What happens when you use string interpolation in ruby?

Question

I thought that ruby just call method to_s but I can't explain how this works:

class Fake
  def to_s
    self
  end
end

"#{Fake.new}"

By the logic this should raise stack level too deep because of infinity recursion. But it works fine and seems to call #to_s from an Object.

=> "#<Fake:0x137029f8>"

But why?

ADDED:

class Fake
  def to_s
    Fake2.new
  end
end

class Fake2
  def to_s
    "Fake2#to_s"
  end
end

This code works differently in two cases:

puts "#{Fake.new}" => "#<Fake:0x137d5ac4>"

But:

puts Fake.new.to_s => "Fake2#to_s"

I think it's abnormal. Can somebody suggest when in ruby interpreter it happens internally?

FYI, if you def to_s; "#{self}"; end then you do get stack level too deep. def to_s; self; end is not recursive though. — Ajedi32
– Ajedi32, Commented Aug 25, 2014 at 15:17
You are mixing two different things in your "Added section" you are just modifying the to_s of Fake class by ouputing the sentence "Fake2#to_s" using another class who have a name close to the first one. I think you should try to understand more about how object oriented work. — tomsoft
– tomsoft, Commented Aug 25, 2014 at 17:41
My question is not about OO but about ruby internals - what happens in string interpolation. — abonec
– abonec, Commented Aug 26, 2014 at 10:07

Frederick Cheung · Accepted Answer · 2014-08-25 20:32:38Z

Short version

Ruby does call to_s, but it checks that to_s returns a string. If it doesn't, ruby calls the default implementation of to_s instead. Calling to_s recursively wouldn't be a good idea (no guarantee of termination) - you could crash the VM and ruby code shouldn't be able to crash the whole VM.

You get different output from Fake.new.to_s because irb calls inspect to display the result to you, and inspect calls to_s a second time

Long version

To answer "what happens when ruby does x", a good place to start is to look at what instructions get generated for the VM (this is all MRI specific). For your example:

puts RubyVM::InstructionSequence.compile('"#{Foo.new}"').disasm

outputs

0000 trace            1                                               (   1)
0002 getinlinecache   9, <is:0>
0005 getconstant      :Foo
0007 setinlinecache   <is:0>
0009 opt_send_simple  <callinfo!mid:new, argc:0, ARGS_SKIP>
0011 tostring         
0012 concatstrings    1
0014 leave

There's some messing around with the cache, and you'll always get trace, leave but in a nutshell this says.

get the constant Foo
call its new method
execute the tostring instruction
execute the concatstrings instruction with the result of the tostring instruction (the last value on the stack (if you do this with multiple #{} sequences you can see it building up all the individual strings and then calling concatstrings once on all consuming all of those strings)

The instructions in this dump are defined in insns.def: this maps these instructions to their implementation. You can see that tostring just calls rb_obj_as_string.

If you search for rb_obj_as_string through the ruby codebase (I find http://rxr.whitequark.org useful for this) you can see it's defined here as

VALUE
rb_obj_as_string(VALUE obj)
{
    VALUE str;

    if (RB_TYPE_P(obj, T_STRING)) {
    return obj;
    }
    str = rb_funcall(obj, id_to_s, 0);
    if (!RB_TYPE_P(str, T_STRING))
    return rb_any_to_s(obj);
    if (OBJ_TAINTED(obj)) OBJ_TAINT(str);
    return str;
}

In brief, if we already have a string then return that. If not, call the object's to_s method. Then, (and this is what is crucial for your question), it checks the type of the result. If it's not a string it returns rb_any_to_s instead, which is the function that implements the default to_s

Good answer. To see this behavior with a simple example: class D; def to_s; Object.new; end; end; puts "#{D.new}"; puts D.new.to_s. Note that printing "#{D.new}" prints info about a D and not about an Object

Collectives™ on Stack Overflow

What happens when you use string interpolation in ruby?

1 Answer 1

Short version

Long version

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

Short version

Long version

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related