NetNews Usenet Archive 1992 #20/NN_1992_20.iso/spool/comp/std/c/2632

Internet Message Format | 1992-09-15 | 6.3 KB

Path: sparky!uunet!mcsun!uknet!acorn!ixi!clive
From: [e-mail] (Clive Feather)
Newsgroups: comp.std.c
Subject: Re: struct hack, and other out-of-array references
Message-ID: <[e-mail]>
Date: 15 Sep 92 20:11:37 GMT
References: <[e-mail] <[e-mail]> <[e-mail]>
Organization: IXI Limited, Cambridge, UK
Lines: 140


[<> = Norman Diamond;  o' = me (Clive Feather) previously;
 ^^ = Stephen R. van den Berg; ~~ = Mark Brader;
 || = Interpretation Ruling;  ## = the standard;  no indent = me now.]

[Sorry for so much quoting, but I can't summarise the previous
discussion any more than I have.]

The question:
=============
In the code:

o'      struct fred { int i; char s [1]; } *f;
o'      char *s;
o'
o'      /* ... */
o'      f = malloc (sizeof f + strlen (ss));
o'      if (f != NULL)
o'          strcpy (f->s, ss);

If strlen (ss) is greater than 0, is the access to f->s [1] allowed in a
strictly conforming program, or does it invoke undefined behaviour
(access outside the bounds of an array) ?

The story so far:
=================
<>  You can't go past the end of an array object.  But if malloc() or some
<>  other variable has defined the end of the actual array object, then the
<>  + operator can get you that far, regardless of the declared type that
<>  some other array variable had before getting flattened to a pointer.

o'  But there is an interpretation that says that, given
o'     int a [5][5];
o'  the access "a [1][6]" is illegal, because it goes past the bounds of the
o'  array "a [1]". In other words, the declared type of the array does
o'  restrict what can happen to a pointer derived from it.

[RFI 17 item 16]

||  For an array of arrays, the permitted pointer arithmetic in Standard
||  ##3.3.6 Semantics (page 48, lines 12-40) is to be understood by
||  interpreting the use of the word "object" as denoting the specific
||  object determined directly by the pointer's type and value, *not* other
||  objects related to that one by contiguity. For example, the following
||  code has undefined behaviour:
||      int a [4][5];
||      a [1][7] = 0;  /* undefined */
||  Some conforming implementations may choose to diagnose an "array bounds
||  violation", while others may choose to interpret such attempted accesses
||  successfully with the "obvious" extended semantics.

[Standard section 6.3.6 (ANSI 3.3.6)]

##  If both the pointer operand and the result point to elements of the
##  same array object, or one past the last element of the same array
##  object, the evaluation shall not produce an overflow; otherwise, the
##  behavior is undefined.

In "a [1][7]", the pointer operand is "a [1]".

~~ Now, the important thing is that nowhere in the standard is there any
~~ license to treat the object a as an array of 20 ints.  It is an array
~~ of 4 arrays of 5 ints each.  If you write a[1][7], you are calling for
~~ the computation of a[1]+7.  a[1] here decays to a pointer to int, which
~~ points to the first element of the array a[1].  a[1]+7 does not point
~~ to that array.

o'  This RFT applies to slices of arrays, but, in *my* opinion, it is
o'  extendable to this case:
[The code at the top of this article]

o'  The pointer f->s points to an object with type "char [1]", and so, if
o'  strlen (ss) > 0, the access to (f->s)[1] required by strlen is undefined
o'  according to this RFI, even though the array has already decayed to a
o'  pointer.

~~ Clive's opinion here is wrong.  The difference
~~ between this and the other case is that the standard provides (in ANSI
~~ section 4.10.3, ISO 7.10.3) specific dispensation for the value returned by
~~ malloc() to:

##  ... be assigned to a pointer to any type of object and then used to
##  access such an object or an array of such objects in the space
##  allocated ...

~~ The pointer f->s does point to an object with type char[1], but it is
~~ also a pointer into the space returned by malloc(), which may be treated
~~ as an array of chars.  That is, when strcpy computes the equivalent of
~~ (f->s)[1], and therefore (f->s)+1, both the pointer operand of +, i.e.
~~ f->s, and the computed result *are* within the same array, namely the one
~~ returned by malloc(), and all is well.

Now read on:
============
I don't think Mark is correct here (obviously, or I wouldn't be
writing :-). The intent of the section about malloc quoted is clearly to
indicate that the space returned can be used for any purpose, and is not
restricted to (say) char arrays only. If the pointer returned from
malloc is cast to a pointer to a character type, then of course any part
of the space can be accessed. However, once the pointer has been cast to
a "struct fred *", then it can only be used to access the contents of a
"struct fred".

I have to say that the Standard seems very confused when talking about
pointers to objects within objects. For example, the above ruling
implies that a single pointer cannot walk through the array "a", because
there is an "iron curtain" between a[0] and a[1], and so on. On the
other hand, it is easy to show that a[0][4] and a[1][0] must be adjacent
in memory, which implies that the pointer can walk through, in code
like:

    for (i = 0, p = &(a [0][0]); i < sizeof a / sizeof a [0][0]; i++)
        printf ("%d\n", *p++);

I think that the only way we are going to decide this question is by
putting together a set of RFIs which cover all the relevant issues
and submitting them.

~~ I would also argue that the Interpretation Ruling cited above would not
~~ apply if the type of the example array "a" had been char instead of int.
~~ The definitions of "object" and "byte" (in ANSI section 1.6, ISO 3.14 and
~~ 3.4) in effect require that it be possible to treat any object as an
~~ array of any character type.

But even then "a" is an array of type "char [5]", not of type "char", so
that doesn't apply. However, to avoid doubt, change my code to:

    struct fred { double d; int arr [1]; } *f;
    int i, n;

    /* ... */
    f = malloc (sizeof f + (n - 1) * sizeof (int));
    if (f != NULL)
        for (i = 0; i < n; i++)
            f->arr [i] = 0;
--
Clive D.W. Feather     | IXI Limited         | If you lie to the compiler,
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