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Home Backend Development PHP Tutorial PHP数组占用内存分析

PHP数组占用内存分析

Jun 20, 2016 pm 12:29 PM

下面的做法会占用多大的内存?

list($appid,$openid) = ["testcontent","test"];
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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";
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上面运行输出的结果如下:

27.9404MB51.3041MB9.1553MB
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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*/  }; 
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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;  
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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; 
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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;  
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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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