
    #1
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  2. Eric Sosman

    BartC Guest

    "Russell Shaw" <rjshawN_o@s_pam.netspace.net.au> wrote in message
    news:...

    >>> sb = sa;
    >>>
    >>> error: incompatible types when assigning to type 'struct <anonymous>'
    >>> from type 'struct <anonymous>'


    > It's easy to write a compiler to do that assignment deterministically, so
    > why's it not allowed?


    In general there may be incompatibilities between two structs, so simple
    assignment is not possible or may be dangerous.

    You're suggesting the compiler should make a special case of certain pairs
    of structs, and allow assignment, while forbidding it for others. That's an
    untidy way of doing things. It's simpler to insist the structs are exactly
    the same. (And makes maintenance simpler: make a minor change to one of the
    two structs, or use a different pack() value, and suddenly those hundreds of
    sa=sb assignments are illegal. Make the same change to the one struct, and
    all those sa=sb will still work.)

    In any case an easy workaround is to use memcpy to copy them, whether they
    are compatible (or even the same size) or not.

    --
    Bartc
    BartC, Jan 19, 2013
    #2
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  3. Eric Sosman

    Eric Sosman Guest

    On 1/19/2013 9:18 AM, Russell Shaw wrote:
    > On 20/01/13 00:27, Eric Sosman wrote:
    >> [...]
    >> Have you ever run into the (in)famous "struct declaration
    >> in parameter list" problem (FAQ Question 2.5)? That's very much
    >> like what's happening here: Having written a type declaration
    >> without a name (tag) or alias (typedef), you have no way to
    >> designate that same type again.

    >
    > Why should it matter that the second struct is "declared" in a separate
    > translation unit? To be able to do the assignment, you have to #include
    > that second declaration in to the first file anyway.


    The point of separate compilation is that the compilations
    are, well, separate. The compiler processes one file on Tuesday
    and the other on Wednesday, perhaps even on different machines.
    The two declarations are not visible to the compiler at the same
    time, so the compiler has no way to connect them. The compiler
    cannot even tell that the declarations were #include'd from the
    same file -- or, quite likely, from distinct files with identical
    content (Tuesday's compilation on Machine A, Wednesday's on
    Machine B with files copied via FTP, say).

    That's why there's a special dispensation for textually
    similar type declarations in different translation units: Without
    such a rule, separate compilation simply wouldn't work.

    Within a single translation unit, though, things are different.
    Two "anonymous" type declarations are treated as distinct types,
    even if they bear a strong resemblance. Yes, this can lead to a
    paradox, where two types in Module X are mutually incompatible,
    but are both compatible with a single type in Module Y. Don't
    get too hung up over it; just laugh it off and maybe use it as
    a trick question to win bets down at the nerd bar.

    > It's easy to write a compiler to do that assignment deterministically,
    > so why's it not allowed?


    I'm not entirely sure what you mean by this -- but if "it's
    easy," go ahead and do it. Get the gcc source, modify the compiler
    to treat duplicated anonymous declarations the way you want, try it
    out on assorted code bases, and see where it leads. If the outcome
    is good, your modification can be "prior art" to support a proposal
    to the next revision of the Standard.

    > p.6.2.7 describes the compatability for structs in separate translation
    > units, but says nothing (ie, does not forbid) about assignments in the
    > same translation unit.


    6.2.7 doesn't allow or forbid any kind of assignment; the
    string "assign" does not even appear there.

    Assignment is covered in 6.5.16, and in 6.5.16.1p1 (second
    bullet) you'll find the requirement that struct assignment must
    be between compatible types. *Then* we go back to 6.2.7 to read
    the rules on compatibility, and learn that identical anonymous
    struct types are compatible only if declared in separate
    translation units. Your pair are not in separate translation
    units so that language doesn't apply; the structs are therefore
    incompatible and do not satisfy the "shall" of 6.5.16.1.

