lua 5.2 C api中的语法更改

Question

我试图编译" Lua编程 "一书中提供的示例

但仅适用于lua 5.1,在5.2上执行此操作的步骤是什么？

这是我正在使用的代码

#include <stdio.h>
#include <string.h>
#include <lua.h>
#include <lauxlib.h>
#include <lualib.h>

int main (void) {
  char buff[256];
  int error;
  lua_State *L = lua_open();   /* opens Lua */
  luaL_openlibs(L);  
  while (fgets(buff, sizeof(buff), stdin) != NULL) {
    error = luaL_loadbuffer(L, buff, strlen(buff), "line") ||
      lua_pcall(L, 0, 0, 0);
    if (error) {
      fprintf(stderr, "%s", lua_tostring(L, -1));
      lua_pop(L, 1);  /* pop error message from the stack */
    }
  }
  lua_close(L);
  return 0;
}

编译后gcc test01.c -I/usr/include/lua5.2 -L/usr/lib/x86_64-linux-gnu -llua5.2我得到以下错误:

test01.c: In function ‘main’:
test01.c:10:18: warning: initialization makes pointer from integer without a cas
t [enabled by default]                                                         
   lua_State *L = lua_open();   /* opens Lua */
                  ^
/tmp/ccyPRlV3.o: In function `main':
test01.c:(.text+0x21): undefined reference to `lua_open'
collect2: error: ld returned 1 exit status

先感谢您.

Answer 1

luaopen()不再使用,它​​被替换为luaL_newstate,您可以使用luaL_newstate创建具有标准分配函数的状态:

lua_State *L = luaL_newstate();    /* opens Lua */
luaL_openlibs(L);                  /* opens the standard libraries */

自Lua 5.1以来,此API已更改