Is it possible to use the ++ operator inside a string interpolation? I've attempted the following:
my $i = 0;
foreach my $line (@lines) {
print "${i++}. $line\n";
}
but I get Compile error: Can't modify constant item in postincrement (++)
Bareword i is equivalent to "i", so you are doing "i"++.
You want:
print($i++, ". $line\n");
Simpler:
print("$i. $line\n");
++$i;
A good way to embed values into a string is sprintf/printf.
printf("%d. %s\n", $i++, $line);
Note that use strict disallows barewords, so you'll also get
Bareword "i" not allowed while "strict subs" in use
That error oddly comes after the error you mentioned.
You can use ${\($var++)} to increment the variable while interpolating it.
use strict ;
use warnings ;
my $var = 5 ;
print "Before: var=$var\n" ;
print "Incremented var=${\($var++)}\n" ;
print "After: var=$var\n" ;
This will print
Before: var=5
Incremented var=6
After: var=6
But I would suggest as mentioned in the comments not to use this code because using printf is easier to write and read.
printf():printf("%d. %s\n", $i++, $line);print "@{[ $i++ ] }. $line\n"(it's similar to @dgw's one, which has been written while I was editing this comment)