Fixnum <=> confusion

ruby-talk
1

Hi,

I am recently learning Ruby, and came across an interesting problem that I do not understand. I was trying to count how many times the sort method was called on an object by overridding <=> to count the number of calls.

I decided to extend Fixnum as follows, but it didn't work. Can anyone tell me why? Thanks.

Bob Evans

Code:

class Fixnum

    @@count = 0

    alias originalComparator <=>

    def <=>(o)
      @@count += 1
      originalComparator(o)
    end

    def Fixnum.count
      @@count
    end
end

Running against it:

>array = [5, 4, 3, 2, 1]
=> [5,4,3,2,1]

>array.sort
=> [1,2,3,4,5]

>Integer.count
=> 0

Expected => 6 > result > 0

Now, if I call 1<=>2 then count increments. Does this mean that Fixnum sorting does not call those operators? Thanks for any help.

2

Hi,

···

In message "Re: Fixnum <=> confusion" on Wed, 6 Oct 2004 16:43:05 +0900, Robert Evans <robert.evans@acm.org> writes:

Now, if I call 1<=>2 then count increments. Does this mean that Fixnum
sorting does not call those operators? Thanks for any help.

sort skips <=> calls for fixnums to gain performance.

              matz.

3

Use The Source, Luke :slight_smile:

$ cd /v/build/ruby/ruby-1.8.1-2004.05.02/
$ less array.c

static int
sort_2(ap, bp)
...
    if (FIXNUM_P(a) && FIXNUM_P(b)) {
        if (a > b) return 1;
        if (a < b) return -1;
        return 0;
    }
    if (TYPE(a) == T_STRING && TYPE(b) == T_STRING) {
        return rb_str_cmp(a, b);
    }
...
static VALUE
sort_internal(ary)
    VALUE ary;
{
    qsort(RARRAY(ary)->ptr, RARRAY(ary)->len, sizeof(VALUE),
          rb_block_given_p()?sort_1:sort_2);
    return ary;
}

In other words: Array#sort, when called without a block, special-cases the
comparison of Fixnum to Fixnum and String to String, doing a hard-coded test
instead of calling the spaceship operator.

Try doing:
  array.sort { |a,b| a<=>b }
to force it to do the method call.

Regards,

Brian.

···

On Wed, Oct 06, 2004 at 04:43:05PM +0900, Robert Evans wrote:

Now, if I call 1<=>2 then count increments. Does this mean that Fixnum
sorting does not call those operators? Thanks for any help.

4

Aha. Thanks for the answer and the friendly reminder about the source.

···

On Oct 6, 2004, at 2:38 AM, Brian Candler wrote:

On Wed, Oct 06, 2004 at 04:43:05PM +0900, Robert Evans wrote:

Now, if I call 1<=>2 then count increments. Does this mean that Fixnum
sorting does not call those operators? Thanks for any help.

Use The Source, Luke :slight_smile:

$ cd /v/build/ruby/ruby-1.8.1-2004.05.02/
$ less array.c

static int
sort_2(ap, bp)
...
    if (FIXNUM_P(a) && FIXNUM_P(b)) {
        if (a > b) return 1;
        if (a < b) return -1;
        return 0;
    }
    if (TYPE(a) == T_STRING && TYPE(b) == T_STRING) {
        return rb_str_cmp(a, b);
    }
...
static VALUE
sort_internal(ary)
    VALUE ary;
{
    qsort(RARRAY(ary)->ptr, RARRAY(ary)->len, sizeof(VALUE),
          rb_block_given_p()?sort_1:sort_2);
    return ary;
}

In other words: Array#sort, when called without a block, special-cases the
comparison of Fixnum to Fixnum and String to String, doing a hard-coded test
instead of calling the spaceship operator.

Try doing:
  array.sort { |a,b| a<=>b }
to force it to do the method call.

Regards,

Brian.