How does chomp change the contents of the string without the need to pass it in by reference?

Aside from the fact that chomp is an operator (so special rules could apply), arguments are always passed by reference in Perl.

sub foo {
   $_[0] = 'def';
}

my $x = 'abc';
foo($x);
print("$x\n");   # def
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You seem to be confusing passing by reference (the term you used and the reason it works) and passing references to variables (something unrelated).

>perl -wMstrict -le
"my $i = 42;
 inc($i);
 print $i;
 sub inc { ++$_[0] }
"
43
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If you check the perl source, you'll see that it is implemented as Perl_do_chomp in doop.c and it takes a pointer to an SV (which is fed to it from the top of the stack down since chomp can take multiple arguments) and that pointer is the C equivalent of a reference.

St. Larry be praised, there is much magic to make our lives easier in Perl.

HTH,
SSF

It will be great if you can post the code snippet of the c code..

Here's an analogy:

my $bar = 41;
foo($bar);
print $bar;

use Inline C => <<'EOC';
void foo(SV* bar) {
   int b = (int)SvIV(bar);
   b++;
   sv_setiv(bar,b);
}
EOC
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The chomp function works in combination with the $/ variable when reading from filehandles.

The $/ variable is the input record separator that is attached to the records you read from a filehandle and it is by default set to the newline character "\n".

The chomp function works by removing the last character from a string only if it matches the value of $/.

See Chomp

The newline character is \n, note the \.