Just reread your comment about "In real code". If you have an array instead of the 1..10, the ternary operator is the way to go. print for @a ? @a : 1..11 will have the first @a in scalar context (hence checking if there are any elements in the array) and the second @a in list context, returning the elements of the array. The 1..11 will properly have list context and generate the list (1,2,...,11).

Array slices are different. They don't have a scalar context per se; if scalar context is imposed (i.e. by being the first operand to ?:) the last element of the slice will be returned (and checked for truth by ?:). So assuming your desired-but-not-working-as-is-code was print for @a[@indices] or 1..11 you need to decide what you mean to do by checking @a[@indices] for truth. That could be any of a number of things. Perhaps one of these would be what you meant: 

print for grep($a[$_], @indices) ? @a[@indices] : 1..11;
print for grep(defined $a[$_], @indices) ? @a[@indices] : 1..11;
print for grep(exists $a[$_], @indices) ? @a[@indices] : 1..11;
print for !grep(!$a[$_], @indices) ? @a[@indices] : 1..11;
print for !grep(!defined $a[$_], @indices) ? @a[@indices] : 1..11;
print for !grep(!exists $a[$_], @indices) ? @a[@indices] : 1..11;
# update: actually use slices after ?

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I.e. use the array if (any/all) of the elements are (true/defined/existing), otherwise fall back on 1..11.

In reply to Re: for loops and 'and' by ysth
in thread for loops and 'and' by Anonymous Monk

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