Question

I do not get to understand how the Perl read($buf) function is able to modify the content of the $buf variable. $buf is not a reference, so the parameter is given by copy (from my c/c++ knowledge). So how come the $buf variable is modified in the caller ?

Is it a tie variable or something ? The C documentation about setbuf is also quite elusive and unclear to me

# Example 1
$buf=''; # It is a scalar, not a ref
$bytes = $fh->read($buf);
print $buf; # $buf was modified, what is the magic ?

# Example 2
sub read_it {
    my $buf = shift;
    return $fh->read($buf);
}
my $buf;
$bytes = read_it($buf);
print $buf; # As expected, this scope $buf was not modified

Answer 1

No magic is needed — all perl subroutines are call-by-alias, if you will. Quoth perlsub:

The array @_ is a local array, but its elements are aliases
for the actual scalar parameters. In particular, if an element $_[0]
is updated, the corresponding argument is updated (or an error occurs
if it is not updatable).

For example:

sub increment {
  $_[0] += 1;
}

my $i = 0;
increment($i);  # now $i == 1

In your “Example 2”, your read_it sub copies the first element of @_ to the lexical $buf, which copy is then modified “in place” by the call to read(). Pass in $_[0] instead of copying, and see what happens:

sub read_this {
  $fh->read($_[0]);  # will modify caller's variable
}
sub read_that {
  $fh->read(shift);  # so will this...
}