keil 调试时为什么RAM区数据修改不了？

只看该作者 · 2013-8-3 08:47

这是小弟的调试C程序；
include <reg51.h>
typedef unsigned char  uchar;
typedef  unsigned int uint ;
uchar data *  point;
main()
{
uchar a,b;
point=0x5A; //指向RAM区的0x5a单元
  *point++=0x55;//向0x5a单元写入数据0行0x55，指针加1
  *point=0xaa;
point=0x5A;
a= *point++; //将0x5a单元的内容读到a中
b=*point; //将0x5b单元的内容读到b中
while(1);
}
这是自动转换的汇编程序：
C:0x0000 020025 LJMP    STARTUP1(C:0025)
   9: main()
10: {
11:    uchar a,b;
12:    point=0x5A;
C:0x0003 750A5A MOV    point(0x0A),#0x5A
13: *point++=0x55
C:0x0006 AF0A    MOV    R7,point(0x0A)
C:0x0008 050A    INC    point(0x0A)
C:0x000A A807    MOV    R0,0x07
C:0x000C 7655    MOV    @R0,#0x55
14: *point=0xaa;
C:0x000E A80A    MOV    R0,point(0x0A)
C:0x0010 76AA    MOV    @R0,#0xAA
15: point=0x5A;
C:0x0012 750A5A MOV    point(0x0A),#0x5A
16: a= *point++;
C:0x0015 AF0A    MOV    R7,point(0x0A)
C:0x0017 050A    INC    point(0x0A)
C:0x0019 A807    MOV    R0,0x07
C:0x001B E6    MOV    A,@R0
C:0x001C F508    MOV    0x08,A
17: b=*point; //将0x5b单元的内容读到b中
C:0x001E A80A    MOV    R0,point(0x0A)
C:0x0020 E6    MOV    A,@R0
C:0x0021 F509    MOV    0x09,A
18: while(1);
C:0x0023 80FE    SJMP    C:0023
133:                MOV    R0,#IDATALEN - 1
C:0x0025 787F    MOV    R0,#0x7F
134:                CLR    A
C:0x0027 E4    CLR    A
135: IDATALOOP:    MOV    @R0,A
C:0x0028 F6    MOV    @R0,A
136:                DJNZ R0,IDATALOOP
C:0x0029 D8FD    DJNZ    R0,IDATALOOP(C:0028)
185:                MOV    SP,#?STACK-1

为什么在调试窗口中memory中查看0x5a单元的值根本没有发生改变，还有a，b的值也没有。？求大神指教