Re: interpolating a variable into an operator

I am trying to store operators in a variable and then use the variable as the operator

Lisp can do this; Perl cannot. It can, however, store an anonymous *subroutine* in a variable. This is syntactically different from an operator, but it can accomplish the same effect:

$op = sub { $_[0] eq $_[1] };
if($op->($somvar, "some text")){
  #do stuff
}
[download]

Resist the urge to use string eval for this.

;$;=sub{$/};@;=map{my($a,$b)=($_,$;);$;=sub{$a.$b->()}} split//,".rekcah lreP rehtona tsuJ";$\=$;[-1]->();print

Comment on Re: interpolating a variable into an operator Download Code

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Re^2: interpolating a variable into an operator by diotalevi (Canon) on Jul 20, 2004 at 19:16 UTC
Bah. We just need more people to treat perl like lisp. This isn't impossible - it is just difficult.	[reply]
Re^3: interpolating a variable into an operator by ihb (Deacon) on Jul 20, 2004 at 20:31 UTC
Perhaps a filter that changes `EXPR1 ´$binop´ EXPR2` [download] to `$The::Module::binops{$binop}->(EXPR1, EXPR2)` [download] `:-)` Some ops won't be fully supported though, like `x` (see Re: Using x to build data structures considered harmful). `ihb`	[reply] [d/l] [select]
Re^4: interpolating a variable into an operator by diotalevi (Canon) on Jul 20, 2004 at 20:44 UTC
Perl is best treated like lisp by being lisp-like. Go after the optree directly and both cases look like `multiply( EXPR1, EXPR2 )` and `repeat( EXPR1, EXPR2 )`. I don't know how you thought to solve this where * is overridable but x isn't. Care to share?	[reply] [d/l] [select]