Re^2: How can we interpolate an expression??

What I understand from this code is: -

First we are converting the expression into an array or list.. Now, had we printed out that list or array, it would have printed out its address.. So, we have added an '@' before it to get an array.. Now, because we are not using a reference variable to this list, we have used a curly braces before using '@'

Have I interpreted it correct.. Is this what happens in Perl?? Or can I get a better explanation of what is happening there?

Comment on Re^2: How can we interpolate an expression??

Replies are listed 'Best First'.
Re^3: How can we interpolate an expression?? by Kenosis (Priest) on Sep 20, 2012 at 18:42 UTC
Great understanding! `say "Product is @{[$number * $multiplier]}"; ^ ^ ^ ^ \| \| \| \| \| \| \| + - EXPRESSION \| \| + - reference \| + - dereference + - interpolate` [download]	[reply] [d/l]
Re^4: How can we interpolate an expression?? by tobyink (Canon) on Sep 20, 2012 at 20:37 UTC
It's such a common combination of operations that it's even got a name: the baby cart. `perl -E'sub Monkey::do{say$_,for@_,do{($monkey=[caller(0)]->[3])=~s{::}{ }and$monkey}}"Monkey say"->Monkey::do'`	[reply]
Re^5: How can we interpolate an expression?? by Kenosis (Priest) on Sep 21, 2012 at 02:46 UTC
Excellent, tobyink! I enjoy what it says there: Discovered by Larry Wall, 1994. It's the Discovered part that's intriguing...	[reply]
Re^6: How can we interpolate an expression?? by tobyink (Canon) on Sep 21, 2012 at 05:49 UTC