Does Lua make use of 64-bit integers?

问题

Does Lua make use of 64-bit integers? How do I use it?

回答1:

Compile it yourself. Lua uses double-precision floating point numbers by default. However, this can be changed in the source (luaconf.h, look for LUA_NUMBER).

回答2:

require "bit"

-- Lua unsigned 64bit emulated bitwises
-- Slow. But it works.

function i64(v)
 local o = {}; o.l = v; o.h = 0; return o;
end -- constructor +assign 32-bit value

function i64_ax(h,l)
 local o = {}; o.l = l; o.h = h; return o;
end -- +assign 64-bit v.as 2 regs

function i64u(x)
 return ( ( (bit.rshift(x,1) * 2) + bit.band(x,1) ) % (0xFFFFFFFF+1));
end -- keeps [1+0..0xFFFFFFFFF]

function i64_clone(x)
 local o = {}; o.l = x.l; o.h = x.h; return o;
end -- +assign regs

-- Type conversions

function i64_toInt(a)
  return (a.l + (a.h * (0xFFFFFFFF+1)));
end -- value=2^53 or even less, so better use a.l value

function i64_toString(a)
  local s1=string.format("%x",a.l);
  local s2=string.format("%x",a.h);
  local s3="0000000000000000";
  s3=string.sub(s3,1,16-string.len(s1))..s1;
  s3=string.sub(s3,1,8-string.len(s2))..s2..string.sub(s3,9);
  return "0x"..string.upper(s3);
end

-- Bitwise operators (the main functionality)

function i64_and(a,b)
 local o = {}; o.l = i64u( bit.band(a.l, b.l) ); o.h = i64u( bit.band(a.h, b.h) ); return o;
end

function i64_or(a,b)
 local o = {}; o.l = i64u( bit.bor(a.l, b.l) ); o.h = i64u( bit.bor(a.h, b.h) ); return o;
end

function i64_xor(a,b)
 local o = {}; o.l = i64u( bit.bxor(a.l, b.l) ); o.h = i64u( bit.bxor(a.h, b.h) ); return o;
end

function i64_not(a)
 local o = {}; o.l = i64u( bit.bnot(a.l) ); o.h = i64u( bit.bnot(a.h) ); return o;
end

function i64_neg(a)
 return i64_add( i64_not(a), i64(1) );
end  -- negative is inverted and incremented by +1

-- Simple Math-functions

-- just to add, not rounded for overflows
function i64_add(a,b)
 local o = {};
 o.l = a.l + b.l;
 local r = o.l - 0xFFFFFFFF;
 o.h = a.h + b.h;
 if( r>0 ) then
   o.h = o.h + 1;
   o.l = r-1;
 end
 return o;
end

-- verify a>=b before usage
function i64_sub(a,b)
  local o = {}
  o.l = a.l - b.l;
  o.h = a.h - b.h;
  if( o.l<0 ) then
    o.h = o.h - 1;
    o.l = o.l + 0xFFFFFFFF+1;
  end
  return o;
end

-- x n-times
function i64_by(a,n)
 local o = {};
 o.l = a.l;
 o.h = a.h;
 for i=2, n, 1 do
   o = i64_add(o,a);
 end
 return o;
end
-- no divisions   

-- Bit-shifting

function i64_lshift(a,n)
 local o = {};
 if(n==0) then
   o.l=a.l; o.h=a.h;
 else
   if(n<32) then
     o.l= i64u( bit.lshift( a.l, n) ); o.h=i64u( bit.lshift( a.h, n) )+ bit.rshift(a.l, (32-n));
   else
     o.l=0; o.h=i64u( bit.lshift( a.l, (n-32)));
   end
  end
  return o;
end

function i64_rshift(a,n)
 local o = {};
 if(n==0) then
   o.l=a.l; o.h=a.h;
 else
   if(n<32) then
     o.l= bit.rshift(a.l, n)+i64u( bit.lshift(a.h, (32-n))); o.h=bit.rshift(a.h, n);
   else
     o.l=bit.rshift(a.h, (n-32)); o.h=0;
   end
  end
  return o;
end

-- Comparisons

function i64_eq(a,b)
 return ((a.h == b.h) and (a.l == b.l));
end

function i64_ne(a,b)
 return ((a.h ~= b.h) or (a.l ~= b.l));
end

function i64_gt(a,b)
 return ((a.h > b.h) or ((a.h == b.h) and (a.l >  b.l)));
end

function i64_ge(a,b)
 return ((a.h > b.h) or ((a.h == b.h) and (a.l >= b.l)));
end

function i64_lt(a,b)
 return ((a.h < b.h) or ((a.h == b.h) and (a.l <  b.l)));
end

function i64_le(a,b)
 return ((a.h < b.h) or ((a.h == b.h) and (a.l <= b.l)));
end


-- samples
a = i64(1);               -- 1
b = i64_ax(0x1,0);        -- 4294967296 = 2^32
a = i64_lshift(a,32);     -- now i64_eq(a,b)==true
print( i64_toInt(b)+1 );  -- 4294967297

X = i64_ax(0x00FFF0FF, 0xFFF0FFFF);
Y = i64_ax(0x00000FF0, 0xFF0000FF);

-- swap algorithm
X = i64_xor(X,Y);
Y = i64_xor(X,Y);
X = i64_xor(X,Y);

print( "X="..i64_toString(X) ); -- 0x00000FF0FF0000FF
print( "Y="..i64_toString(Y) ); -- 0x00FFF0FFFFF0FFFF

回答3:

Lua 5.3 introduces the integer subtype, which uses 64-bit integer by default.

From Lua 5.3 reference manual

The type number uses two internal representations, one called integer and the other called float. Lua has explicit rules about when each representation is used, but it also converts between them automatically as needed (see §3.4.3). Therefore, the programmer may choose to mostly ignore the difference between integers and floats or to assume complete control over the representation of each number. Standard Lua uses 64-bit integers and double-precision (64-bit) floats, but you can also compile Lua so that it uses 32-bit integers and/or single-precision (32-bit) floats. The option with 32 bits for both integers and floats is particularly attractive for small machines and embedded systems. (See macro LUA_32BITS in file luaconf.h.)

来源：https://stackoverflow.com/questions/3104722/does-lua-make-use-of-64-bit-integers