我觉得不会。另外，我在编程中也似乎从来没考虑过重入的问题。
证明如下：
1
// testtimer.cpp : 定义控制台应用程序的入口点。
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 一次执行结果如下：
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