如何在C ++中调用package.preload内部的函数 [英] How to call functions inside package.preload in C++

问题描述

我正在尝试调用 A:update(x)并在C ++中获得返回值 x + 3 .

I'm trying to call A:update(x) and get a returned value x + 3 in C++.

这是我的代码:

#include <lua.hpp>

void main()
{
    lua_State *L = luaL_newstate();
    luaL_openlibs(L);
    lua_settop(L, 0);
    luaL_dostring(L, "package.preload['A'] = function () local A = {}\n"
                     "function A:update(x) return x + 3 end \n"
                     "return A end");
    //call function
    lua_getglobal(L, "require");
    lua_pushstring(L, "A");
    if (lua_pcall(L, 1, LUA_MULTRET, 0) != 0) {
        std::cerr << "lua:" << lua_tostring(L, 1) << '\n';
        lua_pop(L, 1);
    }
    int top = lua_gettop(L);
    lua_getfield(L, -1, "update");
    if (!lua_isfunction(L, -1))
    {
        std::cerr << "lua:" << lua_tostring(L, 1) << '\n';
        lua_pop(L, 1);
    }
    lua_pushnumber(L, 5); //pass the argument 5
    if (lua_pcall(L, 1, LUA_MULTRET, 0))
    {
        std::cerr << "lua:" << lua_tostring(L, 1) << '\n';
        lua_pop(L, 1);
    }
    if (lua_gettop(L) - top)
    {
        if (lua_isnumber(L, -1))
        {
            std::cout << "RETURNED : " << lua_tonumber(L, -1) << std::endl;
        }
    }
    lua_pop(L, 1); // pop 'update'
    lua_pop(L, 1); // pop 'A'
    lua_close(L);
}

我希望它打印出 RETURNED:8 ，但是出现以下错误:

I expect it to print RETURNED : 8 but I get the following error:

Thread 1:EXC_BAD_ACCESS (code=1, address=0x0)

我应该如何纠正我的代码才能正常工作?

How should I correct my code to work?

已我将 A:update(x)更改为 A.update(x)后，它立即起作用.我以为它们的工作原理相同，只是我可以在使用:的函数中使用 self .有人可以向我解释为什么会发生这种情况吗?

EDITED: It worked as soon as I change A:update(x) to A.update(x). I thought they work identically except I can use self in a function that uses :. Could someone please explain to me why this happens?

推荐答案

符号 A:update(x)是 A.update(A，x).这意味着您必须使用两个参数调用函数 update .您缺少两个参数中的第一个.

The notation A:update(x) is syntactic sugar for A.update(A,x). That means you have to call the function update with two parameters. You are lacking the first of the two parameters.

第一个参数 A 已经在堆栈上，但是位于 update 函数的下方".使用 lua_pushvalue ，我们可以将表的副本推入堆栈.

The first parameter A is already on the stack but is located "below" the update function. Using lua_pushvalue we can push a copy of the table onto the stack.

因此，您必须像这样调用函数(省略错误处理位)

Thus you have to call the function like this (omitting the error handling bits)

lua_getfield(L, -1, "update");
lua_pushvalue(L, -2); // push a copy of "A" onto the stack
lua_pushnumber(L, 5); //pass the argument 5
lua_pcall(L, 2, LUA_MULTRET, 0);

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