如何传递指向LuaJIT ffi的指针用作out参数?

本文介绍了如何传递指向LuaJIT ffi的指针用作out参数?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 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