From: Ryan Davis <ryand-ruby@...>
Date: 2011-03-18T05:59:07+09:00
Subject: Re: What is p's real behavior?


On Mar 17, 2011, at 10:24 , Jesús Gabriel y Galán wrote:

> 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‘s standard output.
>> 
>> But my $\ is surely nil, it still append a newline after output.
>> ("output record separator" is $\, right?)
>> Why does it do this?
> 
> Looking into Ruby 1.8.7's source code I see this in io.c:
> 
> 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);
> }
> 
> 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:
> 
> rb_rs = rb_default_rs = rb_str_new2("\n");
> 
> 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?

Couple lines down:

    rb_define_hooked_variable("$\\", &rb_output_rs, 0, rb_str_setter);

That means that rb_output_rs is hooked up to $\ and changing it in ruby will bridge to C.