lua 调用 c 的例子 编译时出错 - V2EX
yhhy168
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lua 调用 c 的例子 编译时出错

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  •   yhhy168 May 29, 2013 5662 views
    This topic created in 4735 days ago, the information mentioned may be changed or developed.
    最近在学 lua,今天在网上找了个 lua 调用 C 的例子学习,例子的代码如下:


    #include <math.h>
    #include "lua.h"
    #include "lualib.h"
    #include "lauxlib.h"


    static int l_sin (lua_State *L)
    {
    double d = luaL_checknumber(L, 1);
    lua_pushnumber(L, sin(d));
    return 1;
    }

    static const struct luaL_reg mylib [] = {
    {"lsin", l_sin},
    {NULL, NULL}
    };

    int luaopen_mylib(lua_State *L)
    {
    luaL_openlib(L, "mylib1", mylib, 0);
    return 1;
    }


    编译时报“mylib.c:14: 错误:数组元素的类型不完全”的错,提示“static const struct luaL_reg mylib [] = {” 这一行的代码出错,网上找了下说 GCC 4.0 以上版本,结构要先定义再用。不过本人 C 盲,想问下这个错该怎样改呢,先多谢了。

    测试环境:
    Lua 5.2.2 Copyright (C) 1994-2013 Lua.org, PUC-Rio
    gcc 版本 4.4.7 20120313 (Red Hat 4.4.7-3) (GCC)
    3 replies    1970-01-01 08:00:00 +08:00
    xatest
        1
    xatest  
       May 29, 2013
    这是C调用Lua,不是Lua调用C吧
    static const struct luaL_reg mylib [] = {
    改成
    static const luaL_reg mylib [] = {
    因为luaL_reg本来就是struct。
    yhhy168
        2
    yhhy168  
    OP
       May 29, 2013
    @xatest
    改了也没编译通过,报另外一个错了:


    mylib.c:14: 错误:expected ‘=’, ‘,’, ‘;’, ‘asm’ or ‘__attribute__’ before ‘mylib’
    mylib.c: 在函数‘luaopen_mylib’中:
    mylib.c:21: 错误:‘mylib’未声明(在此函数内第一次使用)
    mylib.c:21: 错误:(即使在一个函数内多次出现,每个未声明的标识符在其
    mylib.c:21: 错误:所在的函数内也只报告一次。)


    我是这样编译的:
    libmylib.so
    yhhy168
        3
    yhhy168  
    OP
       May 30, 2013
    自己顶下~
    没有人遇到这种情况吗?
    不懂 C 是要学得坎坷些了:(
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