Question

I keep hearing that PHP has overhead. For example a C++ int uses 4 Bytes on a 32 bit system but a PHP int uses more. What is this value?

Answer 1

I need more space than a comment to expand on mario’s findings so I’ll add an answer instead.

The size of a C union will be the size of its largest member (possibly with extra bytes to satisfy alignment constraints). For zvalue_value, that would be the obj which has the size of three pointers (not including the memory required for what those pointers point to):

typedef struct _zend_object {
    zend_class_entry *ce;
    HashTable *properties;
    HashTable *guards; /* protects from __get/__set ... recursion */
} zend_object;

On a 32bit system, a zend_object will take 24 bytes while on a 64bit system it will take 48 bytes. So, every zvalue_value will take at least 24 or 48 bytes regardless of what data you store in it. There’s also the name of the variable which consumes more memory; compiled languages generally discard the names once the compiler is done and treat values as simple sequences of bytes (so a double takes eight bytes, a char takes one byte, etc…).

With regards to your recent questions about PHP booleans, a simple boolean value will consume 24 or 48 bytes for the value, plus a few more bytes for the name, plus four or eight for the zend_unit, plus four (or eight) for the two zend_uchars in this:

struct _zval_struct {
    /* Variable information */
    zvalue_value value;     /* value */
    zend_uint refcount__gc;
    zend_uchar type;    /* active type */
    zend_uchar is_ref__gc;
};

The zend_uchar members will chew up four (or eight) bytes due to alignment constraints, almost every CPU wants to access memory on natural address boundaries and that means that a single byte sized member of a struct will take up four bytes or eight bytes of memory (depending on the CPUs natural word size and alignment constraints). So, a boolean will take somewhere between 36 and 72 bytes of memory.