Defect Report #282

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Submitter: Douglas Walls (US)
Submission Date: 2002-06-11
Source: J11
Version: 1.4
Date: 2004-03-06
Subject: flexible array members & struct padding
Summary Structure and union specifiers, paragraphs 15 and 16 require that any padding for
alignment of a structure containing a flexible array member must preceed the flexible
array member.  This contradicts existing implementations.  We do not believe this was the intent of the C99 specification.


If a struct contains a flexible array member and also requires padding for alignment, then the current C99 specification requires the implementation to put this padding before the flexible array member.  However, existing implementations, including at least GNU C, Compaq C, and Sun C, put the padding after the flexible array member.

The layout used by existing implementations can be more efficient. Furthermore, requiring these existing implementations to change their layout would break binary backwards compatibility with previous versions.

Suggested Technical Corrigendum

Change the wording such that it is implementation defined as to whether the padding is before or after the flexible array member.  

Proposed Technical Corrigendum

In paragraph 16, replace the second and third sentences ("With two ... 106)" with the following text:.

In most situations, the flexible array member is ignored. In particular, the size of the structure is as if the flexible array member were omitted except that it may have more trailing padding than the omission would imply.
replace "Second" with "However" at the start of the following sentence, and delete footnote 106.
Replace the examples (paragraphs 17 to 20) with:

[#17] EXAMPLE After the declaration:

            struct s { int n; double d[]; };
the structure struct s has a flexible array member d. A typical way to use this is:
           int m = /* some value */;
            struct s *p = malloc (sizeof (struct s) + sizeof (double [m]));
and assuming that the call to malloc succeeds, the object pointed to by p behaves, for most purposes, as if p had been declared as:
            struct { int n; double d[m]; } *s1;
(there are circumstances in which this equivalence is broken; in particular, the offsets of member d might not be the same).

[#18] Following the above declaration:

            struct s t1 = { 0 };           // valid
            struct s t2 = { 1, { 4.2 }};   // invalid
            t1.n = 4;                      // valid
            t1.d[0] = 4.2;                 // might be undefined behavior
The initialization of t2 is invalid (and violates a constraint) because struct s is treated as if it does not contain member d. The assigment to t1.d[0] is probably undefined behaviour, but it is possible that
            sizeof (struct s) >= offsetof (struct s, d) + sizeof (double)
in which case the assignment would be legitimate. Nevertheless it cannot appear in strictly conforming code.

[#19] After the further declaration:

            struct ss { int n; };
the expressions:
            sizeof (struct s) >= sizeof (struct ss)
            sizeof (struct s) >= offsetof (struct s, d)
are always equal to 1.
[#20] If sizeof (double) is 8, then after the following code is executed:
            struct s *s1;
            struct s *s2;
            s1 = malloc(sizeof (struct s) + 64);
            s2 = malloc(sizeof (struct s) + 46);
and assuming that the calls to malloc succeed, the objects pointed to by s1 and s2 behave, for most purposes, as if the identifiers had been declared as:
            struct { int n; double d[8]; } *s1;
            struct { int n; double d[5]; } *s2;
[#21] Following the further successful assignments:
            s1 = malloc(sizeof (struct s) + 10);
            s2 = malloc(sizeof (struct s) +  6);
they then behave as if the declarations were:
            struct { int n; double d[1]; } *s1, *s2;
            double *dp;
            dp = &(s1->d[0]);       // valid
            *dp = 42;               // valid
            dp = &(s2->d[0]);       // valid
            *dp = 42;               // undefined behavior
[#22] The assignment:
            *s1 = *s2;
only copies the member n; if any of the array elements are within the first sizeof (struct s) bytes of the structure, these might be copied or simply overwritten with indeterminate values.

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