C調用lua腳本的效率測試
以下代碼以C語言為基準,測試了C調用Lua循環和循環調用Lua的效率。結論是不要頻繁地穿越C/Lua邊界.
代碼整理自: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解釋器指針 */
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 腳本里面的函數由C調用
int use_lua_add(lua_State *L, const char *func_name, int x, int y)
{
int sum; /* 通過名字得到Lua函數 */
lua_getglobal(L, func_name); /* 第一個參數 */
lua_pushnumber(L, x); /* 第二個參數 */
lua_pushnumber(L, y); /* 調用函數,告知有兩個參數,一個返回值 */
lua_call(L, 2, 1); /* 得到結果 */
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;
}
運行結果:
C++: 31ms.
The sum is 20000000.
Lua loop_add: 437ms.
The sum is 20000000.
Loop lua add: 2360ms.
The sum is 20000000.


