Re: why can't I shift a split? (List vs Array)

[LanX]

The fundamental problem here is that lists and arrays are two different things. That's a FAQ

Lists are temporary data sequences which can't be changed!

But Arrays are variable(s) with @sigils and they are changed.

Shift can only operate on arrays, because they are shortened afterwards.

Once you've understood this...

> why can't I shift a split?

Because split returns a list!

Unless you assign it to an array it can't be shifted.

Addendum

Others have shown solutions with (lists)[slices] which work similar to @array[slices] . That's only possible because a slice doesn't change its object, unlike shift.

Further reading

• List's terminology and Parentheses in Perl
• Examples for "LIST", "list context", "list value", "list assigment", "list operator"

[misterperl]

Rolf yes, I know that- but it should be EZ to coerce a list into an array?

[LanX]

> should be EZ to coerce a list into an array

sure:

  DB<2> p shift @{[42..59]}
42

or

  DB<3> $shift = sub { shift @{$_[0]} }

  DB<4> p [42..59]->$shift
42

but it doesn't make sense since the resulting shortened array is destroyed right away.

Using a list slice is the logical° and easiest approach, without wasting operations.

  DB<5> p (42..59)[0]
42

compare an array slice which wastes resources again, by allocating an array in memory which is destroyed again.

  DB<6> p [42..59]->[0]
42

°) and "implementing" pop list is symmetrical (42..59)[-1]