Even simpler: 
my $c = 2;
my @a = ($c, $c+=1);
print "@a"; # prints 3 3

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Similarly: 
my @a = ($c, ++$c); # @a is now (3, 3)

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But: 
 @a = ($c, $c++) # @a is now (3, 2) !!!

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Do you get it? In the first two cases, the value of $c is modified before the array gets populated. In the last case, because of the use of the post-increment operator, $c is incremented only after the second value is stored into the array, but before the first value is stored into the array.

In reply to Re: Why does the first $c evaluate to the incremented value in [$c, $c += $_] ? by Laurent_R
in thread Why does the first $c evaluate to the incremented value in [$c, $c += $_] ? by smls

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