Re: Lua Error

[Martijn van Buul <martijn.van.buul@...>]

On 8/17/2012 11:29 AM, Edward Scerri wrote:
> Dear All,
>
> I am trying to output a numeric value using in a 4digit hexadecimal
> format in the following way:
> " io.write(string.format("%04x", printchars)) "
>
> I am getting an error as follows:
>
> ./pulse_parser.lua 2012-08-16
> /usr/bin/lua: ./pulse_parser.lua:16: bad argument #2 to 'format'
> (integer expected, got number)

This, in combination with

> Copyright (C) 1994-2008 Lua.org, PUC-Rio (double int32)"
.                                          ^^^^^^^^^^^^^^^

makes me think you're using the LNUM patch, which introduces a 
difference between integers and floating point. In this case, the 
string.format expects an integer, but it got a floating point number.

 AFAIK, there area two options:

1) Don't use Lua with LNUM patch; it is mostly compatible but not 
entirely so

2) Try to explicitly convert the number to an integer. I haven't used 
the LNUM patch much, but ISTR you can "cast" a number to integer by 
doing something like

 tonumber(math.floor( printchars + .5))

as 'tonumber' will convert a floating point number which can be 
represented by an integer to an integer.