c++ - 从 C++ 调用未知(按名称)lua 函数

Question

这是我想做的:

1) 有一个用户定义的lua函数，不知道名字。说是:

function f(x) return 2*x; end

2) 然后用户将从 Lua 中调用函数(在步骤 3 中设计)，例如:

a=foo(f,3) --expecting a=6

3) foo 的 C++ 函数是:

int lua_foo(lua_State *L)
{
    int nargs = lua_gettop(L);
    if(nargs<2) throw "ERROR: At least two arguments i) Function ii) number must be supplied";


    int type = lua_type(L, 1);
    if(type!=LUA_TFUNCTION) throw "ERROR: First argument must be a function";

    double arg2=lua_tonumber(L,2);
    lua_pushnumber(L,arg2);

    lua_pcall(L, 1, 1, 0) ; //trying to call the function f

    double result=lua_tonumber(L,-1); //expecting the result to be 6
    lua_pushnumber(L,result);

    lua_pop(L,nargs);

    return 1;
}

在 C++ 代码中，我知道第一个参数是一个函数，第二个参数是一个数字。我正在尝试使用第二个参数(数字)作为其参数来调用第一个参数(函数)。

Answer 1

如果函数设计如下:

/* avoid `lua_` (or `luaL_`) prefix for your own functions */
static int l_foo(lua_State *L)
{
    /* `lauxlib.h` contains a lot of useful helper functions, e.g. for
     * argument type checking: */
    luaL_checktype(L, 1, LUA_TFUNCTION);
    luaL_checknumber(L, 2);
    /* discard any extra arguments to `foo`; ignoring extra arguments
     * is customary for Lua functions */
    lua_settop(L, 2);
    /* the `lua_settop()` above ensures that the two topmost elements
     * of the stack are the function `f` and its argument, so
     * everything is set for the `lua_call()` */
    lua_call(L, 1, 1);
    /* return the topmost value on the Lua stack (the result); all
     * other stack values are removed by Lua automatically (but in
     * this case the result is the only value on the stack as the
     * `lua_call()` popped the function and the argument, and pushed
     * one result) */
    return 1;
}

它按预期工作。