我覺得不會。另外，我在編程中也似乎從來沒考慮過重入的問題。
證明如下：
?1
//?testtimer.cpp?:?定義控制臺應(yīng)用程序的入口點。
?2//
?3
?4#include?"stdafx.h"
?5#include?<windows.h>
?6#include?<conio.h>
?7
?8static?UINT?idTimer?=?0;
?9static?int?reentry?=?0;
10static?int?call_cnt?=?0;
11
12void?LengthyWork(void)
13
{
14
????//Sleep(3000);
15????int?i?=?0,j?=?0;
16
????for(i;i?<?50000;)?{
17????????i++;
18????????for(j?=?i;j?>?0;)?{
19????????????j--;
20
????????}
21????}
22
23
}
24VOID?CALLBACK?OnTimer(HWND?hwnd,
25????UINT?uMsg,
26????UINT_PTR?idEvent,
27????DWORD?dwTime
28)
29{
30????++call_cnt;
31????printf("entry(%d)??reentry:%d\n",call_cnt,reentry);
32????++reentry;
33????LengthyWork();
34????--reentry;
35????printf("exit(%d)???reentry:%d\n",call_cnt,reentry);
36}
37int?_tmain(int?argc,?_TCHAR*?argv[])
38{
39????idTimer?=?SetTimer(NULL,0,1000,OnTimer);
40????int?ret?=?0;
41????MSG?msg;
42????while(1)?{
43????????if(kbhit())?{
44????????????return?0;
45????????}
46????????ret?=?GetMessage(&msg,NULL,0,0);
47????????if(ret)?{
48????????????TranslateMessage(&msg);
49????????????DispatchMessage(&msg);
50????????}
51????}
52????return?0;
53}
54 一次執(zhí)行結(jié)果如下：
entry(1)? reentry:0
exit(1)?? reentry:0
entry(2)? reentry:0
exit(2)?? reentry:0
entry(3)? reentry:0
exit(3)?? reentry:0
entry(4)? reentry:0
exit(4)?? reentry:0