lua 5.2 C api中的语法更改

Question

I was trying to compile the example provided in the book Programming in Lua

But only works for lua 5.1, What are the steps to do it on 5.2?

This is the code I am using

#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;
}

After compiling with gcc test01.c -I/usr/include/lua5.2 -L/usr/lib/x86_64-linux-gnu -llua5.2 I get the following errors:

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

Thank you in advance.

Answer 1

luaopen() is not used anymore, it's replaced by luaL_newstate , you can use luaL_newstate to create a state with a standard allocation function:

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

This API is changed since Lua 5.1

Answer 2

尝试：

lua_State *L = lua_newstate();