    --
    Eric Sosman
    d
    Eric Sosman, Jan 19, 2013
    #3
  4. Eric Sosman

    Shao Miller Guest

    On 1/19/2013 09:18, Russell Shaw wrote:
    > On 20/01/13 00:27, Eric Sosman wrote:
    >> On 1/19/2013 7:09 AM, Russell Shaw wrote:
    >>> Hi,
    >>> In gcc-4.7 C99, i get an error (in a function scope):
    >>>
    >>> struct {
    >>> int a;
    >>> } sa;
    >>>
    >>> struct {
    >>> int a;
    >>> } sb;
    >>>
    >>> sb = sa;
    >>>
    >>> error: incompatible types when assigning to type 'struct <anonymous>'
    >>> from type 'struct <anonymous>'
    >>>
    >>>
    >>> ISO/IEC 9899:201x Committee Draft -- August 11, 2008 WG14/N1336
    >>>
    >>> 6.5.16.1 Simple assignment
    >>> Para 1
    >>> -- the left operand has a qualified or unqualified version of a
    >>> structure or union type compatible with the type of the right;
    >>>
    >>>
    >>> 6.2.7 Compatible type and composite type
    >>> Para 1
    >>> Two types have compatible type if their types are the same. Additional
    >>> rules for determining whether two types are compatible are described in
    >>> 6.7.2 for type specifiers, in 6.7.3 for type qualifiers, and in 6.7.5
    >>> for declarators.48) Moreover, two structure, union, or enumerated types
    >>> declared in separate translation units are compatible if their tags and
    >>> members satisfy the following requirements: If one is declared with a
    >>> tag, the other shall be declared with the same tag. If both are complete
    >>> types, then the following additional requirements apply: there shall be
    >>> a one-to-one correspondence between their members such that each pair of
    >>> corresponding members are declared with compatible types, and such that
    >>> if one member of a corresponding pair is declared with a name, the other
    >>> member is declared with the same name. For two structures, corresponding
    >>> members shall be declared in the same order. For two structures or
    >>> unions, corresponding bit-fields shall have the same widths. For two
    >>> enumerations, corresponding members shall have the same values.
    >>>
    >>>
    >>> The above mentions "separate translation units", but i don't see why the
    >>> error should happen.

    >>
    >> Um, er, because your `sa' and `sb' are not in separate
    >> translation units? And hence aren't compatible?
    >>
    >> The fact that two types look the same (as yours do) doesn't
    >> mean they actually are the same; it's just a coincidence of
    >> their representation. On many systems `int' and `long' have
    >> the same representation and even the same behavior, yet they
    >> are distinct types nonetheless. (Incompatible types, too,
    >> though implicit conversions often disguise the fact.)
    >>
    >> Have you ever run into the (in)famous "struct declaration
    >> in parameter list" problem (FAQ Question 2.5)? That's very much
    >> like what's happening here: Having written a type declaration
    >> without a name (tag) or alias (typedef), you have no way to
    >> designate that same type again.

    >
    > Why should it matter that the second struct is "declared" in a separate
    > translation unit?


    The usual tradition for two translation units to use the same
    programmer-defined type is for them to both #include a header that
    defines that type. But during translation, it's not really the header
    that defines the type, it's the translation unit. So we need to ask,
    "Is the type used in TU A the same as in TU B?" The discussion about
    compatible types answers this question.

    > To be able to do the assignment, you have to #include
    > that second declaration in to the first file anyway.
    >


    You don't have to #include anything. You can copy and paste the type's
    definition manually, or the object's or function's declaration/definition.

    > It's easy to write a compiler to do that assignment deterministically,
    > so why's it not allowed?
    >
    > p.6.2.7 describes the compatability for structs in separate translation
    > units, but says nothing (ie, does not forbid) about assignments in the
    > same translation unit.


    How can you argue that their types are the same in this circumstance?
    Since the type has no associated identifier, how can you be sure?
    Please consider another simple case (which presents the same issue):

    int main(void) {
    if ((struct { int x; } *) 0 == (struct { int x; } *) 0)
    return 0;
    return 0;
    }

    How do you know that these types are the same? How do you know how many
    padding bytes these two referenced (pointed-to) types have after their
    'x' members? How do you know the alignment requirements of the
    referenced types?

    Suppose you have:

    typedef int tni;
    struct { int size; } apple;
    struct { tni size; } orange;

    Should these two objects be considered to have the same type, if we saw
    these in the same translation unit? It could be that the _intention_ is
    for each to belong to a different domain, and the discussion of
    compatible types helps to prevent against accidental crossing of domains.

    --
    - Shao Miller
    --
    "Thank you for the kind words; those are the kind of words I like to hear.

