Python内存管理模块的一个奇技淫巧
jinghangop
8年前
<p>最近在读Python源码中有关内存管理的部分。Python在分配小块内存(小于256字节)时,采用了内存池,以降低对内核内存分配程序的调用频次。在内存池的设计上,采用了一个分层的设计,由上到下依次是arena、pool、block。这次我看到的这个比较费解的结构,就来自于分配内存时,对于pool的处理。</p> <h2>谜团</h2> <p>在最主要的分配内存的函数_PyObject_Alloc中,我看到了这么一段代码:</p> <pre> <code class="language-python">pool = usable_arenas->freepools; // some other code ... init_pool: /* Frontlink to used pools. */ next = usedpools[size + size]; /* == prev */ pool->nextpool = next; pool->prevpool = next; next->nextpool = pool; next->prevpool = pool; </code></pre> <p>从注释和函数命名中也能看出来,这段代码的意思,大概是从arena的freepools中拿出一个pool,插入到usedpools的开头。但是,这里有一个奇怪的地方:</p> <p>如果说usedpools中存储的不同size的pool链表,那么为什么index要用size+size来表示呢?</p> <h2>初探</h2> <p>首先,我在代码中搜索了一下usedpools,找到了这么一段注释:</p> <p>This is involved. For an index i, usedpools[i+i] is the header for a list of</p> <p>all partially used pools holding small blocks with “size class idx” i. So</p> <p>usedpools[0] corresponds to blocks of size 8, usedpools[2] to blocks of size</p> <p>16, and so on: index 2*i <-> blocks of size (i+1)<<ALIGNMENT_SHIFT.</p> <p>哦,就是说,我要找到保存8字节block的pool,就去usedpools[0+0];找16字节的pool,就去usedpools[1+1]。再看一下上面代码中的size的由来:</p> <pre> <code class="language-python">size = (uint)(nbytes - 1) >> ALIGNMENT_SHIFT; </code></pre> <p>果然可以和注释里写的对应起来(注意,注释里的2*i是usedpools的index,而这里的nbytes是实际需要分配的block的字节大小,所以一个需要左移位,一个需要右移位,互相转换)。</p> <p>但是目前为止,好像处在一个“知其然,而不知其所以然”的状态:这里没有解释为什么是这样一个奇葩的size+size的index。</p> <h2>深入</h2> <p>往下,又有一段注释(Python源码的注释真详细):</p> <p>Major obscurity: While the usedpools vector is declared to have poolp</p> <p>entries, it doesn’t really. It really contains two pointers per (conceptual)</p> <p>poolp entry, the nextpool and prevpool members of a pool_header. The</p> <p>excruciating initialization code below fools C so that</p> <p>usedpool[i+i]</p> <p>“acts like” a genuine poolp, but only so long as you only reference its</p> <p>nextpool and prevpool members. The “- 2 <em>sizeof(block</em> )” gibberish is</p> <p>compensating for that a pool_header’s nextpool and prevpool members</p> <p>immediately follow a pool_header’s first two members:</p> <p>union { block *_padding;</p> <p>uint count; } ref;</p> <p>block *freeblock;</p> <p>each of which consume sizeof(block <em> </em></p> <p>) bytes. So what usedpools[i+i] really</p> <p>if* C believes it’s a poolp</p> <p>pointer, then p->nextpool and p->prevpool are both p (meaning that the headed circular list is empty).</p> <p>这一大段注释说了什么呢?大意就是,你以为usedpools里放的是poolp类型?其实里面是两两一对的poolp,一个是nextpool指针,一个是prevpool指针。这里还使用了一些trick,让C语言认为usedpools里放的是pool_header变量,只要使用者能保证只使用它的nextpool、prevpool成员。</p> <p>直接看usedpools的定义吧:</p> <pre> <code class="language-python">#define PTA(x) ((poolp )((uint8_t *)&(usedpools[2*(x)]) - 2*sizeof(block *))) #define PT(x) PTA(x), PTA(x) static poolp usedpools[2 * ((NB_SMALL_SIZE_CLASSES + 7) / 8) * 8] = { PT(0), PT(1), PT(2), PT(3), PT(4), PT(5), PT(6), PT(7) #if NB_SMALL_SIZE_CLASSES > 8 , PT(8), PT(9), PT(10), PT(11), PT(12), PT(13), PT(14), PT(15) …… }; </code></pre> <p>展开宏定义:</p> <pre> <code class="language-python">static poolp usedpools[2 * ((NB_SMALL_SIZE_CLASSES + 7) / 8) * 8] = { PTA(0), PTA(0),PTA(1),PTA(1), PTA(2),PTA(2), PTA(3),PTA(3),... …… }; </code></pre> <p>而PTA(x)定义为,usedpools[2x]的地址减去2个block*长度的地址。结合pool_header的定义:</p> <pre> <code class="language-python">struct pool_header { union { block *_padding; uint count; } ref; /* number of allocated blocks */ block *freeblock; /* pool's free list head */ struct pool_header *nextpool; /* next pool of this size class */ struct pool_header *prevpool; /* previous pool "" */ uint arenaindex; /* index into arenas of base adr */ uint szidx; /* block size class index */ uint nextoffset; /* bytes to virgin block */ uint maxnextoffset; /* largest valid nextoffset */ }; </code></pre> <p>nextpool和pool_header的地址之间,正好差了2个block*的偏移量。所以,在初始化usedpools之后,当把usedpools[size+size]当作pool_header时,那么usedpools[size+size]->nextpool将正好取到usedpools[size+size]的地址,也就是指向usedpools[size+size];而usedpools[size+size]->prevpool将也正好取到usedpools[size+size]的地址。因此:</p> <pre> <code class="language-python">next = usedpools[size + size]; /* == prev */ </code></pre> <p>这段代码+注释就说得通了。</p> <p>然后,经过:</p> <pre> <code class="language-python">pool->nextpool = next; pool->prevpool = next; next->nextpool = pool; next->prevpool = pool; </code></pre> <p>这么一折腾,usedpools中存储的一对地址(nextpool, prevpool)就指向了真正的pool_header变量pool。</p> <h2>知其所以然</h2> <p>但是,为什么要这么绕的设计这样的usedpools结构呢,直接存储一个完整的pool_header不简单明了吗?下面这段注释给出了答案:</p> <p>It’s unclear why the usedpools setup is so convoluted. It could be to</p> <p>minimize the amount of cache required to hold this heavily-referenced table</p> <p>(which only <em>needs</em> the two interpool pointer members of a pool_header).</p> <p>就是说,这样的一个usedpools表,会经常的被使用,使用这种结构,就可以减少放入缓存中的数据量(毕竟,我们只需要放入两个指针,而不是整个pool_header),从而也就提高了缓存的效率。为了提高效率,还真是无所不用其极啊。</p> <p>到这里,obmalloc.c中我认为最费解的一个数据结构,就弄清楚啦。</p> <p> </p> <p> </p> <p>来自:http://blog.guoyb.com/2017/03/15/python-obmalloc-trick/</p> <p> </p>