Question

我开始学习汇编语言。我只想澄清一件事：

通过调试改变数据会导致我的操作系统/应用程序失败吗？

我已经通过cmd运行调试并完成了E命令

E 100

在这里，我使用space来查看存储在这些位置的内存中的内容：

0B79:0100 BE.  90.   02.    ... etc

覆盖这些内存空间会导致当前正在运行的系统出错吗？

启动时，Debug将65k字节的内存分配为工作空间，前100个字节为PSP。我的问题可能更直接：

我正在查看未分配内存的工作空间，或os /应用程序正在使用的内存。

再次请原谅我的新手，这毕竟是我对汇编语言的看法。

Answer 1

我认为它是64 KB未分配的内存段。通过调试，我们可以简单地构建* .com应用程序，但通常不是* .exe。

使用汇编程序，我们可以构建* .exe应用程序。在* .exe中，我们可以以固体形式保留一个代码段，一个数据段和一个堆栈段。如果DOS应用程序正在启动，那么DOS会将所有可用内存提供给应用程序，但是我们不知道有多少内存段是免费的，并且位于该地址上。为了分配更多内存，我们可以在第一步中释放所有未使用的内存，在第二步中我们可以分配更多的内存。

SETFREE:  mov     bx, ss   ; Using for *.exe:
          mov     ax, es   ; calculating the number of paragraps from
          sub     bx, ax   ; the PSP to the beginning of the stack
          mov     ax, sp   ; + adding the length of the stack
          add     ax, 0Fh
          shr     ax, 4
          add     bx, ax
          mov     ah, 4Ah
          int   21h
          ret
;---------------------------------------------------------------------------
GETSPACE: mov     ah, 48h ; BX = number of paragraphs
          int   21h
          ret             ; Return: AX = segment

RBIL-＆GT; inter61b.zip-＆GT; INTERRUP.G

--------D-2148-------------------------------
INT 21 - DOS 2+ - ALLOCATE MEMORY
AH = 48h
BX = number of paragraphs to allocate
Return: CF clear if successful
    AX = segment of allocated block
CF set on error
    AX = error code (07h,08h) (see #01680 at AH=59h/BX=0000h)
    BX = size of largest available block
Notes:  DOS 2.1-6.0 coalesces free blocks while scanning for a block to
  allocate
.COM programs are initially allocated the largest available memory
  block, and should free some memory with AH=49h before attempting any
  allocations
under the FlashTek X-32 DOS extender, EBX contains a protected-mode
  near pointer to the allocated block on a successful return
SeeAlso: AH=49h,AH=4Ah,AH=58h,AH=83h

--------D-2149-------------------------------
INT 21 - DOS 2+ - FREE MEMORY
AH = 49h
ES = segment of block to free
Return: CF clear if successful
CF set on error
    AX = error code (07h,09h) (see #01680 at AH=59h/BX=0000h)
Notes:  apparently never returns an error 07h, despite official docs; DOS 2.1+
code contains only an error 09h exit
DOS 2.1-6.0 does not coalesce adjacent free blocks when a block is
freed, only when a block is allocated or resized
the code for this function is identical in DOS 2.1-6.0 except for
calls to start/end a critical section in DOS 3.0+
SeeAlso: AH=48h,AH=4Ah

--------D-214A-------------------------------
INT 21 - DOS 2+ - RESIZE MEMORY BLOCK
AH = 4Ah
BX = new size in paragraphs
ES = segment of block to resize
Return: CF clear if successful
 CF set on error
    AX = error code (07h,08h,09h) (see #01680 at AH=59h/BX=0000h)
    BX = maximum paragraphs available for specified memory block
Notes:  under DOS 2.1-6.0, if there is insufficient memory to expand the block
as much as requested, the block will be made as large as possible
DOS 2.1-6.0 coalesces any free blocks immediately following the block
to be resized
SeeAlso: AH=48h,AH=49h,AH=83h

程序集 - Debug.exe - 通过调试改变数据会导致os / apps失败吗？

1 个答案: