Re: Lua 5.3 Integer

[v <v19930312@...>]

On Wed, 2019-10-09 at 11:48 -0400, Katie Volz wrote:
> For all the arithmetic operators(+/*-), if your inputs are integers,
> then your output must be an integer (unless you go outside the
> representable range).  The exception is division, so integer division
> handles that case.  So if you convert all your inputs to integers and
> use integer division instead of float division, you'll never have a
> non-integer.
> 
> On Wed, Oct 9, 2019, 11:44 AM Russell Haley <russ.haley@gmail.com>
> wrote:
> > 
> > On Wed, Oct 9, 2019 at 6:02 AM Roberto Ierusalimschy <
> > roberto@inf.puc-rio.br> wrote:
> > > > Is there any way to enforce integer arithmetic in Lua 5.3
> > > > 
> > > > e.g.
> > > > 
> > > > x = 5 / 2 -- => 2
> > > > where x is an integer?
> > > 
> > > 1) Do not use '/' or '^'.
> > > 
> > > 2) For any value that enters the computation that you cannot be
> > > sure
> > > it is an integer, apply a conversion to it. How to convert
> > > depends
> > > on you:
> > > 
> > > - x | 0   -- fast, raises an error if x is not an integer value
> > > 
> > > - math.tointeger(x)      -- returns falsy if x is not an integer
> > > value
> > > 
> > > - math.floor, math.ceil, math.modf
> > >      -- rounds x (but the result may not be in the proper range)
> > > 
> > > 
> > > -- Roberto
> > > 
> > 
> > For my edification:  The answer is then "No, there is no way 'in
> > general' to perform integer math"? It seems to me that all the
> > suggestions above are value conversions/coercion and have nothing
> > to do with actually performing the integer math equation requested
> > by the OP? 
> > 

If you go outside of integer range, result is still an integer, just
not one you may expect.