Question

简单的版本是我希望我在C中构建的__tostring方法在Lua中打印对象时输出“glxy_object（100）”。如果我注释掉“问题代码”注释（从new（）方法读取表数据的位）之间的代码，它就能完美地实现这一点。很明显，我没有将堆栈返回到需要的位置，但是我一直在盯着这段代码看看错误的位置。

的main.c

#include "lua.h"
#include "lualib.h"
#include "lauxlib.h"

typedef struct {
  double x;
  double y;
  double z;
} glxy_vector3;

typedef struct {
  int id;
  glxy_vector3 position;
  glxy_vector3 rotation;
} glxy_object;

#define OBJNAME "glxy_object"

static int glxy_object_new(lua_State *L) {
  int n = lua_gettop(L);
  if (n != 2)
    return luaL_error(L, "Got %d arguments expected 2", n);

  size_t nbytes = sizeof(glxy_object);
  glxy_object *o = (glxy_object *)lua_newuserdata(L, nbytes);

  luaL_getmetatable(L, OBJNAME);
  lua_setmetatable(L, -2);

  // id
  o->id = luaL_checknumber(L, 1);

  // TROUBLE CODE
  lua_pop(L, 1);

  lua_pushstring(L, "position");
  lua_gettable(L, -2);
  lua_getfield(L, -1, "x");
  o->position.x = lua_tonumber(L, -1);
  lua_pop(L, 1);
  lua_getfield(L, -1, "y");
  o->position.y = lua_tonumber(L, -1);
  lua_pop(L, 1);
  lua_getfield(L, -1, "z");
  o->position.z = lua_tonumber(L, -1);
  lua_pop(L, 2);

  lua_pushstring(L, "rotation");
  lua_gettable(L, -2);
  lua_getfield(L, -1, "x");
  o->rotation.x = lua_tonumber(L, -1);
  lua_pop(L, 1);
  lua_getfield(L, -1, "y");
  o->rotation.y = lua_tonumber(L, -1);
  lua_pop(L, 1);
  lua_getfield(L, -1, "z");
  o->rotation.z = lua_tonumber(L, -1);
  lua_pop(L, 2);
  // END TROUBLE CODE

  // This actually prints the data just fine, so I am reading it correctly at least...
  printf("new() - id: %d - position: (%0.2f, %0.2f, %0.2f), rotation: (%0.2f, %0.2f, %0.2f)\n",
    o->id, o->position.x, o->position.y, o->position.z, o->rotation.x, o->rotation.y, o->rotation.z
  );

  return 1;
}

static glxy_object *glxy_object_check(lua_State *L) {
  luaL_checktype(L, 1, LUA_TUSERDATA);
  glxy_object *o = (glxy_object *)luaL_checkudata(L, 1, OBJNAME);
  if (o == NULL) luaL_typerror(L, 1, OBJNAME);
  return o;
}

static int glxy_object_tostring(lua_State *L) {
  glxy_object *o = (glxy_object *)lua_touserdata(L, 1);
  if (o == NULL) luaL_typerror(L, 1, OBJNAME);
  lua_pushfstring(L, "glxy_object(%d)", o->id);
  return 1;
}

static const struct luaL_reg glxy_object_f[] = {
  { "new", glxy_object_new },
  { NULL, NULL }
};

static const struct luaL_reg glxy_object_m[] = {
  { "__tostring", glxy_object_tostring },
  { NULL, NULL }
};

int glxy_register_object(lua_State *L) {
  luaL_openlib(L, OBJNAME, glxy_object_f, 0);
  luaL_newmetatable(L, OBJNAME);

  luaL_openlib(L, 0, glxy_object_m, 0);
  lua_pushliteral(L, "__index");
  lua_pushvalue(L, -3);
  lua_rawset(L, -3);
  lua_pushliteral(L, "__metatable");
  lua_pushvalue(L, -3);
  lua_rawset(L, -3);

  lua_pop(L, 1);
  return 1;
}

int main(void) {
  // setup lua
  L = luaL_newstate();
  luaL_openlibs(L);

  // register Lua accessable C objects
  glxy_register_object(L);
  lua_pop(L, 1);

  luaL_dofile(L, "main.lua");

  return 0;
}

main.lua

local o = glxy_object.new(100, { position={ x=1.0, y=2.0, z=3.0 }, rotation={ x=4.0, y=5.0, z=6.0 } })
print(o)

当前输出

new() - id: 100 - position: (1.00, 2.00, 3.00), rotation: (4.00, 5.00, 6.00)
table: 0x7fe702510b70

注释掉'故障代码'时的

预期输出/输出（减去不正确的pos / rot值）

new() - id: 100 - position: (0.00, 0.00, 0.00), rotation: (0.00, 0.00, 0.00)
glxy_object(100)

为了便于阅读，我修剪/组合了我的代码库。

Answer 1

“故障代码”的问题出现在评论后面的第一行：lua_pop(L, 1);。这会弹出新创建的用户数据，因此glxy_object_new最终会返回其第二个参数而不是创建的glxy_object。

有两种快速解决方法。第一个是删除lua_pop(L, 1);行并将故障代码中的两行更改为lua_gettable(L, -2);到lua_gettable(L, -3)。第二个是在故障代码之后的某处添加另一个lua_pushuserdata(L, o)。

从C中的args读取Lua表 - 不正确的堆栈级别？

1 个答案: