Re: oct and vec

[Zaxo]

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 $foo,$/'
qaq

Further 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.