    Cheerily," -- Richard Harter
    Shao Miller, Jan 19, 2013
    #4
  5. Eric Sosman

    Shao Miller Guest

    On 1/19/2013 21:21, Russell Shaw wrote:
    > On 20/01/13 03:45, Shao Miller wrote:
    >> On 1/19/2013 09:18, Russell Shaw wrote:
    >>> On 20/01/13 00:27, Eric Sosman wrote:
    >>>> On 1/19/2013 7:09 AM, Russell Shaw wrote:
    >>>>> Hi,
    >>>>> In gcc-4.7 C99, i get an error (in a function scope):
    >>>>>
    >>>>> struct {
    >>>>> int a;
    >>>>> } sa;
    >>>>>
    >>>>> struct {
    >>>>> int a;
    >>>>> } sb;
    >>>>>
    >>>>> sb = sa;
    >>>>>
    >>>>> error: incompatible types when assigning to type 'struct <anonymous>'
    >>>>> from type 'struct <anonymous>'
    >>>>>
    >>>>>
    >>>>> ISO/IEC 9899:201x Committee Draft -- August 11, 2008 WG14/N1336
    >>>>>
    >>>>> 6.5.16.1 Simple assignment
    >>>>> Para 1
    >>>>> -- the left operand has a qualified or unqualified version of a
    >>>>> structure or union type compatible with the type of the right;
    >>>>>
    >>>>>
    >>>>> 6.2.7 Compatible type and composite type
    >>>>> Para 1
    >>>>> Two types have compatible type if their types are the same. Additional
    >>>>> rules for determining whether two types are compatible are
    >>>>> described in
    >>>>> 6.7.2 for type specifiers, in 6.7.3 for type qualifiers, and in 6.7.5
    >>>>> for declarators.48) Moreover, two structure, union, or enumerated
    >>>>> types
    >>>>> declared in separate translation units are compatible if their tags
    >>>>> and
    >>>>> members satisfy the following requirements: If one is declared with a
    >>>>> tag, the other shall be declared with the same tag. If both are
    >>>>> complete
    >>>>> types, then the following additional requirements apply: there
    >>>>> shall be
    >>>>> a one-to-one correspondence between their members such that each
    >>>>> pair of
    >>>>> corresponding members are declared with compatible types, and such
    >>>>> that
    >>>>> if one member of a corresponding pair is declared with a name, the
    >>>>> other
    >>>>> member is declared with the same name. For two structures,
    >>>>> corresponding
    >>>>> members shall be declared in the same order. For two structures or
    >>>>> unions, corresponding bit-fields shall have the same widths. For two
    >>>>> enumerations, corresponding members shall have the same values.
    >>>>>
    >>>>>
    >>>>> The above mentions "separate translation units", but i don't see
    >>>>> why the
    >>>>> error should happen.
    >>>>
    >>>> Um, er, because your `sa' and `sb' are not in separate
    >>>> translation units? And hence aren't compatible?
    >>>>
    >>>> The fact that two types look the same (as yours do) doesn't
    >>>> mean they actually are the same; it's just a coincidence of
    >>>> their representation. On many systems `int' and `long' have
    >>>> the same representation and even the same behavior, yet they
    >>>> are distinct types nonetheless. (Incompatible types, too,
    >>>> though implicit conversions often disguise the fact.)
    >>>>
    >>>> Have you ever run into the (in)famous "struct declaration
    >>>> in parameter list" problem (FAQ Question 2.5)? That's very much
    >>>> like what's happening here: Having written a type declaration
    >>>> without a name (tag) or alias (typedef), you have no way to
    >>>> designate that same type again.
    >>>
    >>> Why should it matter that the second struct is "declared" in a separate
    >>> translation unit?

    >>
    >> The usual tradition for two translation units to use the same
    >> programmer-defined
    >> type is for them to both #include a header that defines that type. But
    >> during
    >> translation, it's not really the header that defines the type, it's the
    >> translation unit. So we need to ask, "Is the type used in TU A the
    >> same as in TU
    >> B?" The discussion about compatible types answers this question.

    >
    > When both sa and sb are declared in the one common header, sb = sa still
    > gives the same error.
    >


    Agreed.

    > I just don't see why the point about being compatible between different
    > translation units is made, when the same header declaration is going to
    > be used in both anyway.
    >


    This doesn't quite make sense to me. If you have one .h file that
    declares two objects, can you please explain:

    - In which .c file will 'sa' be defined?
    - In which .c file will 'sb' be defined?
    - In which .c file will the assignment occur?

