Method `&()

Method `&

mixed `&(mixed arg1)
mixed `&(mixed arg1, mixed arg2, mixed ... extras)
mixed `&(object arg1, mixed arg2)
mixed `&(mixed arg1, object arg2)
int `&(int arg1, int arg2)
string `&(string arg1, string arg2)
array `&(array arg1, array arg2)
mapping `&(mapping arg1, mapping arg2)
mapping `&(mapping arg1, array arg2)
mapping `&(mapping arg1, multiset arg2)
multiset `&(multiset arg1, multiset arg2)
type `&(type|program arg1, type|program arg2)

Description

Bitwise and/intersection.

Every expression with the & operator becomes a call to this function, i.e. a&b is the same as predef::`&(a,b).

Returns

If there's a single argument, that argument is returned.

If there are more than two arguments the result is: `&(`&(arg1, arg2), @extras).

Otherwise, if arg1 is an object with an lfun::`&(), that function is called with arg2 as argument, and its result is returned.

Otherwise, if arg2 is an object with an lfun::``&(), that function is called with arg1 as argument, and its result is returned.

Otherwise the result depends on the argument types:

arg1 can have any of the following types:

int	Bitwise and of arg1 and arg2.
string	The result is a string where each character is the bitwise and of the characters in the same position in arg1 and arg2. The arguments must be strings of the same length.
array|mapping|multiset	The result is like arg1 but only with the elements/indices that match any in arg2 (according to `>, `<, `== and, in the case of mappings, hash_value).
type|program	Type intersection of arg1 and arg2.

The function is not destructive on the arguments - the result is always a new instance.

Note

If this operator is used with arrays or multisets containing objects which implement lfun::`==() but not lfun::`>() and lfun::`<(), the result will be undefined.

See also

`|(), lfun::`&(), lfun::``&()