我正在使用名为doug lea内存分配器的动态内存分配器，它使用最佳拟合方法在堆上分配内存。该算法是更多其他算法的基础，但我发现在分配的块的情况下，该块的头部包含前一个块的最后4个字节的数据。我检查了算法解释，但找不到原因。我想知道这个前一个块的4个字节的分配目的是什么？ 我还想出了一个解释，就是在其他块中分配.dtors部分以进行同步和正确使用空间但是想知道细节。
this is the figure of chunks of dlmalloc algorithm
上图包含已分配块和空闲块的结构。在空闲块中，前4个字节包含前一个块的大小，但在分配的块中，前四个字节包含前一个已分配块的用户数据的最后四个字节，这对我来说似乎有点混乱我想知道在当前块中仅分配先前分配的块的四个字节的目的是什么。
答案 0 :(得分：2)
是块重叠。曾几何时，记忆非常昂贵。 这是dlmalloc，ptmalloc和glibc malloc中的一个功能。
代码中有一个相当不错的解释：
This struct declaration is misleading (but accurate and necessary).
It declares a "view" into memory allowing access to necessary
fields at known offsets from a given base. See explanation below.
struct malloc_chunk {
INTERNAL_SIZE_T prev_size; /* Size of previous chunk (if free). */
INTERNAL_SIZE_T size; /* Size in bytes, including overhead. */
struct malloc_chunk* fd; /* double links -- used only if free. */
struct malloc_chunk* bk;
};
malloc_chunk详情：
(The following includes lightly edited explanations by Colin Plumb.)
Chunks of memory are maintained using a `boundary tag' method as
described in e.g., Knuth or Standish. (See the paper by Paul
Wilson ftp://ftp.cs.utexas.edu/pub/garbage/allocsrv.ps for a
survey of such techniques.) Sizes of free chunks are stored both
in the front of each chunk and at the end. This makes
consolidating fragmented chunks into bigger chunks very fast. The
size fields also hold bits representing whether chunks are free or
in use.
An allocated chunk looks like this:
chunk->+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| Size of previous chunk, if allocated | |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| Size of chunk, in bytes |P|
mem-> +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| User data starts here... .
. .
. (malloc_usable_space() bytes) .
. |
next ->+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| Size of chunk |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
Where "chunk" is the front of the chunk for the purpose of most of
the malloc code, but "mem" is the pointer that is returned to the
user. "Nextchunk" is the beginning of the next contiguous chunk.
Chunks always begin on even word boundries, so the mem portion
(which is returned to the user) is also on an even word boundary, and
thus at least double-word aligned.
Free chunks are stored in circular doubly-linked lists, and look like this:
chunk-> +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| Size of previous chunk |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
`head:' | Size of chunk, in bytes |P|
mem-> +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| Forward pointer to next chunk in list |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| Back pointer to previous chunk in list |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| Unused space (may be 0 bytes long) .
. .
. |
next-> +-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
`foot:' | Size of chunk, in bytes |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
The P (PREV_INUSE) bit, stored in the unused low-order bit of the
chunk size (which is always a multiple of two words), is an in-use
bit for the *previous* chunk. If that bit is *clear*, then the
word before the current chunk size contains the previous chunk
size, and can be used to find the front of the previous chunk.
The very first chunk allocated always has this bit set,
preventing access to non-existent (or non-owned) memory. If
prev_inuse is set for any given chunk, then you CANNOT determine
the size of the previous chunk, and might even get a memory
addressing fault when trying to do so.
Note that the `foot' of the current chunk is actually represented
as the prev_size of the NEXT chunk. This makes it easier to
deal with alignments etc but can be very confusing when trying
to extend or adapt this code.
The two exceptions to all this are
1. The special chunk `top' doesn't bother using the
trailing size field since there is no next contiguous chunk
that would have to index off it. After initialization, `top'
is forced to always exist. If it would become less than
MINSIZE bytes long, it is replenished.
2. Chunks allocated via mmap, which have the second-lowest-order
bit (IS_MMAPPED) set in their size fields. Because they are
allocated one-by-one, each must contain its own trailing size field.
答案 1 :(得分：0)
我没有具体研究过dlmalloc，但这里有一个可能的解释：
在具有需要16字节对齐的对象的架构上（与英特尔SSE一样），返回的地址必须是16的倍数。如果标头包含12个字节的信息，包含块的大小，以及一些链接信息到将块与前一个块合并，可以将头定义为具有16的长度，前四个字节用于前一个分配的块的末尾。如果此前一个块是空闲的，则分配器可以使用此空间进行优化。