汇编新手一个很郁闷的题，小弟我自己试过了很多遍，结果都是一样的郁闷，不知道自己错哪了
汇编新手一个很郁闷的题，我自己试过了很多遍，结果都是一样的郁闷，不知道自己哪里错了
下面是我写的代码：
****************************
assume cs:code,ds:data,ss:stack
data segment
dw 0123h,0456h,0789h,0abch,0defh,0fedh,0cbah,0987h
data ends
stack segment
dw 0,0,0,0,0,0,0,0
stack ends
code segment
start: mov ax,stack
mov ss,ax
mov sp,16
mov ax,data
mov ds,ax
push ds:[0]
push ds:[2]
pop ds:[2]
pop ds:[0]
mov ax,4c00h
int 21h
code ends
end start
********************
上面的代码能够顺利的通过编译以及连接，最后我通过debug来运行的时候，发现了一个问题。就是，我通过ds查看数据段的内容，结果竟然不是 0123h,0456h,0789h,0abch,0defh,0fedh,0cbah,0987h
但是我通过ss，最后发现数据段的内容竟然在栈段里面去了。我真的是百思不得其解。
看一下下面是在debug中运行的情况
D:\ac>debug 12.exe
-r
AX=0000 BX=0000 CX=0042 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000
DS=13DD ES=13DD SS=13ED CS=13EF IP=0000 NV UP EI PL NZ NA PO NC
13EF:0000 B8EE13 MOV AX,13EE
-d 13dd:0
13DD:0000 CD 20 FF 9F 00 9A F0 FE-1D F0 4F 03 E8 0D 8A 03 . ........O.....
13DD:0010 E8 0D 17 03 E8 0D D7 0D-01 01 01 00 02 FF FF FF ................
13DD:0020 FF FF FF FF FF FF FF FF-FF FF FF FF 95 13 4C 01 ..............L.
13DD:0030 A8 12 14 00 18 00 DD 13-FF FF FF FF 00 00 00 00 ................
13DD:0040 05 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................
13DD:0050 CD 21 CB 00 00 00 00 00-00 00 00 00 00 20 20 20 .!...........
13DD:0060 20 20 20 20 20 20 20 20-00 00 00 00 00 20 20 20 .....
13DD:0070 20 20 20 20 20 20 20 20-00 00 00 00 00 00 00 00 ........
-d 13ed:0
13ED:0000 23 01 56 04 89 07 BC 0A-EF 0D ED 0F BA 0C 87 09 #.V.............
13ED:0010 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................
13ED:0020 B8 EE 13 8E D0 BC 10 00-B8 ED 13 8E D8 FF 36 00 ..............6.
13ED:0030 00 FF 36 02 00 8F 06 02-00 8F 06 00 00 B8 00 4C ..6............L
13ED:0040 CD 21 00 00 00 00 00 00-00 00 00 00 00 00 00 00 .!..............
13ED:0050 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................
13ED:0060 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................
13ED:0070 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................
------解决方案--------------------
是王爽老师书上例题吧,自己能找出问题那最好的.
------解决方案--------------------
这是王爽的《汇编语言》里的例题
13ED:0000 23 01 56 04 89 07 BC 0A-EF 0D ED 0F BA 0C 87 09 #.V............. 这段就是data
13ED:0010 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................这段就是stack
因为你是直接查看的，所以程序并没有执行start里的代码
data段里的数23 01 56 04 89 07 BC 0A-EF 0D ED 0F BA 0C 87 09 看的时候要记得高位和低位
所以这段就是 0123h,0456h,0789h,0abch,0defh,0fedh,0cbah,0987h
------解决方案--------------------
下面是我写的代码：
****************************
assume cs:code,ds:data,ss:stack
data segment
dw 0123h,0456h,0789h,0abch,0defh,0fedh,0cbah,0987h
data ends
stack segment
dw 0,0,0,0,0,0,0,0
stack ends
code segment
start: mov ax,stack
mov ss,ax
mov sp,16
mov ax,data
mov ds,ax
push ds:[0]
push ds:[2]
pop ds:[2]
pop ds:[0]
mov ax,4c00h
int 21h
code ends
end start
********************
上面的代码能够顺利的通过编译以及连接，最后我通过debug来运行的时候，发现了一个问题。就是，我通过ds查看数据段的内容，结果竟然不是 0123h,0456h,0789h,0abch,0defh,0fedh,0cbah,0987h
但是我通过ss，最后发现数据段的内容竟然在栈段里面去了。我真的是百思不得其解。
看一下下面是在debug中运行的情况
D:\ac>debug 12.exe
-r
AX=0000 BX=0000 CX=0042 DX=0000 SP=0000 BP=0000 SI=0000 DI=0000
DS=13DD ES=13DD SS=13ED CS=13EF IP=0000 NV UP EI PL NZ NA PO NC
13EF:0000 B8EE13 MOV AX,13EE
-d 13dd:0
13DD:0000 CD 20 FF 9F 00 9A F0 FE-1D F0 4F 03 E8 0D 8A 03 . ........O.....
13DD:0010 E8 0D 17 03 E8 0D D7 0D-01 01 01 00 02 FF FF FF ................
13DD:0020 FF FF FF FF FF FF FF FF-FF FF FF FF 95 13 4C 01 ..............L.
13DD:0030 A8 12 14 00 18 00 DD 13-FF FF FF FF 00 00 00 00 ................
13DD:0040 05 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................
13DD:0050 CD 21 CB 00 00 00 00 00-00 00 00 00 00 20 20 20 .!...........
13DD:0060 20 20 20 20 20 20 20 20-00 00 00 00 00 20 20 20 .....
13DD:0070 20 20 20 20 20 20 20 20-00 00 00 00 00 00 00 00 ........
-d 13ed:0
13ED:0000 23 01 56 04 89 07 BC 0A-EF 0D ED 0F BA 0C 87 09 #.V.............
13ED:0010 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................
13ED:0020 B8 EE 13 8E D0 BC 10 00-B8 ED 13 8E D8 FF 36 00 ..............6.
13ED:0030 00 FF 36 02 00 8F 06 02-00 8F 06 00 00 B8 00 4C ..6............L
13ED:0040 CD 21 00 00 00 00 00 00-00 00 00 00 00 00 00 00 .!..............
13ED:0050 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................
13ED:0060 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................
13ED:0070 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................
------解决方案--------------------
是王爽老师书上例题吧,自己能找出问题那最好的.
------解决方案--------------------
这是王爽的《汇编语言》里的例题
13ED:0000 23 01 56 04 89 07 BC 0A-EF 0D ED 0F BA 0C 87 09 #.V............. 这段就是data
13ED:0010 00 00 00 00 00 00 00 00-00 00 00 00 00 00 00 00 ................这段就是stack
因为你是直接查看的，所以程序并没有执行start里的代码
data段里的数23 01 56 04 89 07 BC 0A-EF 0D ED 0F BA 0C 87 09 看的时候要记得高位和低位
所以这段就是 0123h,0456h,0789h,0abch,0defh,0fedh,0cbah,0987h
------解决方案--------------------