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

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

Cheers Rolf
_{(addicted to the Perl Programming Language :)

Wikisyntax for the Monastery}

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Re^2: why can't I shift a split? (List vs Array) by misterperl (Friar) on Aug 26, 2022 at 14:07 UTC
Rolf yes, I know that- but it should be EZ to coerce a list into an array?	[reply]
Re^3: why can't I shift a split? (List vs Array) by LanX (Saint) on Aug 26, 2022 at 23:15 UTC
> should be EZ to coerce a list into an array sure: `DB<2> p shift @{[42..59]} 42` [download] or `DB<3> $shift = sub { shift @{$_[0]} } DB<4> p [42..59]->$shift 42` [download] 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` [download] 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` [download] Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery} °) and "implementing" `pop` list is symmetrical `(42..59)[-1]`	[reply] [d/l] [select]