如何将指针传递给LuaJIT ffi以用作out参数？

Question

假设有以下C代码：

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

...如何在LuaJIT中执行以下操作？

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