隐式声明 luaL_openlibs。

我正在编写一个将Lua嵌入到C程序中的简单测试。

我在Windows/Mingw和Linux上都遇到了同样的问题。 这是Linux上我使用的gcc命令:

gcc -Wall -o test_lua lua_test.c -I/usr/include/lua5.1 -llua5.1

在Windows上:

gcc -Wall -o test_lua.exe lua_test.c -llua5.1

在两种情况下,我都会得到以下警告:

warning: implicit declaration of function
              'luaL_openlibs' [-Wimplicit-function-declaration]

程序可以正常工作,但我可能没有在其中使用任何标准的Lua库? 为什么我会收到这个警告?我在lauxlib.h中看到了luaL_openLibs的定义!

这是C部分:

#include <lua.h>
#include <lauxlib.h>
#include <stdlib.h>
#include <stdio.h>

int main(int argc, char *argv[]) {

  int status;
  lua_State *L;

  L = luaL_newstate();

  // Init lua
  luaL_openlibs(L);

  // Load script
  status = luaL_loadfile(L,"lua_test.lua");
  if (status) {
    fprintf(stderr,"Couldn't load file\n");
    exit(1);
  }

  // Push data
  lua_pushnumber(L, 5000);
  lua_setglobal(L, "clife");
  lua_pushnumber(L, 6000);
  lua_setglobal(L, "ttime");
  lua_pushnumber(L, 3000);
  lua_setglobal(L, "atime");

  // Run script
  int result = lua_pcall(L, 0, LUA_MULTRET, 0);
  if (result) {
    fprintf(stderr,"Failed to run script: %s\n", lua_tostring(L,-1));
    exit(1);
  }

  // Value at top of the stack is the result
  const char *schedule = lua_tostring(L,-1);

  fprintf(stdout,"Computed schedule is: %s\n", schedule);

  // Close lua
  lua_pop(L, 1);
  lua_close(L);

  return 0;

}

这是Lua部分:

io.write("lua_test.lua -- will generate schedule\n")

io.write("Wizard life is " .. clife .. "\n")

schedule = ""
ctime = ttime - atime
if clife > 4500 then
   schedule = schedule .. "[" .. ctime .. ",p]"
   schedule = schedule .. "[" .. ctime+500 .. ",a]"
   schedule = schedule .. "[" .. ctime+1000 .. ",i]"
   schedule = schedule .. "[" .. ctime+1500 .. ",n]\n"
else
   schedule = schedule .. "[" .. ctime .. ",d]"
   schedule = schedule .. "[" .. ctime+500 .. ",r]"
   schedule = schedule .. "[" .. ctime+1000 .. ",a]"
   schedule = schedule .. "[" .. ctime+1500 .. ",i]"
   schedule = schedule .. "[" .. ctime+1500 .. ",n]\n"
end

io.write("Returning " .. schedule .. "\n");

return schedule

原文链接 https://stackoverflow.com/questions/8386352

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stackoverflow用户549472
stackoverflow用户549472

据我所知,在我的 5.1.4 版本安装中,该函数位于 lualib.h,而不是 lauxlib.h。

2011-12-05 14:04:00
stackoverflow用户1593299
stackoverflow用户1593299
May be luaL_openlibs defines in a ifdef block.

Use `-E` with gcc to get the source after preprocessing. Pipe. Grep.

可能在 ifdef 块中定义了 luaL_openlibs

使用 gcc-E 选项以获取预处理后的源代码。使用管道和 grep 命令。

2012-08-12 10:32:48