将带有字段的表作为参数传递给C ++的Lua函数? [英] Passing table with fields as an argument to Lua function from C++?

问题描述

我想知道如何用字段和值组成一个Lua表，以便可以将它作为参数传递给C ++的Lua函数.

I would like to know how to form a Lua table with fields and values so I can pass it as an argument to a Lua function from C++.

我知道如何使用索引来构成表格，但是我不知道如何从由字段和值组成的表格中构成.

I know how to form a table using indices but I don't know how to from a table made of fields and values.

例如，我想将此表作为C ++的参数发送给Lua函数.

For example, I want to send this table to a Lua function as an argument from C++.

t = {xpos = 50, ypos = 80, message = 'hello'}

下面的代码是我能得到的最接近的代码，但这只是没有字段名的索引表.

The below code is the closest I could get, but it's just indexed table with no field name.

lua_getglobal(L, "myLuaFunc");
if (lua_type(L, -1) == LUA_TFUNCTION)
{
    lua_newtable(L);
    lua_pushinteger(L, 1);
    lua_pushnumber(L, 50);
    lua_pushinteger(L, 2);
    lua_pushnumber(L, 80);
    lua_pushinteger(L, 3);
    lua_pushstring(L, 'hello');   
    lua_settable(L, -3);
    if (lua_pcall(L, 1, 0, 0))
        std::cout << "Error : " << lua_tostring(L, -1) << std::endl;   
}
lua_pop(L, 1);

推荐答案

您还可以使用lua_setfield，它使代码更短并且更容易阅读:

You can also use lua_setfield, which makes the code shorter and probably easier to read:

    lua_newtable(L);
    lua_pushinteger(L, 50);         // xpos = 50
    lua_setfield(L, -2, "xpos");
    lua_pushinteger(L, 80);         // ypos = 80
    lua_setfield(L, -2, "ypos");
    lua_pushstring(L, "hello");     // message = "hello"
    lua_setfield(L, -2, "message");

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