luaL_openlib 替代 Lua 5.2 [英] luaL_openlib replacement for Lua 5.2

问题描述

我正在改编一个为 Lua 编写的库 <5.2 接到一个电话，我不知道相当于:

I am adapting a library written for Lua < 5.2 and got to a call I don't know the equivalent of:

luaL_openlib(L, "Polycore", polycoreLib, 0);

其中 polycoreLib 是一个

static const struct luaL_Reg polycoreLib []

如何替换对 luaL_openlib 的调用?

How can I replace the call to luaL_openlib?

lua wiki 仅声明:

The lua wiki only states:

诸如luaL_openlib(L, name, lreg, x); 之类的调用应该仔细重写，因为将搜索并可能创建具有给定名称的全局表.

Calls such as luaL_openlib(L, name, lreg, x); should be carefully rewritten because a global table with the given name will be searched and possibly created.

推荐答案

对此有两个答案:一个用于复制早期版本的行为(在此处创建全局表)，另一个用于实现现在的行为常规(即创建并返回一个匿名表).

There's two answers to this: one for replicating the behaviour of earlier versions here (where a global table is created), and one for implementing the behaviour that is now conventional (which is to create and return an anonymous table).

对于前者:

lua_newtable(L);
luaL_setfuncs(L, polycoreLib, 0);
lua_setglobal(L, "Polycore");

这与luaL_openlib 完全不一样，因为如果存在全局表Polycore，它将覆盖它而不是合并用它.如果合并是一个问题，请先使用 lua_getglobal，然后如果它推送了一个表，请重新使用它，而不是创建一个新的:

This isn't quite the same as luaL_openlib, because if there is an existing global table Polycore it will overwrite it rather than merging with it. If merging is a concern, use lua_getglobal first, then if it pushed a table re-use that rather than creating a new one:

lua_getglobal(L, "Polycore");
if (lua_isnil(L, -1)) {
  lua_pop(L, 1);
  lua_newtable(L);
}
luaL_setfuncs(L, polycoreLib, 0);
lua_setglobal(L, "Polycore");

后者更容易，因为您不需要关心合并:

The latter is easier because you don't need to care about merging:

lua_newtable(L);
luaL_setfuncs(L, polycoreLib, 0);
return 1;

使用这种方法，绑定表是调用者的责任，如下所示:

With this approach, it is the caller's reponsibility to bind the table, as in:

local Polycore = require "Polycore"

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