std::align
[ptr.align]Section: 20.2.5 [ptr.align] Status: New Submitter: Melissa Mears Opened: 2014-08-06 Last modified: 2014-11-03
Priority: 3
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Discussion:
The specification of std::align
does not appear to specify what happens when the value of the size
parameter is 0. (The question of what happens when alignment
is 0 is mentioned in another Defect Report, 2377;
it would change the behavior to be undefined rather than potentially implementation-defined.)
size
being 0 is interesting because the result is ambiguous. Consider the following code's output:
#include <cstdio> #include <memory> int main() { alignas(8) char buffer[8]; void *ptr = &buffer[1]; std::size_t space = sizeof(buffer) - sizeof(char[1]); void *result = std::align(8, 0, ptr, space); std::printf("%d %td\n", !!result, result ? (static_cast<char*>(result) - buffer) : std::ptrdiff_t(-1)); }
There are four straightforward answers as to what the behavior of std::align
with size 0 should be:
The behavior is undefined because the size is invalid.
The behavior is implementation-defined. This seems to be the status quo, with current implementations using #3.
Act the same as size == 1
, except that if size == 1
would fail but would be defined and succeed
if space were exactly 1 larger, the result is a pointer to the byte past the end of the ptr
buffer. That is, the
"aligned" version of a 0-byte object can be one past the end of an allocation. Such pointers are, of course, valid when not
dereferenced (and a "0-byte object" shouldn't be), but whether that is desired is not specified in the Standard's definition
of std::align
, it appears. The output of the code sample is "1 8
" in this case.
Act the same as size == 1
; this means that returning "one past the end" is not a possible result. In this case,
the code sample's output is "0 -1
".
The two compilers I could get working with std::align
, Visual Studio 2013 and Clang 3.4, implement #3. (Change %td
to
%Id
on Visual Studio 2013 and earlier. 2014 and later will have %td
.)
Proposed resolution: