To reproduce:
>>> '%i' % 12345678901.0
TypeError: int argument required

Contrast with:
>>> '%i' % 1234567890.0
'1234567890'

Previous experience led me to expect that the '%i' format code would
work for all numerical types.  Admittedly, there's nothing in the docs
promising this, but it works *almost* all the time.

In fact, the operator fails to convert floats which are too big to fit
into a machine long.  The code for the '%i' operator handles objects of
type long specially, and then uses PyInt_AsLong to handle all other
objects (in formatint).

The ideal solution would be to ask the object to convert itself into an
int *before* the special test for type long; this would give the
expected behavior.

An acceptable solution would be to make the behavior consistent by
refusing all floats passed as the argument to '%i', but I expect this
would break a lot of code.  Of course, right now most of that code is
broken anyway, since it will throw a TypeError when the input float
happens to be large.

On the one hand, if you want to format a float like that, you should use
"%.0f".  On the other hand, this should either consistently fail or
succeed, not depending on how large the float is.

I think in 2.6 we can't change this, but in 3.0, I propose that all
these accept only integers.  I find the behavior of %x and %o with
floats particularly anti-intuitive.

On the third hand, we're expecting % to be phased out in favor of
.format(), which allows floats of any value with the i, d, o, x format
specifiers.  So perhaps we should leave this alone?

> I think in 2.6 we can't change this,

I'm not sure when it happened, but it looks like this
*was* fixed for 2.6.

Closing.