Clever Geek Handbook

Calling & (x, y) bitwise operator

After exploring Why does the + function work on tuples? I have the following question.

Can anyone explain to me why it Base.:&(1,2)

works but &(1,2)

fails? At the same time, Base.:|(1,2)

and |(1,2)

.

+3


source to share


1 answer


The reason is that &

as a unary operator it is a special form, as it is used in syntax ccall

(although this syntax is now deprecated ). Hence, it is &(1, 2)

parsed as Expr(:&, :(1, 2))

.



  • |

    is not a unary operator, so it is |(1, 2)

    parsed as 1 | 2

    a function call.
  • +

    and -

    have special parsing rules so that +(1, 2)

    u -(1, 2)

    can be parsed as function calls with two arguments (otherwise they would be a function call with one argument in tuples, so at runtime). &

    does not obey these rules as it is a special form and not a normal operator.
  • Base.:&

    is not parsed as an operator at all, but rather just a normal field-to-identifier reference. So there is no ambiguity here and it is parsed like a normal function call. Likewise, it is (&)(1, 2)

    parsed like a normal function call because it is (&)

    parsed like a normal identifier.
+7


source






More articles:


All Articles