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

[Dirk Laurie <dpl@...>]

On Sat, Jul 30, 2011 at 07:39:49PM +0200, Lorenzo Donati wrote:
> On 30/07/2011 12.17, liam mail wrote:
> > vm_details = function()
> > local env = string.dump(function()end)
> > assert(env:byte(6) == 0x00)--if hit then it is not an official Lua release
> > local number_type
> > local int_width
> > if env:byte(12) == 0x0 then
> > if env:byte(11) == 0x08 then
> > number_type = 'double'
> > elseif env:byte(11) == 0x04 then
> > number_type = 'float'
> > else
> > number_type = 'unknown'
> > end
> > else
> > number_type = 'int type of size ' .. tonumber(env:byte(11))
> > end
> > return number_type, tonumber(env:byte(8))
> > end
> >
> 
> I don't understand it fully. It seems that this relies on the internal 
> VM or bytecode format, which is not formalized anywhere, so you cannot 
> rely on that - moreover it could change from version to version, so you 
> would have to rewrite the function for each new Lua release. (BTW 
> int_width is not used, a typo or what?)
> 

function fix_format ()
    local a = 2^53-1  -- all mantissa bits are 1
    local format = string.format
    -- test for presence of %d bug
    if ("%d"):format(a) == "9007199254740991" then return 1 end
    -- replaces string.format by one that works around the bug
    -- BUG: unsophisticated replacement of %d by %.0f, e.g
    -- =string.format("Fixes the %%d %s","bug")) --> Fixes the %.0f bug
    local format = string.format
    string.format = function (f,...)
        return format(string.gsub(f,"%%([%+%-]?%d*)d","%%%1.0f"),...)
        end
    end

print(("%+020d   %d"):format(2^53-1, 2^53-1))
--> -0000000000000000001   -1
fix_format()
print(("%+020d   %d"):format(2^53-1, 2^53-1))
--> +0009007199254740991   9007199254740991

The output shown above comes from the Lua interpreter in
lua-5.2.0-beta-rc3-mingw.zip, running under "wine cmd" on
my Linux machine.

Of course, this does not work around all the other places where
the failure of a C long to contain a Lua integer is a problem.

Dirk