frange() question

[Stargaming]

On Thu, 20 Sep 2007 09:08:17 -0400, George Trojan wrote:

> A while ago I found somewhere the following implementation of frange():
> 
> def frange(limit1, limit2 = None, increment = 1.):
>      """
>      Range function that accepts floats (and integers). Usage:
>      frange(-2, 2, 0.1)
>      frange(10)
>      frange(10, increment = 0.5)
>      The returned value is an iterator.  Use list(frange) for a list.
>      """
>      if limit2 is None:
>          limit2, limit1 = limit1, 0.
>      else:
>          limit1 = float(limit1)
>      count = int(math.ceil(limit2 - limit1)/increment) return (limit1 +
>      n*increment for n in range(count))
> 
> I am puzzled by the parentheses in the last line. Somehow they make
> frange to be a generator:
>  >> print type(frange(1.0, increment=0.5))
> <type 'generator'>
> But I always thought that generators need a keyword "yield". What is
> going on here?
> 
> George

Consider the following:

    def foo():
        yield 1
    def bar():
        return foo()

Still, ``type(bar())`` would be a generator.

I don't want to tell you anything wrong because I don't know how 
generators are implemented on the C level but it's more like changing 
foo's (or frange's, in your example) return value.

HTH,
Stargaming