gdb 调试内嵌汇编c程序时，好象不能单步显示执行结果

今天调试了一个内嵌汇编c程序，好郁闷，用gdb跟踪了一中午，始终没有结果，单步执行，直接执行到汇编后面的程序语句了，好奇怪。在cu发贴没人鸟我，上完课后，又仔细的看了代码，原来有一条汇编语句让我搞错了，
在编译执行，程序可以得到正确的结果，但是gdb跟踪还是老问题：

Administrator@wzt-u44op82g4jw ~/asm
$ gcc -o 4 4.c -g

Administrator@wzt-u44op82g4jw ~/asm
$ gdb 4
GNU gdb 6.3.50_2004-12-28-cvs (cygwin-special)
Copyright 2004 Free Software Foundation, Inc.
GDB is free software, covered by the GNU General Publi
welcome to change it and/or distribute copies of it un
Type "show copying" to see the conditions.
There is absolutely no warranty for GDB.  Type "show w
This GDB was configured as "i686-pc-cygwin"...
(gdb) l
1       int main(void)
2       {
3           int m,n,sum = 0;
4
5           scanf("%d%d",&m,&n);
6           if ( m != 0 && n != 0 ) {
7       /*
8               while(m != 0) {
9                   if ( m & 1 )
10                      sum += n;
(gdb)
11                  m = m >> 1;
12                  n = n << 1;
13              }
14      */
15              __asm__("movl $0,%%eax\n"
16                      "1:\tcmpl $0,%1\n\t"
17                      "je 3f\n\t"
18                      "testl $1,%1\n\t"
19                      "jz 2f\n\t"
20                      "addl %2,%0\n"
(gdb) b 15
Breakpoint 1 at 0x4010a8: file 4.c, line 15.
(gdb) r
Starting program: /home/Administrator/asm/4.exe
1 4

Breakpoint 1, main () at 4.c:15
15              __asm__("movl $0,%%eax\n"
(gdb) n
29          printf("%d * %d = %d\n",m,n,sum);
(gdb) n
1 * 4 = 4
31          return 0;
(gdb) n
32      }

难道gdb不能单步跟踪内嵌汇编吗？后来又查下gcc手册，也没看到类似的调试选项。如果哪位兄弟知道的话，告诉我啊。。。