    I'm not sure which scenario you're imagining.

    >>> To be able to do the assignment, you have to #include
    >>> that second declaration in to the first file anyway.
    >>>

    >>
    >> You don't have to #include anything. You can copy and paste the type's
    >> definition manually, or the object's or function's
    >> declaration/definition.
    >>


    Did you spot this part, above?

    >>> It's easy to write a compiler to do that assignment deterministically,
    >>> so why's it not allowed?
    >>>
    >>> p.6.2.7 describes the compatability for structs in separate translation
    >>> units, but says nothing (ie, does not forbid) about assignments in the
    >>> same translation unit.

    >>
    >> How can you argue that their types are the same in this circumstance?
    >> Since the
    >> type has no associated identifier, how can you be sure? Please
    >> consider another
    >> simple case (which presents the same issue):
    >>
    >> int main(void) {
    >> if ((struct { int x; } *) 0 == (struct { int x; } *) 0)
    >> return 0;
    >> return 0;
    >> }
    >>
    >> How do you know that these types are the same? How do you know how
    >> many padding
    >> bytes these two referenced (pointed-to) types have after their 'x'
    >> members? How
    >> do you know the alignment requirements of the referenced types?

    >
    > Both types are structurally the same because they're both pointers to
    > structs that are built the same way.
    >


    There are plenty of types whose representation and alignment
    requirements match. Does that mean that they should be interchangeable?
    That was the point of my "apples" and "oranges" example, just below.
    Type is something more than representation and alignment. It means you
    can say "these things are different than these other things, and should
    not be mixed up." How would you ensure that things don't get mixed up
    inappropriately if representation and alignment was good enough for
    compatibility?

    > I'm assuming those two structs are built the same way with the same
    > padding, because if they weren't, the compiler ABI is then randomly
    > inconsistant.
    >


    Yes, I'd agree that you're probably right close to 100% of the time,
    especially due to the "common initial sequence" discussion for unions
    and due to the compatibility requirement across translation units. I
    think a problem is your use of "ABI", which doesn't have a meaning in
    the Standard.

    >> Suppose you have:
    >>
    >> typedef int tni;
    >> struct { int size; } apple;
    >> struct { tni size; } orange;
    >>
    >> Should these two objects be considered to have the same type, if we
    >> saw these in
    >> the same translation unit? It could be that the _intention_ is for
    >> each to
    >> belong to a different domain, and the discussion of compatible types
    >> helps to
    >> prevent against accidental crossing of domains.

    >
    > That's what i'd like to see,


    That's not what _I'd_ like to see! I want type safety that allows me to
    specify different domains of values, without any care for internal
    representational co-incidences. I don't want to accidentally compare
    apples and oranges.

    > but i was trying to figure out how
    > assignments between them can be done if they come from different
    > translation units.
    >


    Translation unit #1:

    struct { int i; double d; } foo;

    Translation unit #2:

    extern struct { int i; double d; } foo;

    Now the type of 'foo' in TU #1 can be said to be compatible with the
    type in TU #2.

    > I cannot devise how sb = sa can be found in any source, where sb and sa
    > have separate struct declarations.
    >


    Why you are searching for such a case? Type compatibility is used in a
    lot of places other than assignment. As just one example, suppose you have:

    Translation unit #1:

    static struct { int i; double d; } foo;
    void * foo_ptr = &foo;

    void callme(void) {
    if (foo.i == 42 && foo.d == 3.14159)
    /* Success */
    return;
    }

    Translation unit #2:

    extern void callme(void);
    extern void * foo_ptr;

    static struct { int i; double d; } bar;

    static void somefunc(void) {
    bar.i = 42;
    bar.d = 3.14159;
    memcpy(foo_ptr, &bar, sizeof bar);
    callme();
    }

    If 'foo' and 'bar' didn't have compatible types, then when 'somefunc'
    calls 'callme' and 'callme' tries to read the value of 'foo.i' and
    'foo.d', it'd be undefined behaviour. Because they have compatible
    types, things are ok.

    >
    > The only use case i can think of is:
    >
    >
    > Translation unit1:
    >
    > // tu1.h header:
    > struct {
    > int a;
    > } sb;
    >
    > struct sb *str_new(void);
    >
    >
    > // tu1.c
    > struct {
    > int a;
    > } sa;
    >
    > struct sa *str_new(void)
    > {
    > struct sa *str = malloc(sizeof(struct sa));
    > return str;
    > }


    I don't quite follow the point of this code.

