如何让最小的程序占的资源最少-----100分求解
我用另一个进程打开测试进程,在测试进程里只是等待一个信号,然后退出,请问如何能让这个进程占的资源最少?我想测系统可以最多开多少进程,在笔记本上测了下,不到160个,在台式机上测了下,没有超过3600个,我把汇编版和C++版的代码贴出来,希望高手指点,谢谢!
.386
.Model Flat, StdCall
Option Casemap :None ; 不区分大小写(对API与API常数无效)
;___________________________________________________________________________
Include kernel32.inc
IncludeLib kernel32.lib
.CODE
szName db "tp_Exit",0,0
START:
push esi
invoke OpenEvent,00100000h,0,offset szName
mov esi,eax
invoke WaitForSingleObject,esi,-1
invoke CloseHandle,esi
pop esi
invoke ExitProcess,0
END START
----------------------------------
#include "stdafx.h"
int APIENTRY WinMain(HINSTANCE hInstance,
HINSTANCE hPrevInstance,
LPSTR lpCmdLine,
int nCmdShow)
{
// TODO: Place code here.
HANDLE heExit=OpenEventW(SYNCHRONIZE,FALSE,L"tp_Exit");
WaitForSingleObject(heExit,INFINITE);
CloseHandle(heExit);
return 0;
}