微机原理与接口技术课后习题答案(郭兰英)

dispa

dispa

dldisp

dldisp1:

dldisp ⑵

wdata wordtemp

参考

. . . .

proc push cx push dx mov cl,4 mov dl,ah shr dl,cl call dldisp mov dl,ah and dl,0fh call dldisp mov dl,al shr dl,cl call dldisp mov dl,al and dl,0fh call dldisp pop dx pop cx ret endp ; proc push ax or dl,30h cmp dl,39h jbe dldisp1 add dl,7 mov ah,2 int 21h pop ax ret endp end

.model small .stack .data dw 34abh dw ? .code .startup mov ax,wdata mov wordtemp,ax call dispa

dispa

dispa

dldisp

dldisp1:

dldisp ⑶

wdata

参考

. . . .

.exit 0 ; proc push cx push dx mov cl,4

mov dl,byte ptr wordtemp+1 shr dl,cl call dldisp

mov dl,byte ptr wordtemp+1 and dl,0fh call dldisp

mov dl,byte ptr wordtemp shr dl,cl call dldisp

mov dl,byte ptr wordtemp and dl,0fh call dldisp pop dx pop cx ret endp ; proc push ax or dl,30h cmp dl,39h jbe dldisp1 add dl,7 mov ah,2 int 21h pop ax ret endp end

.model small .stack .data dw 34abh .code .startup push wdata call dispa

. . . .

pop ax ;add sp,2 .exit 0 ; dispa proc push bp mov bp,sp push ax push cx push dx mov ax,[bp+4] mov cl,4 mov dl,ah shr dl,cl call dldisp mov dl,ah and dl,0fh call dldisp mov dl,al shr dl,cl call dldisp mov dl,al and dl,0fh call dldisp pop dx pop cx pop ax pop bp ret dispa endp ; dldisp proc push ax or dl,30h cmp dl,39h jbe dldisp1 add dl,7 dldisp1: mov ah,2 int 21h pop ax ret dldisp endp end 3.36 解:

如果利用共享变量传递函数,且变量定义和使用不

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