Assigning a number to a scalar doesn't free the string buffer, it just marks it as unused. You "solution" can actually increase the memory used.

$ perl -MDevel::Size=total_size -E'
   $a[5]="5678";
   say total_size($a[5]);

   $x=int($a[5]);
   say total_size($a[5]);
'
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undef $s (but not $s = undef;) will free the string buffer.

$ perl -MDevel::Size=total_size -E'
   $a[5]="5678";
   say total_size($a[5]);

   $t=int($a[5]);
   undef $a[5];
   $a[5]=$t;
   say total_size($a[5]);
'
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It doesn't downgrade the scalar from PVIV to IV, though. You'd need to start with a new scalar to do that.

$ perl -MDevel::Peek -MDevel::Size=total_size -E'
   $a[5]="5678";
   say total_size($a[5]);

   $t=int($a[5]);
   delete $a[5];
   $a[5]=$t;
   say total_size($a[5]);
'
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Yeah sorry, bad example.

I was thinking in terms of using it within in your $a=[@$a], e.g. $a= [ @$a[0..4], int($a->[5]) ], where it *would* get a new SV.

Dave.