    (Minor note: The pattern:

    obj_type * ptr = malloc(sizeof *ptr);

    says "Allocate enough storage for the pointee" whereas your version says
    "Allocate enough storage for obj_type", which happens to match the right
    size for the pointee.)

    Here is a thought experiment:

    Imagine that 9 out of 10 people want:

    struct { int i; } root;

    to represent a one-of-a-kind object, that should never be compared with
    any other object, even if there is an object like:

    struct { int i; } starting_point;

    Now imagine that you need to be able to access 'root' from any other
    translation unit. You can make a header for those _other_ translation
    units:

    extern struct { int i; } root;

    Because the header is actually part of each of the other translation
    units, type compatibility says that the all of the translation units
    think that 'root' is the same type.

    Here is a different thought experiment:

    Imagine that 1 out of 10 people want:

    struct { int i; } count;

    and:

    struct { int i; } number;

    to have the same type because they have the same composition, and want
    to be able to assign the values of these objects to one another because
    both are objects that might make sense in the same domain.

    Now imagine that 1 out of 10 people see a strange rule that seems to
    interfere with this goal, but seems to allow it when the objects are in
    different translation units. In fact, if the two objects are declared
    in two translation units (as a header #inclusion would do), the
    allowance doesn't even have any benefit for the goal!

    But the two sets of people in these thought experiments have different
    goals. These thought experiments aren't offered as an actual
    representation of the world, but as a possible way to understand how
    people with different goals can view the rules very differently.

    (The 9 people might say to the 1 person, "so identify the type so it can
    be re-used, then we both win, to an extent.")

    --
    - Shao Miller
    --
    "Thank you for the kind words; those are the kind of words I like to hear.

    Cheerily," -- Richard Harter
    Shao Miller, Jan 20, 2013
    #5
  6. On Sun, 20 Jan 2013 12:46:51 +1100, Russell Shaw wrote:

    > The problem is that if you #include the declaration of an anonymous
    > struct sa from a separate translation unit, then the compiler complains
    > about sb = sa.


    I get the feeling that you misunderstand what a translation unit is.
    As defined in the C standard (C99 clause 5.1.1.1/1), a translation unit
    consists of a source file *after preprocessing*. This means that all
    #include directives have been expanded into the text of the file they
    reference in a translation unit.
    Therefor, it is impossible (by definition) to #include a declaration from
    another translation unit.

    >
    > If you *don't* #include the declaration of an anonymous struct sa from a
    > separate translation unit, then the compiler complains that sa isn't
    > declared.
    >
    > In what use-case can sb = sa be done without complaint?


    You would need three translation units.

    //TU1.c
    struct {
    int x;
    } sa = { 0 };

    //TU2.c
    struct {
    int x;
    } sb = { 1 };

    //TU3.c
    extern struct {
    int x;
    } sa, sb;

    void foo(void)
    {
    sa = sb;
    }

    Now, although each of the three anonymous structs is a distinct type,
    they must be treated as compatible types because the compiler never sees
    two of them together in the same translation unit.

    Bart v Ingen Schenau
    Bart van Ingen Schenau, Jan 20, 2013
    #6
  7. Eric Sosman

    Eric Sosman Guest

    On 1/19/2013 8:46 PM, Russell Shaw wrote:
    > [... two "identical" struct types are incompatible ...]
    >
    > The problem is that if you #include the declaration of an anonymous
    > struct sa from a separate translation unit, then the compiler complains
    > about sb = sa.


    I don't understand this. A "translation unit" is all of
    the source the compiler sees at one time: The original source
    file, all the files and headers it #includes, all the files and
    headers they in turn #include, all macros defined on compiler
    command lines or whatever, ... If you #include a declaration,
    that declaration is part of *the* translation unit; it is not
    an interloper from "a separate translation unit."

    > If you *don't* #include the declaration of an anonymous struct sa from a
    > separate translation unit, then the compiler complains that sa isn't
    > declared.


    Same disconnect; I still don't understand you.

    > In what use-case can sb = sa be done without complaint?


