如何通过一个指针LuaJIT FFI用作out参数？(How to pass a pointer t

假设有下列的C代码：

struct Foo { int dummy; }
int tryToAllocateFoo(Foo ** dest);

...如何做LuaJIT以下？

Foo * pFoo = NULL;
tryToAllocateFoo(&pFoo);

Answer 1:

local ffi = require 'ffi'

ffi.cdef [[
  struct Foo { int dummy; };
  int tryToAllocateFoo(Foo ** dest);
]]

local theDll = ffi.load(dllName)

local pFoo = ffi.new 'struct Foo *[1]'
local ok = theDll.tryToAllocateFoo(pFoo)

if ok == 0 then -- Assuming it returns 0 on success
  print('dummy ==', pFoo[0].dummy)
end

文章来源: How to pass a pointer to LuaJIT ffi to be used as out argument?