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 witharg2
as argument, and its result is returned.Otherwise, if
arg2
is an object with an lfun::``&(), that function is called witharg1
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
andarg2
.string
The result is a string where each character is the bitwise and of the characters in the same position in
arg1
andarg2
. 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 inarg2
(according to `== and, in the case of mappings, hash_value).type
|program
Type intersection of
arg1
andarg2
.The function is not destructive on the arguments - the result is always a new instance.
- See also