PHP数组占用内存分析
下面的做法会占用多大的内存？
list($appid,$openid) = ["testcontent","test"];
测试
$m0 = memory_get_usage();$k = range(1,200000);$m1 = memory_get_usage();echo round(($m1-$m0)/pow(1024,2),4) ."MB\n";foreach ($k as $i){ $n1 = "kk$i"; $n2 = "tt$i"; list($$n1,$$n2) = [$i,$i*3];}$m2 = memory_get_usage();echo round(($m2-$m1)/pow(1024,2),4) ."MB\n";$m1 = memory_get_usage();foreach ($k as $i){ $n1 = "kk$i"; $n2 = "tt$i"; $$n1 = $i+time(); $$n2 = 2*time();}$m2 = memory_get_usage();echo round(($m2-$m1)/pow(1024,2),4) ."MB\n";上面运行输出的结果如下：
27.9404MB51.3041MB9.1553MB
可见数组占用的内存远大于正常分配的内容
原理
在PHP中都使用long类型来代表数字，没有使用int类型。大家都明白PHP是一种弱类型的语言，它不会去区分变量的类型，没有int float char *之类的概念。我们看看php在zend里面存储的变量，PHP中每个变量都有对应的 zval，Zval结构体定义在Zend/zend.h里面，其结构:
typedef struct _zval_struct zval; struct _zval_struct { /* Variable information */ zvalue_value value; /* The value 1 12字节(32位机是12，64位机需要8+4+4=16) */ zend_uint refcount__gc; /* The number of references to this value (for GC) 4字节 */ zend_uchar type; /* The active type 1字节*/ zend_uchar is_ref__gc; /* Whether this value is a reference (&) 1字节*/ }; PHP使用一种UNION结构来存储变量的值,即zvalue_value 是一个union，UNION变量所占用的内存是由最大成员数据空间决定。
typedef union _zvalue_value { long lval; /* long value */ double dval; /* double value */ struct { /* string value */ char *val; int len; } str; HashTable *ht; /* hash table value */ zend_object_value obj; /*object value */ } zvalue_value; 最大成员数据空间是struct str，指针占*val用4字节，INT占用4字节，共8字节。struct zval占用的空间为8+4+1+1 = 14字节，其实呢，在zval中数组，字符串和对象还需要另外的存储结构，数组则是一个 HashTable:
HashTable结构体定义在Zend/zend_hash.h.
typedef struct _hashtable { uint nTableSize;//4 uint nTableMask;//4 uint nNumOfElements;//4 ulong nNextFreeElement;//4 Bucket *pInternalPointer; /* Used for element traversal 4*/ Bucket *pListHead;//4 Bucket *pListTail;//4 Bucket **arBuckets;//4 dtor_func_t pDestructor;//4 zend_bool persistent;//1 unsigned char nApplyCount;//1 zend_bool bApplyProtection;//1 #if ZEND_DEBUG int inconsistent;//4 #endif } HashTable; HashTable 结构需要 39 个字节，每个数组元素存储在 Bucket 结构中:
typedef struct bucket { ulong h; /* Used for numeric indexing 4字节 */ uint nKeyLength; /* The length of the key (for string keys) 4字节 */ void *pData; /* 4字节*/ void *pDataPtr; /* 4字节*/ struct bucket *pListNext; /* PHP arrays are ordered. This gives the next element in that order4字节*/ struct bucket *pListLast; /* and this gives the previous element 4字节 */ struct bucket *pNext; /* The next element in this (doubly) linked list 4字节*/ struct bucket *pLast; /* The previous element in this (doubly) linked list 4字节*/ char arKey[1]; /* Must be last element 1字节*/ } Bucket; Bucket 结构需要 33 个字节，键长超过四个字节的部分附加在 Bucket 后面，而元素值很可能是一个 zval 结构，另外每个数组会分配一个由 arBuckets 指向的 Bucket 指针数组， 虽然不能说每增加一个元素就需要一个指针，但是实际情况可能更糟。这么算来一个数组元素就会占用 54 个字节，与上面的估算几乎一样。
一个空数组至少会占用 14(zval) + 39(HashTable) + 33(arBuckets) = 86 个字节，作为一个变量应该在符号表中有个位置，也是一个数组元素，因此一个空数组变量需要 118 个字节来描述和存储。从空间的角度来看，小型数组平均代价较大，当然一个脚本中不会充斥数量很大的小型数组，可以以较小的空间代价来获取编程上的快捷。但如果将数组当作容器来使用就是另一番景象了，实际应用经常会遇到多维数组，而且元素居多。比如10k个元素的一维数组大概消耗540k内存，而10k x 10 的二维数组理论上只需要 6M 左右的空间，但是按照 memory_get_usage 的结果则两倍于此，[10k,5,2]的三维数组居然消耗了23M，小型数组果然是划不来的。
参考
php数组占用内存大小分析
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