Detailed description C calls lua script Efficiency Test

CCallLua scriptEfficiencyTestIs the content to be introduced in this article. The following code usesCLanguage as the benchmark,TestNowCCallLuaLoop and loop callLua. The conclusion is that do not traverse frequentlyC/LuaBoundary.

 
 

  
  
# Include <time. h>
  
  
 
  
  
Extern "C"
  
  
{
  
  
# Include "lua. h"
  
  
# Include "lualib. h"
  
  
# Include "lauxlib. h"
  
  
}/* Lua interpreter pointer */
  
  
 
  
  
Const char LUA_SCRIPT [] =
  
  
"Function loop_add (a, B )"
  
  
"Local sum = 0"
  
  
"For I = 1, 10000000 do"
  
  
"Sumsum = sum + a + B"
  
  
"End"
  
  
"Return sum"
  
  
"End"
  
  
""
  
  
"Function add (a, B )"
  
  
"Return a + B"
  
  
"End"
  
  
;
  
  
 
  
  
// The functions in the lua script are called by C.
  
  
Int use_lua_add (lua_State * L, const char * func_name, int x, int y)
  
  
{
  
  
Int sum;/* obtain the Lua function by name */
  
  
Lua_getglobal (L, func_name);/* The first parameter */
  
  
Lua_pushnumber (L, x);/* second parameter */
  
  
Lua_pushnumber (L, y);/* call the function to inform that there are two parameters, one returned value */
  
  
Lua_call (L, 2, 1);/* get the result */
  
  
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;
  
  
}
 
 

Running result:

 
 

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

Summary: DetailsCCallLua scriptEfficiencyTestI hope this article will help you!