Re: problem with string.format %d and very large integers

[Dimiter 'malkia' Stanev <malkia@...>]

Yup, but one can do it inlined

(sizeof(int)<sizeof(lua_Number)) ? "%0.f" : "%d"

or (but requires some non-header change)

const char* formatters[2] = { "%d", "%0.f" }
formatters[ sizeof(int) < sizeof( lua_Number) ]

Or this stupid trick, without requiring above:

("%d\0\0%0.f" + ((sizeof(int) < sizeof(lua_Number))<<2))

\0 is supposed to be 0 (I'm never sure how to encode this in "C")

#define LUA_STRFORMAT_FLOATINGINT sizeof(int) < sizeof(lua_Number)
#if LUA_STRFORMAT_FLOATINGINT
#define LUA_STRFORMAT_INTEGER = "%.0f"
#else
#define LUA_STRFORMAT_INTEGER = "%d"

On 7/28/2011 11:20 AM, Luiz Henrique de Figueiredo wrote:
> > I'd like to see this solved with defines in a config, with something like
> >
> > #define LUA_STRFORMAT_FLOATINGINT sizeof(int)<  sizeof(lua_Number)
> > #if LUA_STRFORMAT_FLOATINGINT
>
> That cannot work because sizeof is not evaluated at preprocessing time.