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上遇到同样的问题。 这是我在Linux上使用的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
在这两种情况下,我都有以下警告:
warning: implicit declaration of function
'luaL_openlibs' [-Wimplicit-function-declaration]
该程序有效,但也许我不使用任何标准的Lua库?
为什么我会收到此警告?我在luaL_openLibs
中看到了lauxlib.h
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
答案 0 :(得分:11)
AFAIK&amp;在我的5.1.4安装中,函数驻留在lualib.h中,而不是lauxlib.h
答案 1 :(得分:2)
可能是luaL_openlibs在ifdef块中定义。
使用-E
和gcc在预处理后获取源代码。管。 grep的。