assembly - 没有 32 位寄存器的 32 位/16 位有符号整数除法？

Question

我正在尝试将 32 位有符号整数除以 16 位有符号整数，以获得有符号 32 位商和 16 位余数。

我的目标是没有 FPU 的 286。

我过去已经编写过代码来进行 32 位无符号除法:

DIV32 PROC

;DIVIDES A 32-BIT VALUE BY A 16-BIT VALUE.

;ALTERS AX
;ALTERS BX
;ALTERS DX

;EXPECTS THE 32-BIT DIVIDEND IN DX:AX
;EXPECTS THE 16-BIT DIVISOR IN BX

;RETURNS THE 32-BIT QUOTIENT IN DX:AX
;RETURNS THE 16-BIT REMAINDER IN BX

    push di
    push si


    mov di, ax ;di -> copy of LSW of given dividend
    mov ax, dx ;ax -> MSW of given dividend
    xor dx, dx ;dx:ax -> 0:MSW  
    div bx     ;ax:dx -> ax=MSW of final quotient, dx=remainder

    mov si, ax ;si -> MSW of final quotient
    mov ax, di ;dx:ax -> dx=previous remainder, ax=LSW of given dividend
    div bx     ;ax:dx -> ax=LSW of final quotient, dx=final remainder  
    mov bx, dx ;bx -> final remainder
    mov dx, si ;dx:ax -> final quotient


    pop si
    pop di
    ret

DIV32 ENDP 

到目前为止，我已经尝试做显而易见的事情，只是通过将 XOR DX, DX 替换为 CWD 和 DIV 来修改现有代码 与 IDIV:

IDIV32 PROC

;DIVIDES A SIGNED 32-BIT VALUE BY A SIGNED 16-BIT VALUE.

;ALTERS AX
;ALTERS BX
;ALTERS DX

;EXPECTS THE SIGNED 32-BIT DIVIDEND IN DX:AX
;EXPECTS THE SIGNED 16-BIT DIVISOR IN BX

;RETURNS THE SIGNED 32-BIT QUOTIENT IN DX:AX
;RETURNS THE 16-BIT REMAINDER IN BX

    push di
    push si


    mov di, ax ;di -> copy of LSW of given dividend
    mov ax, dx ;ax -> MSW of given dividend
    cwd        ;dx:ax -> 0:MSW, or ffff:MSW  
    idiv bx    ;ax:dx -> ax=MSW of final quotient, dx=remainder

    mov si, ax ;si -> MSW of final quotient
    mov ax, di ;dx:ax -> dx=previous remainder, ax=LSW of given dividend
    idiv bx    ;ax:dx -> ax=LSW of final quotient, dx=final remainder  
    mov bx, dx ;bx -> final remainder
    mov dx, si ;dx:ax -> final quotient


    pop si
    pop di
    ret

IDIV32 ENDP 

这适用于某些计算，例如 -654,328/2=-327164 (0xfff60408/2=fffb0204)。但它不适用于某些输入，-131,076/2 返回 -2 余数 0 的错误结果。除数 1 或 -1 似乎会导致除法错误，无论被除数如何。

我测试了许多不同的正负股息和除数，试图找到某种正确和错误结果的模式，我注意到它无法正确返回 -65538 的结果。

我有预感，我应该根据输入有条件地使用 CWD，但似乎 XOR DX, DX 更频繁地返回不正确的结果。当除数和被除数均为正且商小于 0x7fffffff 时，两者都有效。

Answer 1

我不知道有什么算法可以将大负数分成几部分并为 IDIV 做好准备。我会计算被除数和除数的绝对值，使用函数 DIV32，最后根据存储的符号处理结果:

IDIV32 PROC      ; DX:AX / BX = DX/AX rem BX
    ; 99 / 5   =  19 rem 4
    ; 99 / -5  = -19 rem 4
    ; -99 / 5  = -19 rem -4
    ; -99 / -5 =  19 rem -4

    mov ch, dh          ; Only the sign bit counts!
    shr ch, 7           ; CH=1 means negative dividend
    mov cl, bh          ; Only the sign bit counts!
    shr cl, 7           ; CL=1 means negative divisor

    cmp ch, 1           ; DX:AX negative?
    jne J1              ; No: Skip the next two lines
    not dx              ; Yes: Negate DX:AX
    neg ax              ; CY=0 -> AX was NULL
    cmc
    adc dx, 0           ; Adjust DX, if AX was NULL
    J1:

    cmp cl, 1           ; BX negative?
    jne J2              ; No: Skip the next line
    neg bx              ; Yes: Negate BX
    J2:

    push cx             ; Preserve CX
    call DIV32
    pop cx              ; Restore CX

    cmp ch, cl          ; Had dividend and divisor the same sign?
    je J3               ; Yes: Skip the next two lines
    not dx
    neg ax              ; CY=0 -> AX was NULL
    cmc
    adc dx, 0           ; Adjust DX, if AX was NULL
    J3:

    cmp ch, 1           ; Was divisor negative?
    jnz J4              ; No: Skip the next line
    neg bx              ; Negate remainder
    J4:

    ret
IDIV32 ENDP