in reply to oct and vec
The first argument of vec says what variable it acts on. The third says how many bits make a unit in the vector. The second is the offset in units of the third argument where things will happen.
ord applied to a string returns ord of its first character.
Thus, vec($foo,0,8) = ord($foo); is a no-op, replacing the first character of $foo by itself. In more depth:
$ perl -e'my $foo = "qa"; vec($foo,0,8) = ord $foo;print $foo,$/' qa $ perl -e'my $foo = "qa"; vec($foo,1,8) = ord $foo;print $foo,$/' qq $ perl -e'my $foo = "qa"; vec($foo,2,8) = ord $foo;print $foo,$/' qaqFurther increments of the second argument would place NUL in the intermediate positions of the string.
Update: I just noticed that the op is ==, numerical equality. I think that should always be true, but there might be tricky edge cases related to encoding.
After Compline,
Zaxo
In Section
Seekers of Perl Wisdom