如何传递指向LuaJIT ffi的指针用作out参数? [英] How to pass a pointer to LuaJIT ffi to be used as out argument?

问题描述

假设有以下C代码:

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

...如何在LuaJIT中进行关注?

...How to do following in LuaJIT?

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

推荐答案

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

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