    I showed a few elsethread; here's one of them again:

    struct { int a; } sa, sb;
    ...
    sa.a = 42;
    sb = sa;

    --
    Eric Sosman
    d
    Eric Sosman, Jan 20, 2013
    #7
  8. Eric Sosman

    Eric Sosman Guest

    On 1/20/2013 8:02 PM, Russell Shaw wrote:
    > On 21/01/13 01:08, Eric Sosman wrote:
    >> On 1/19/2013 8:46 PM, Russell Shaw wrote:
    >>> [... two "identical" struct types are incompatible ...]
    >>>
    >>> The problem is that if you #include the declaration of an anonymous
    >>> struct sa from a separate translation unit, then the compiler complains
    >>> about sb = sa.

    >>
    >> I don't understand this. A "translation unit" is all of
    >> the source the compiler sees at one time: The original source
    >> file, all the files and headers it #includes, all the files and
    >> headers they in turn #include, all macros defined on compiler
    >> command lines or whatever, ... If you #include a declaration,
    >> that declaration is part of *the* translation unit; it is not
    >> an interloper from "a separate translation unit."

    >
    > Precisely. Because of that problem, i could not see how declarations
    > from separate translation units should be compatible, when including a
    > header with the second declaration now becomes a part of the first
    > translation unit anyway, thus making the point mute.


    (s/mute/moot/)

    That's what the special dispensation in 6.2.7 is for: It
    says that "lexically identical" [my phrase] structs in separate
    translation units are compatible. You #include the header from
    a source file that is part of translation unit A, and again from
    a source in TU B, and you then know that the types in A and B
    are compatible. (As mentioned earlier, you don't even need to
    #include the very same file in both TU's: you can #include a copy,
    or you can copy-and-paste the text, or you can even re-type the
    declaration from scratch. As long as the two declarations in A
    and B satisfy 6.2.7, the types are compatible.)

    In a sense, 6.2.7's special dispensation can be viewed as a
    hack, a way to get around the limitations of tool sets. Note the
    precision of the language: The types in A and B are "compatible,"
    not "the same." They cannot be "the same" because there is no
    UUID for types, no anchor that allows a compiler in California
    and a compiler in Cameroon to cooperate -- especially if the two
    compilers process their source files in different decades. Yet
    another way of looking at the issue is to realize that types have
    no linkage (6.2.2): If they did, the special dispensation would
    be unnecessary, but a whole lot of additional tool machinery
    would be needed. 6.2.7 is a formalization of "We want separate
    compilation, so we need a way for different compilers on different
    days in different parts of the world to agree on compatible types,
    without inventing a central universal type registry."

    But within a single TU the difficulties of cross-compiler
    coordination vanish: It's one compiler on one day looking at one
    batch of source, and it needs no special dispensation to figure
    out that `struct foo' means the same thing every time it pops up
    in a particular scope. That's why 6.2.7's language doesn't cover
    "lexically identical" declarations in a single TU: The need for
    the hack doesn't exist, and there's some (smallish) benefit to
    type safety in avoiding hacks where hacking isn't required.

    --
    Eric Sosman
    d
    Eric Sosman, Jan 21, 2013
    #8
  9. Eric Sosman

    James Kuyper Guest

    On 01/20/2013 10:08 PM, Eric Sosman wrote:
    ....
    > That's what the special dispensation in 6.2.7 is for: It
    > says that "lexically identical" [my phrase] structs in separate
    > translation units are compatible.


    Just to nit-pick: I would expect things that are "lexically identical"
    to have to match token for token, but 6.2.7p1 is (very slightly) more
    lenient than that. It doesn't require that member types be the same,
    merely "compatible". It doesn't require that alignment specifiers to be
    the same, but merely "equivalent". Bit widths do have to be the "same",
    but I believe that the expressions used for those widths do not; a width
    of 4 is the same as a width of 2*2, even though the first is only one
    token, and the second is three.
    --
    James Kuyper
    James Kuyper, Jan 21, 2013
    #9
  10. Eric Sosman

    James Kuyper Guest

    On 01/20/2013 10:08 PM, Eric Sosman wrote:
    ....
    > That's what the special dispensation in 6.2.7 is for: It
    > says that "lexically identical" [my phrase] structs in separate
    > translation units are compatible.


