Re: Lua 5.3: How to get integer from out of range float?

[Roberto Ierusalimschy <roberto@...>]

> Using 5.3 with #define LUA_INT_INT:
> 
> return math.pow(2,31)>>0
> stdin:1: number has no integer representation
> stack traceback:
>   stdin:1: in main chunk
>   [C]: in ?
> 
> To get integer anyway, with wraparound as it should be, ie equivalent of
> int32_t i = (int64_t)d in C, one can:
> 
> string.format("%.0f", math.pow(2,31))>>0
> -2147483648
> 
> This gives the correct result, but feels somewhat kludgy taking the detour
> via string. Does anyone know of a better way?

I guess this does what you want:

-- convert a float to a 32-bit integer, doing wrap around
function int (x)
  local x = math.floor(x)       -- or ceil or modf or some other rounding
  x = (x + 2^31) % 2^32 - 2^31
  return math.tointeger(x)
end


-- Roberto