On Mar 17, 2011, at 13:59 , Ryan Davis wrote:

>=20
> On Mar 17, 2011, at 10:24 , Jes=FAs Gabriel y Gal=E1n wrote:
>=20
>> On Thu, Mar 17, 2011 at 6:02 PM, Yu-Hsuan Lai <raincolee / gmail.com> =
wrote:
>>> I read the RDoc and it says:
>>> For each object, directly writes *obj*.inspect followed by the =
current
>>> output record separator to the program=91s standard output.
>>>=20
>>> But my $\ is surely nil, it still append a newline after output.
>>> ("output record separator" is $\, right?)
>>> Why does it do this?
>>=20
>> Looking into Ruby 1.8.7's source code I see this in io.c:
>>=20
>> void
>> rb_p(obj)                       /* for debug print within C code */
>>   VALUE obj;
>> {
>>   rb_io_write(rb_stdout, rb_obj_as_string(rb_inspect(obj)));
>>   rb_io_write(rb_stdout, rb_default_rs);
>> }
>>=20
>> This is the function called by the function defined as 'p'. As you =
can
>> see it's outputting the rb_default_rs. Searching for this in the
>> source, it's only assigned to here, in io.c:
>>=20
>> rb_rs =3D rb_default_rs =3D rb_str_new2("\n");
>>=20
>> So it seems it's assigned to "\n". I don't know if this rb_default_rs
>> is assigned to something else somewhere else (a grep -r rb_default_rs
>> * only shows the assignment I showed above), maybe someone with more
>> knowledge can chime in. If this is not the case, then I guess the
>> documentation should say "the default record separator". Anyone?
>=20
> Couple lines down:
>=20
>    rb_define_hooked_variable("$\\", &rb_output_rs, 0, rb_str_setter);
>=20
> That means that rb_output_rs is hooked up to $\ and changing it in =
ruby will bridge to C.

DOH... Wow. I'm tired. Apparently my espresso hasn't hit the bloodstream =
yet. I overlooked rb_output_rs vs rb_default_rs. Xavier is right, 'p' =
doesn't honor $\ at all. The doco is wrong. I'll change it.