C调用lua脚本的效率测试



以下代码以C语言为基准，测试了C调用Lua循环和循环调用Lua的效率。结论是不要频繁地穿越C/Lua边界.

代码整理自：http://blog.csdn.net/Tomorrow/archive/2008/06/11/2536884.aspx



#include <time.h>

extern "C"
{
#include "lua.h"
#include "lualib.h"
#include "lauxlib.h"
}/* Lua解释器指针 */

const char LUA_SCRIPT[] =
"function loop_add(a, b) "
" local sum = 0 "
" for i = 1, 10000000 do "
" sum = sum + a + b "
" end "
" return sum "
"end "
" "
"function add(a, b) "
" return a + b "
"end "
;

// lua 脚本里面的函数由C调用
int use_lua_add(lua_State *L, const char *func_name, int x, int y)
{
int sum; /* 通过名字得到Lua函数 */
lua_getglobal(L, func_name); /* 第一个参数 */
lua_pushnumber(L, x); /* 第二个参数 */
lua_pushnumber(L, y); /* 调用函数，告知有两个参数，一个返回值 */
lua_call(L, 2, 1); /* 得到结果 */
sum = (int)lua_tointeger(L, -1);
lua_pop(L, 1);
return sum;
}

int main()
{
int i, sum = 0;
clock_t tStart, tStop;

lua_State *L = lua_open(); /* opens Lua */
luaL_openlibs(L);
if (luaL_dostring(L, LUA_SCRIPT)) // Run lua script
{
printf("run script failed/n");
lua_close(L);
return -1;
}

sum = 0;
tStart = clock();
for (i = 0; i < 10000000; i++)
{
sum += 1 + 1;
}
tStop = clock();
printf("C++: %dms./nThe sum is %u./n",
(tStop - tStart) * 1000 / CLOCKS_PER_SEC, sum);

sum = 0;
tStart = clock();
sum = use_lua_add(L, "loop_add", 1, 1);
tStop = clock();
printf("Lua loop_add: %dms./nThe sum is %u./n",
(tStop - tStart) * 1000 / CLOCKS_PER_SEC, sum);

sum = 0;
tStart = clock();
for (i = 0; i < 10000000; i++)
{
sum += use_lua_add(L, "add", 1, 1);
}
tStop = clock();
printf("Loop lua add: %dms./nThe sum is %u./n",
(tStop - tStart) * 1000 / CLOCKS_PER_SEC, sum);


lua_close(L);
return 0;
}



运行结果：



C++: 31ms.
The sum is 20000000.
Lua loop_add: 437ms.
The sum is 20000000.
Loop lua add: 2360ms.

The sum is 20000000.