0

我只想为 TASM 编写简单的 .asm 代码,在 C++ 中工作

int t=2;
for(int i=0;i<2;i++)
t=t+(i-1)*7*t;

如何使用 TASM 实现它?

4

1 回答 1

0

这将在 8086 TASM 中从 1 循环到 100:

    .MODEL SMALL

    .STACK 100h

    .DATA   
    Finished DB 10, 13, 'Loop x 100 finished.  Congratulations! $', 10, 13

    .CODE

    MAIN PROC

            MOV AX, @data            ; Required at the start of every program (inside your main procedure, from what I've seen)
            MOV DS, AX

            MOV CX, 100              ; Set CX to 100
            MOV BX, 0                ; Counter (for double-verification, I guess...lol)

    StrtLoop:                        ; When a loop starts, it does CX-- (subtracts 1 from CX)   

            INC BX                    ; This does BX++, which increments BX by 1

    LOOP StrtLoop                     ; Go back to StrtLoop label

            CMP BX, 100              ; Compare BX to 100...
            JE DispMsg               ; Jump-if-Equal...CMP BX, 100 sets flags, and if they are set,
                                     ;  JE will Jump you to DispMsg (to get "congratulations" message).

            JMP SkipMsg              ; Jump to the SkipMsg label (so you don't see the "congratulations" message).

    DispMsg:                         ; If BX = 100, you JE here.
            MOV AH, 09H              ; Displays the message stored in the defined byte "Finished"
            MOV DX, OFFSET Finished
            INT 21H
    SkipMsg:                         ; If BX != 100, you JMP here.
            MOV AL, 0h               ; Op code to exit to DOS from the assembler.
            MOV AH, 4CH
            INT 21H

    MAIN ENDP
    END MAIN

我希望它有所帮助。我做了基本的循环,所以你可以做你的代码的其他部分(我不知道 C++,哈哈)。祝你好运!这很难,但同时也很有趣(至少对我而言)。

于 2012-04-27T22:29:15.467 回答