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 aspredef::`&(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.
- See also