奇怪的宏（TASM）

考虑以下宏：

pixelFast MACRO
    ; This macro draws a pixel, assuming the coordinates are already loaded in cx&dx and the color is in al.
    xor bh, bh
    mov ah, 0ch
    int 10h
ENDM

drawRect MACRO x1, y1, x2, y2, color
    LOCAL @@loop, @@row_loop
    xor cx, cx
    mov dx, y1
    mov al, BYTE PTR [color]

    @@loop:
        mov cx, x1
        @@row_loop:
            pixelFast

            inc cx
            cmp cx, x2
            jna @@row_loop

        inc dx
        cmp dx, y2
        jna @@loop
ENDM

rendToolBar MACRO
    drawRect COLORDISP_X1, COLORDISP_Y1, COLORDISP_X2, COLORDISP_Y2, foreground_color
    mov temp_color, 36h
    drawRect COLORBTN1_X1, COLORBTN1_Y1, COLORBTN1_X2, COLORBTN1_Y2, temp_color
    mov temp_color, 2Eh
    drawRect COLORBTN2_X1, COLORBTN2_Y1, COLORBTN2_X2, COLORBTN2_Y2, temp_color
    mov temp_color, 4h
    drawRect COLORBTN3_X1, COLORBTN3_Y1, COLORBTN3_X2, COLORBTN3_Y2, temp_color
    mov temp_color, 2Bh
    drawRect COLORBTN4_X1, COLORBTN4_Y1, COLORBTN4_X2, COLORBTN4_Y2, temp_color
ENDM

Somewhere in my code, I use the rendToolBar macro. It is supposed to draw a big white canvas, and then a small square and next to it some smaller squares in a certain pattern, which is irrelevant for my question. Notice that rendToolBar calls drawRect 5 times. I followed this code in turbo debugger (because something went awfully wrong) and noticed that in the 4th execution of the drawRect macro, the "int 10h" from pixelFast is not actually "int 10h", but rather "int 2". This causes an NMI, which messes stuff up for my program. I want to know what makes TASM expand the macro differently for that line in the 4th call for that macro, despite the fact that this line "int 10h" does not rely on any macro arguments. If you look at this image, you can see the unexpected "int 2" there, which was supposed to be an "int 10". After it, you can see:

cmp [bx+si], ax
add ch, bh
cmp [bx+03], dx

根据宏的源代码，这3条指令实际上应该是

inc cx
cmp cx, COLORBTN3_X2
jna @@row_loop

还有一些其他指令在中断之前有点偏离，但您明白了。

考虑将逻辑（段：偏移）地址转换为线性地址的数学：

CS:IP = 49ae:03cc = 49eac where 3cc是第一个意外字节的偏移量。

SS:SP = 39ed:fffc = 49ecc.

可视化这两个线性地址，我们有

   |       |
   | 49ecc | <-- Stack pointer, going down
   |       |

 Only 32 bytes below

   |       |
   | 49eac | <-- Execution flow, going up
   |       |

您的堆栈必须在屏幕截图之前的某个时刻与代码段发生冲突。
尝试设置堆栈，使其距离代码足够远。

The maximum stack size in real mode is 64KiB, as this is the size of a segment.
In DOS is safe to assume that the memory after your program is not used¹ and, as long as it exists, so you can use it for the stack. This is not wasting memory as DOS is not multitasking.
Note that the stack segment won't take space on the binary file unless you explicitly define things in it.

有两种方法来管理堆栈：

1. 使用汇编器
   请参阅TASM手册 http://bitsavers.informatik.uni-stuttgart.de/pdf/borland/turbo_assembler/Turbo_Assembler_Version_5_Users_Guide.pdf供参考，第 92 页。

   如果您正在使用STACK指令只是对堆栈的估计大小设置了上限。

   如果您正在编写 EXE，则可以使用 model 修饰符FARSTACK.
   SS:SP应根据链接器在 MZ 标头上写入的值进行设置。
   这样您就可以通过不将堆栈段放入dgroup.

    

2. Manually
   如果您知道不需要完整的 64KiB 堆栈，则可以将其放在数据段的末尾（对于 COM 来说也是代码段）

   ;COM                      ;EXE
   mov ax, cs                mov ax, ds  ;Assume SMALL memory model
   mov ss, ax                mov ss, ax  ;see below
   xor sp, sp                xor sp, sp
   

   这给出了 64KiB -.

   如果您需要完整的 64KiB 堆栈，您可以只使用下一个可用段

   ;COM                      ;EXE
   mov ax, cs                mov ax, ds      ;Assume SMALL memory model, if not          
   add ax, 1000h             add ax, 1000h   ;use the symbol for the last data      
   mov ss, ax                mov ss, ax      ;segment
   xor sp, sp                xor sp, sp
   

   这假设最后一个段已完全使用，但可以使您免于进行某些段/偏移/符号算术。

¹ This is because DOS is not multitasking and programs are .
A non resident part of COMMAND.COM is loaded a the top of the conventional memory but it can be overwritten.