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

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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);

lua ffi luajit

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edited Dec 23 '12 at 16:00 prapin 2,3501622

asked Dec 23 '12 at 14:07 Alexander Gladysh 8,4661049101

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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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answered Dec 23 '12 at 22:01 finnw 22.6k564130

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Aha! You have to use [1] instead of a second *! Makes sense, but takes some getting used to. –  Alexander Gladysh Dec 24 '12 at 3:59