The . operator is the string concatenation operator, so @array . "\n" stringifies @array and adds "\n" to the end, which returns scalar(@array) . "\n", i.e. the count of elements in @array and a "\n".

Not 100% correct ... :)

"scalar(@array)"? Yes!

"stringifies @array"? No!

It's more like stringifies scalar(@a) ...

  DB<26> @a=1..3
 
  DB<27> p scalar @a
3
  DB<28> p "@a"
1 2 3
  DB<29> p @a."\n"
3
 
  DB<30> p "@a"."\n" ; # $" is separator for stringification
1 2 3
 
  DB<31> print "@a\n"
1 2 3
 
  DB<32> p @a,"\n"   ; # $, is separator for print args
123
 
  DB<33> p "@a","\n" 
1 2 3
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UPDATE: From perlop#Operator Precedence and Associativity:

"With very few exceptions, these¹ all operate on scalar values only, not array values."

(1) = perl operators

"stringifies @array"? No!

It stringifies the result of @array.

To be fair, "stringifies @array" could mean "stringifies the result of @array", but most people think of "@array" as a variable rather than code. Clarity is definitely lacking.

The concat operator evaluates its operands in scalar context, to which @a responds by returning the number 3. The concat operator then stringifies that value.

Stringification is defined as

The process of producing a "string" representation of an abstract object.

and "3" is for sure no representation of @array.

The important point to learn is that most operators enforce scalar context of the operands, IMHO the stringification of +3 to "3" is in this case of small importance ...

Just want to clarify the terminology, before people start to think that stringification enforces scalar context!

Cheers Rolf