    Just to nit-pick: I would expect things that are "lexically identical"
    to have to match token for token, but 6.2.7p1 is (very slightly) more
    lenient than that. It doesn't require that member types be the same,
    merely "compatible". It doesn't require that alignment specifiers to be
    the same, but merely "equivalent". Bit widths do have to be the "same",
    but I believe that the expressions used for those widths do not; a width
    of 4 is the same as a width of 2*2, even though the first is only one
    token, and the second is three.
    --
    James Kuyper
    James Kuyper, Jan 21, 2013
    #10
  11. Eric Sosman

    James Kuyper Guest

    On 01/20/2013 08:02 PM, Russell Shaw wrote:
    > On 21/01/13 01:08, Eric Sosman wrote:
    >> On 1/19/2013 8:46 PM, Russell Shaw wrote:
    >>> [... two "identical" struct types are incompatible ...]
    >>>
    >>> The problem is that if you #include the declaration of an anonymous
    >>> struct sa from a separate translation unit, then the compiler complains
    >>> about sb = sa.

    >>
    >> I don't understand this. A "translation unit" is all of
    >> the source the compiler sees at one time: The original source
    >> file, all the files and headers it #includes, all the files and
    >> headers they in turn #include, all macros defined on compiler
    >> command lines or whatever, ... If you #include a declaration,
    >> that declaration is part of *the* translation unit; it is not
    >> an interloper from "a separate translation unit."

    >
    > Precisely. Because of that problem, i could not see how declarations from
    > separate translation units should be compatible, when including a header with
    > the second declaration now becomes a part of the first translation unit anyway,
    > thus making the point mute.


    I'm not sure I understand your objection. #include is a red herring
    here; all that matters is what the translation units contain after all
    #include directives have been processed. Therefore, I'll give an example
    involving no shared header files (which would, in general, be a bad
    coding practice - but then, IMO, so is the use of unnamed struct types).

    file declared.c:
    extern struct {unsigned u:4; _Alignas(long long) double d;} a, b;

    file defined.c:
    struct {unsigned u:4; _Alignas(_Alignof(long long)) double d;} a, b;

    In each file, 'a' is compatible with 'b', despite the fact that they
    both have an anonymous struct type, because the same anonymous struct
    definition was used to declare both objects.
    The one definition rule requires that the external declarations in
    declared.c be compatible with the definitions in defined.c, and they
    are, but for a different reason: the definitions of the
    struct types in the two different translation units meet all of the
    requirements of 6.2.7p1:

    "If one is declared with a tag, the other shall be declared with the
    same tag." - neither one is declared with a tag.

    "If both are completed anywhere within their respective translation
    units, then the following additional requirements apply: there shall
    be a one-to-one correspondence between their members such that each pair
    of corresponding members are declared with compatible types; ..." -
    There is such a correspondence - the types of the corresponding members
    are not merely compatible, but identical.

    "... if one member of the pair is declared with an alignment specifier,
    the other is declared with an equivalent alignment specifier; ..."
    - 'd' is declared with equivalent alignment specifiers in both struct
    definitions.

    "... and if one member of the pair is declared with a name, the other is
    declared with the same name. ..." - for every pair, the names are in
    fact the same.

    "For two structures, corresponding members shall be declared in the same
    order." - they are declared in the same order.

    "For two structures or unions, corresponding bit-fields shall have the
    same widths." - 'u' has the same width in both structure definitions.

    The key point is that these requirements only render compatible struct
    definitions that appear in different translation units. Therefore, even
    though the following line define struct types that also meet all of
    those requirements, because they occur in the same translation unit,
    they define incompatible struct types:

    struct {unsigned u:4; _Alignof(long long) double d;} a;
    struct {unsigned u:4; _Alignof(long long) double d;} b;

    This may seem counter-intuitive, but the reason for this difference is
    that it was not considered necessary to extend this specification to
    include struct types declared in the same translation unit. The standard
    already provides three different ways to define structure objects in the
    same translation unit to have compatible types: struct tags, typedefs,
    or sharing the same anonymous struct definition, as in my example above.
    None of those things would be sufficient to allow declarations of struct
    types in different translation units to be compatible - it's only
    6.2.7p1 that makes that possible.

    Note: I don't agree with this decision; I don't see any advantage to
    restricting the guarantee of compatibility to structs defined in
    different translation units. However, that is what the standard specifies.

    > I'll go with the interpretation as stated in this group, because it makes for
    > less complication in the compiler, and is safer to use.


    I'm not sure what you consider to be "the interpretation as stated in
    this group". Is it compatible with what I've said above?
    --
    James Kuyper

    --
    James Kuyper
    James Kuyper, Jan 21, 2013
    #11
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