How to pass a pointer to LuaJIT ffi to be used as out argument?

流过昼夜 提交于 2020-01-01 05:18:08

问题


Assuming there is following C code:

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

...How to do following in LuaJIT?

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

回答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


来源:https://stackoverflow.com/questions/14011598/how-to-pass-a-pointer-to-luajit-ffi-to-be-used-as-out-argument

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